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Engineering Mechanics - Dynamics Chapter 15

*e vB1y*− cos θ( ) *vB1x *sin θ( )−( ) *vB2n*=

*vB1x
*

*vB1y
*

*vB2n
*

*vB2t
*

*t
*

*R
*

⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

Find *vB1x vB1y*, *vB2n*, *vB2t*, *t*, *R*,( )=
*vB1x
*

*vB1y
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

3.00

6.00− ⎛ ⎜ ⎝

⎞ ⎟ ⎠

ft s

=

*t *0.19 s=

*R *0.79 ft=

*vB2n
*

*vB2t
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

1.70

6.36 ⎛ ⎜ ⎝

⎞ ⎟ ⎠

ft s

=
*vB2n
*

*vB2t
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

6.59 ft s

=

***Problem 15-72
**

The drop hammer *H *has a weight *WH* and falls from rest *h* onto a forged anvil plate *P *that has a
weight *WP*. The plate is mounted on a set of springs that have a combined stiffness *kT*. Determine
(a) the velocity of *P *and *H *just after collision and (b) the maximum compression in the springs
caused by the impact. The coefficient of restitution between the hammer and the plate is *e*.
Neglect friction along the vertical guideposts *A *and *B*.

Given:

*WH *900 lb= *kT *500
lb
ft

=

*WP *500 lb= *g *32.2
ft

s2 =

*h *3 ft= *e *0.6=

Solution:

δ*st
WP
kT
*

= *vH1 *2*g h*=

Guesses

*vH2 *1
ft
s

= *vP2 *1
ft
s

= δ 2 ft=

Given
*WH
g
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

*vH1
WH
*

*g
*⎛
⎜
⎝

⎞ ⎟ ⎠

*vH2
WP
g
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

*vP2*+=

*e vH1 vP2 vH2*−=

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Engineering Mechanics - Dynamics Chapter 15

1 2

*kT*δ*st
*2 1

2

*WP
g
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

*vP2
*2+

1 2

*kT*δ
2 *WP *δ δ*st*−( )−=

*vH2
*

*vP2
*

δ

⎛⎜ ⎜ ⎜ ⎝

⎞⎟ ⎟ ⎟ ⎠

Find *vH2 vP2*, δ,( )=
*vH2
*

*vP2
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

5.96

14.30 ⎛ ⎜ ⎝

⎞ ⎟ ⎠

ft s

= δ 3.52 ft=

**Problem 15-73
**

It was observed that a tennis ball when served horizontally a distance *h* above the ground strikes the
smooth ground at *B* a distance *d* away. Determine the initial velocity *vA* of the ball and the velocity
*vB* (and θ) of the ball just after it strikes the court at *B*. The coefficient of restitution is *e*.

Given:

*h *7.5 ft=

*d *20 ft=

*e *0.7=

*g *32.2
ft

s2 =

Solution:

Guesses *vA *1
ft
s

= *vB2 *1
ft
s

=

*vBy1 *1
ft
s

= θ 10 deg= *t *1 s=

Given *h
*1
2

*g t*2= *d vA t*=

*e vBy1 vB2 *sin θ( )= *vBy1 g t*=

*vA vB2 *cos θ( )=

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Engineering Mechanics - Dynamics Chapter 15

*vA
*

*t
*

*vBy1
*

*vB2
*

θ

⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝

⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

Find *vA t*, *vBy1*, *vB2*, θ,( )= *vA *29.30 fts= *vB2 *33.10
ft
s

= θ 27.70 deg=

**Problem 15-74
**

The tennis ball is struck with a horizontal velocity *vA*, strikes the smooth ground at *B*, and bounces
upward at θ= θ*1*. Determine the initial velocity *vA*, the final velocity *vB*, and the coefficient of
restitution between the ball and the ground.

Given:

*h *7.5 ft=

*d *20 ft=

θ*1 *30 deg=

*g *32.2
ft

s2 =

Solution: θ θ*1*=

Guesses *vA *1
ft
s

= *t *1 s= *vBy1 *1
ft
s

= *vB2 *1
ft
s

= *e *0.5=

Given *h
*1
2

*g t*2= *d vA t*= *vBy1 g t*=

*e vBy1 vB2 *sin θ( )= *vA vB2 *cos θ( )=

*vA
*

*t
*

*vBy1
*

*vB2
*

*e
*

⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝

⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠

Find *vA t*, *vBy1*, *vB2*, *e*,( )= *vA *29.30
ft
s

= *vB2 *33.84
ft
s

= *e *0.77=

**Problem 15-75
**

The ping-pong ball has mass *M*. If it is struck with the velocity shown, determine how high *h* it
rises above the end of the smooth table after the rebound. The coefficient of restitution is *e*.

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Engineering Mechanics - Dynamics Chapter 15

Given:

*M *2 gm= *a *2.25 m=

*e *0.8= *b *0.75 m=

θ 30 deg= *g *9.81
m

s2 =

*v *18
m
s

=

Solution: Guesses *v1x *1
m
s

= *v1y *1
m
s

= *v2x *1
m
s

= *v2y *1
m
s

=

*t1 *1 s= *t2 *2 s= *h *1 m=

Given *v1x v *cos θ( )= *a v *cos θ( )*t1*= *v1y g t1 v *sin θ( )+=

*v2x v1x*= *e v1y v2y*= *b v2x t2*= *h v2y t2
g
*2

⎛⎜ ⎝

⎞⎟ ⎠

*t2
*2−=

*v1x
*

*v1y
*

*v2x
*

*v2y
*

*t1
*

*t2
*

*h
*

⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

Find *v1x v1y*, *v2x*, *v2y*, *t1*, *t2*, *h*,( )=

*v1x
*

*v1y
*

*v2x
*

*v2y
*

⎛ ⎜ ⎜ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

15.59

10.42

15.59

8.33

⎛⎜ ⎜ ⎜ ⎜ ⎝

⎞⎟ ⎟ ⎟ ⎟ ⎠

m s

=
*t1
*

*t2
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

0.14

0.05 ⎛ ⎜ ⎝

⎞ ⎟ ⎠

s=

*h *390 mm=

***Problem 15-76
**

The box *B *of weight* WB *is dropped from rest a distance *d* from the top of the plate *P *of
weight* WP*, which is supported by the spring having a stiffness *k*. Determine the maximum
compression imparted to the spring. Neglect the mass of the spring.

352

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Engineering Mechanics - Dynamics Chapter 15

Given: *WB *5 lb= *WP *10 lb= *g *32.2
ft

s2 =

*k *30
lb
ft

= *d *5 ft= *e *0.6=

Solution:

δ*st
WP
*

*k
*= *vB1 *2*g d*=

Guesses *vB2 *1
ft
s

= *vP2 *1
ft
s

= δ 2 ft=

Given
*WB
g
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

*vB1
WB
g
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

*vB2
WP
g
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

*vP2*+= *e vB1 vP2 vB2*−=

1 2

*k*δ*st
*2 1

2

*WP
g
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

*vP2
*2+

1 2

*k*δ2 *WP *δ δ*st*−( )−=

*vB2
*

*vP2
*

δ

⎛⎜ ⎜ ⎜ ⎝

⎞⎟ ⎟ ⎟ ⎠

Find *vB2 vP2*, δ,( )=
*vB2
*

*vP2
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

1.20−

9.57 ⎛ ⎜ ⎝

⎞ ⎟ ⎠

ft s

= δ 1.31 ft=

**Problem 15-77
**

A pitching machine throws the ball of weight *M* towards the wall with an initial velocity *vA* as
shown. Determine (a) the velocity at which it strikes the wall at *B*, (b) the velocity at which it
rebounds from the wall and (c) the distance *d *from the wall to where it strikes the ground at *C*.

Given:

*M *0.5 kg= *a *3 m=

*vA *10
m
s

= *b *1.5 m=

*e *0.5=θ 30 deg=

*g *9.81
m

s2 =

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Engineering Mechanics - Dynamics Chapter 15

*vBx1 *1
m
s

= *vBx2 *1
m
s

=

*vBy1 *1
m
s

= *vBy2 *1
m
s

=

*h *1 m= *d *1 m=

*t1 *1 s= *t2 *1 s=

Given

*vA *cos θ( )*t1 a*= *b vA *sin θ( )*t1*+
1
2

*g t1
*2− *h*=

*vBy2 vBy1*= *vA *sin θ( ) *g t1*− *vBy1*=

*d vBx2 t2*= *h vBy2 t2*+
1
2

*g t2
*2− 0=

*vA *cos θ( ) *vBx1*= *e vBx1 vBx2*=

*vBx1
*

*vBy1
*

*vBx2
*

*vBy2
*

*h
*

*t1
*

*t2
*

*d
*

⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

Find *vBx1 vBy1*, *vBx2*, *vBy2*, *h*, *t1*, *t2*, *d*,( )=
*vBx1
*

*vBy1
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

8.81 m s

=

*vBx2
*

*vBy2
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

4.62 m s

=

*d *3.96 m=

**Problem 15-78
**

The box of weight *Wb* slides on the surface for which the coefficient of friction is μ*k*. The box has
velocity *v* when it is a distance *d* from the plate. If it strikes the plate, which has weight *Wp* and is
held in position by an unstretched spring of stiffness *k*, determine the maximum compression
imparted to the spring. The coefficient of restitution between the box and the plate is *e*. Assume that
the plate slides smoothly.

354

GuessesSolution:

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Engineering Mechanics - Dynamics Chapter 15

Given:

*Wb *20 lb= *Wp *10 lb=

μ*k *0.3= *k *400
lb
ft

=

*v *15
ft
s

= *e *0.8=

*d *2 ft= *g *32.2
ft

s2 =

Solution:

Guesses *vb1 *1
ft
s

= *vb2 *1
ft
s

= *vp2 *1
ft
s

= δ 1 ft=

Given 1 2

*Wb
g
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

*v*2 μ*k Wb d*−
1
2

*Wb
g
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

*vb1
*2=

*Wb
g
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

*vb1
Wb
g
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

*vb2
Wp
g
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

*vp2*+=

*e vb1 vp2 vb2*−=
1
2

*Wp
g
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

*vp2
*2 1

2
*k*δ2=

*vb1
*

*vb2
*

*vp2
*

δ

⎛ ⎜ ⎜ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

Find *vb1 vb2*, *vp2*, δ,( )=
*vb1
*

*vb2
*

*vp2
*

⎛⎜ ⎜ ⎜ ⎝

⎞⎟ ⎟ ⎟ ⎠

13.65

5.46

16.38

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

ft s

= δ 0.456 ft=

**Problem 15-79
**

The billiard ball of mass *M* is moving with a speed *v* when it strikes the side of the pool table at
*A*. If the coefficient of restitution between the ball and the side of the table is *e*, determine the
speed of the ball just after striking the table twice, i.e., at *A*, then at *B*. Neglect the size of the
ball.

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Engineering Mechanics - Dynamics Chapter 15

Given:

*M *200 gm=

*v *2.5
m
s

=

θ 45 deg=

*e *0.6=

Solution:

Guesses

*v2 *1
m
s

= θ*2 *1 deg= *v3 *1
m
s

= θ*3 *1 deg=

Given *e v *sin θ( ) *v2 *sin θ*2*( )= *v *cos θ( ) *v2 *cos θ*2*( )=
*e v2 *cos θ*2*( ) *v3 *sin θ*3*( )= *v2 *sin θ*2*( ) *v3 *cos θ*3*( )=

*v2
*

*v3
*

θ*2
*

θ*3
*

⎛⎜ ⎜ ⎜ ⎜ ⎜⎝

⎞⎟ ⎟ ⎟ ⎟ ⎟⎠

Find *v2 v3*, θ*2*, θ*3*,( )=
*v2
*

*v3
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

2.06

1.50 ⎛ ⎜ ⎝

⎞ ⎟ ⎠

m s

=
θ*2
*

θ*3
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

31.0

45.0 ⎛ ⎜ ⎝

⎞ ⎟ ⎠

deg=

*v3 *1.500
m
s

=

***Problem 15-80
**

The three balls each have the same mass *m*. If *A *is released from rest at θ, determine the angle
φ to which *C *rises after collision. The coefficient of restitution between each ball is *e*.

Solution:

Energy

0 *l *1 cos θ( )−( )*m g*+ 1
2

*m vA
*2=

*vA *2 1 cos θ( )−( )*g l*=

Collision of ball *A* with *B*:

*m vA *0+ *m v'A m v'B*+= *e vA v'B v'A*−= *v'B
*1
2

1 *e*+( )*v'B*=

Collision of ball *B* with *C*:

356

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Engineering Mechanics - Dynamics Chapter 15

*m v'B *0+ *m v''B m v''C*+= *e v'B v''C v''B*−= *v''C
*1
4

1 *e*+( )2*vA*=

Energy

1 2

*m v''c
*2 0+ 0 *l *1 *cos *φ( )−( )*m g*+= 1

2 1 16

⎛⎜ ⎝

⎞⎟ ⎠

1 *e*+( )4 2( ) 1 cos θ( )−( ) 1 cos φ( )−( )=

1 *e*+
2

⎛⎜ ⎝

⎞⎟ ⎠

4
1 cos θ( )−( ) 1 cos φ( )−= φ acos 1 1 *e*+

2 ⎛⎜ ⎝

⎞⎟ ⎠

4 1 cos θ( )−( )−

⎡ ⎢ ⎣

⎤ ⎥ ⎦

=

**Problem 15-81
**

Two smooth billiard balls *A *and *B
*each have mass *M*. If *A *strikes *B *with
a velocity *vA* as shown, determine their
final velocities just after collision. Ball
*B *is originally at rest and the
coefficient of restitution is *e*. Neglect
the size of each ball.

Given:

*M *0.2 kg=

θ 40 deg=

*vA *1.5
m
s

=

*e *0.85=

Solution: Guesses *vA2 *1
m
s

= *vB2 *1
m
s

= θ*2 *20 deg=

Given *M*− *vA *cos θ( ) *M vB2 M vA2 *cos θ*2*( )+=

*e vA *cos θ( ) *vA2 *cos θ*2*( ) *vB2*−=

*vA *sin θ( ) *vA2 *sin θ*2*( )=

*vA2
*

*vB2
*

θ*2
*

⎛ ⎜ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎟ ⎠

Find *vA2 vB2*, θ*2*,( )= θ*2 *95.1 deg=
*vA2
*

*vB2
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

0.968

1.063− ⎛ ⎜ ⎝

⎞ ⎟ ⎠

m s

=

357

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Engineering Mechanics - Dynamics Chapter 15

The two hockey pucks *A *and *B *each have a mass *M*. If they collide at *O *and are deflected
along the colored paths, determine their speeds just after impact. Assume that the icy surface
over which they slide is smooth. *Hint: *Since the *y'* axis is *not *along the line of impact, apply
the conservation of momentum along the *x'* and *y'* axes.

Given:

*M *250 g= θ*1 *30 deg=

*v1 *40
m
s

= θ*2 *20 deg=

*v2 *60
m
s

= θ*3 *45 deg=

Solution:

Initial Guess:

*vA2 *5
m
s

= *vB2 *4
m
s

=

Given

*M v2 *cos θ*3*( ) *M v1 *cos θ*1*( )+ *M vA2 *cos θ*1*( ) *M vB2 *cos θ*2*( )+=

*M*− *v2 *sin θ*3*( ) *M v1 *sin θ*1*( )+ *M vA2 *sin θ*1*( ) *M vB2 *sin θ*2*( )−=
*vA2
*

*vB2
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

Find *vA2 vB2*,( )=
*vA2
*

*vB2
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

6.90

75.66 ⎛ ⎜ ⎝

⎞ ⎟ ⎠

m s

=

**Problem 15-83
**

Two smooth coins *A* and *B*, each having the same mass, slide on a smooth surface with the
motion shown. Determine the speed of each coin after collision if they move off along the dashed
paths. *Hint:* Since the line of impact has not been defined, apply the conservation of momentum
along the *x* and *y* axes, respectively.

358

**Problem 15-82
**

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Engineering Mechanics - Dynamics Chapter 15

Given:

*vA1 *0.5
ft
s

=

*vB1 *0.8
ft
s

=

α 30 deg=

β 45 deg=

γ 30 deg=

*c *4=

*d *3=

Solution:

Guesses *vB2 *0.25
ft
s

= *vA2 *0.5
ft
s

=

Given

*vA1*−
*c
*

*c*2 *d*2+

⎛ ⎜ ⎝

⎞ ⎟ ⎠

*vB1 *sin γ( )− *vA2*− sin β( ) *vB2 *cos α( )−=

*vA1*−
*d
*

*c*2 *d*2+

⎛ ⎜ ⎝

⎞ ⎟ ⎠

*vB1 *cos γ( )+ *vA2 *cos β( ) *vB2 *sin α( )−=

*vA2
*

*vB2
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

Find *vA2 vB2*,( )=
*vA2
*

*vB2
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

0.766

0.298 ⎛ ⎜ ⎝

⎞ ⎟ ⎠

ft s

=

***Problem 15-84
**

The two disks *A *and *B *have a mass *MA* and *MB*, respectively. If they collide with the initial
velocities shown, determine their velocities just after impact. The coefficient of restitution is *e*.

Given:

*MA *3 kg=

*MB *5 kg=

θ 60 deg=

*vB1 *7
m
s

=

359

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Engineering Mechanics - Dynamics Chapter 15

*vA1 *6
m
s

=

*e *0.65=

Solution: Guesses *vA2 *1
m
s

= *vB2 *1
m
s

= θ*2 *20 deg=

Given

*MA vA1 MB vB1 *cos θ( )− *MA vA2 MB vB2 *cos θ*2*( )+=
*e vA1 vB1 *cos θ( )+( ) *vB2 *cos θ*2*( ) *vA2*−= *vB1 *sin θ( ) *vB2 *sin θ*2*( )=

*vA2
*

*vB2
*

θ*2
*

⎛ ⎜ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎟ ⎠

Find *vA2 vB2*, θ*2*,( )= θ*2 *68.6 deg=
*vA2
*

*vB2
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

3.80−

6.51 ⎛ ⎜ ⎝

⎞ ⎟ ⎠

m s

=

**Problem 15-85
**

Two smooth disks *A* and *B* each have mass *M*. If both disks are moving with the velocities shown
when they collide, determine their final velocities just after collision. The coefficient of restitution is *e*.

Given:

*M *0.5 kg= *c *4= *vA1 *6
m
s

=

*e *0.75= *d *3= *vB1 *4
m
s

=

Solution:

Guesses

*vA2 *1
m
s

= *vB2 *1
m
s

= θ*A *10 deg= θ*B *10 deg=

Given *vA1 *0( ) *vA2 *sin θ*A*( )= *vB1 c
c*2 *d*2+

⎛ ⎜ ⎝

⎞ ⎟ ⎠

*vB2 *sin θ*B*( )=

*M vB1
d
*

*c*2 *d*2+

⎛ ⎜ ⎝

⎞ ⎟ ⎠

*M vA1*− *M vA2 *cos θ*A*( ) *M vB2 *cos θ*B*( )−=

*e vA1 vB1
d
*

*c*2 *d*2+

⎛ ⎜ ⎝

⎞ ⎟ ⎠

+⎡⎢ ⎣

⎤ ⎥ ⎦

*vA2 *cos θ*A*( ) *vB2 *cos θ*B*( )+=

360

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Engineering Mechanics - Dynamics Chapter 15

*vA2
*

*vB2
*

θ*A
*

θ*B
*

⎛⎜ ⎜ ⎜ ⎜ ⎜⎝

⎞⎟ ⎟ ⎟ ⎟ ⎟⎠

Find *vA2 vB2*, θ*A*, θ*B*,( )=
θ*A
*

θ*B
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

0.00

32.88 ⎛ ⎜ ⎝

⎞ ⎟ ⎠

deg=
*vA2
*

*vB2
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

1.35

5.89 ⎛ ⎜ ⎝

⎞ ⎟ ⎠

m s

=

**Problem 15-86
**

Two smooth disks *A* and *B* each have mass *M*. If both disks are moving with the velocities shown
when they collide, determine the coefficient of restitution between the disks if after collision *B* travels
along a line angle θ counterclockwise from the *y* axis.

Given:

*M *0.5 kg= *c *4= *vA1 *6
m
s

=

θ*B *30 deg= *d *3= *vB1 *4
m
s

=

Solution:

Guesses

*vA2 *2
m
s

= *vB2 *1
m
s

= θ*A *10 deg= *e *0.5=

Given *vA1*0 *vA2 *sin θ*A*( )= *vB1 c
c*2 *d*2+

⎛ ⎜ ⎝

⎞ ⎟ ⎠

*vB2 *cos θ*B*( )=

*M vB1
d
*

*c*2 *d*2+

⎛ ⎜ ⎝

⎞ ⎟ ⎠

*M vA1*− *M vA2 *cos θ*A*( ) *M vB2 *sin θ*B*( )−=

*e vA1 vB1
d
*

*c*2 *d*2+

⎛ ⎜ ⎝

⎞ ⎟ ⎠

+⎡⎢ ⎣

⎤ ⎥ ⎦

*vA2 *cos θ*A*( ) *vB2 *sin θ*B*( )+=

*vA2
*

*vB2
*

θ*A
*

*e
*

⎛ ⎜ ⎜ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

Find *vA2 vB2*, θ*A*, *e*,( )=
*vA2
*

*vB2
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

1.75−

3.70 ⎛ ⎜ ⎝

⎞ ⎟ ⎠

m s

= *e *0.0113=

**Problem 15-87
**

Two smooth disks *A* and *B* have the initial velocities shown just before they collide at *O*. If they
have masses *mA* and *mB*, determine their speeds just after impact. The coefficient of restitution is *e*.

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Engineering Mechanics - Dynamics Chapter 15

Given:

*vA *7
m
s

= *mA *8 kg= *c *12= *e *0.5=

*vB *3
m
s

= *mB *6 kg= *d *5=

Solution: θ atan
*d
c
*

⎛⎜ ⎝

⎞⎟ ⎠

= θ 22.62 deg=

Guesses *vA2t *1
m
s

= *vA2n *1
m
s

=

*vB2t *1
m
s

= *vB2n *1
m
s

=

Given *vB *cos θ( ) *vB2t*= *vA*− cos θ( ) *vA2t*=

*mB vB *sin θ( ) *mA vA *sin θ( )− *mB vB2n mA vA2n*+=

*e vB vA*+( ) sin θ( ) *vA2n vB2n*−=

*vA2t
*

*vA2n
*

*vB2t
*

*vB2n
*

⎛ ⎜ ⎜ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

Find *vA2t vA2n*, *vB2t*, *vB2n*,( )=

*vA2t
*

*vA2n
*

*vB2t
*

*vB2n
*

⎛ ⎜ ⎜ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

6.46−

0.22−

2.77

2.14−

⎛⎜ ⎜ ⎜ ⎜ ⎝

⎞⎟ ⎟ ⎟ ⎟ ⎠

m s

=

*vA2 vA2t
*2 *vA2n
*

2+= *vA2 *6.47
m
s

=

*vB2 vB2t
*2 *vB2n
*

2+= *vB2 *3.50
m
s

=

***Problem 15-88
**

The “stone” *A* used in the sport of curling slides over the
ice track and strikes another “stone” *B* as shown. If
each “stone” is smooth and has weight *W*, and the
coefficient of restitution between the “stones” is *e*,
determine their speeds just after collision. Initially *A* has
velocity *vA1* and *B* is at rest. Neglect friction.

Given: *W *47 lb= *vA1 *8
ft
s

=

*e *0.8= θ 30 deg=

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Engineering Mechanics - Dynamics Chapter 15

Solution:

Guesses *vA2t *1
ft
s

= *vA2n *1
ft
s

=

*vB2t *1
ft
s

= *vB2n *1
ft
s

=

Given *vA1 *sin θ( ) *vA2t*= 0 *vB2t*=

*vA1 *cos θ( ) *vA2n vB2n*+=

*e vA1 *cos θ( ) *vB2n vA2n*−=

*vA2t
*

*vA2n
*

*vB2t
*

*vB2n
*

⎛ ⎜ ⎜ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

Find *vA2t vA2n*, *vB2t*, *vB2n*,( )=

*vA2t
*

*vA2n
*

*vB2t
*

*vB2n
*

⎛ ⎜ ⎜ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

4.00

0.69

0.00

6.24

⎛⎜ ⎜ ⎜ ⎜ ⎝

⎞⎟ ⎟ ⎟ ⎟ ⎠

ft s

=

*vA2 vA2t
*2 *vA2n
*

2+= *vA2 *4.06
ft
s

=

*vB2 vB2t
*2 *vB2n
*

2+= *vB2 *6.24
ft
s

=

**Problem 15-89
**

The two billiard balls *A *and *B *are originally in contact
with one another when a third ball *C *strikes each of
them at the same time as shown. If ball *C *remains at
rest after the collision, determine the coefficient of
restitution. All the balls have the same mass. Neglect
the size of each ball.

Solution:

Conservation of “*x*” momentum:

*m v *2*m v' *cos 30 deg( )=

*v *2*v' *cos 30 deg( )= 1( )

Coefficient of restitution:

*e
v'
*

*v *cos 30 deg( )
= 2( )

Substituiting Eq. (1) into Eq. (2) yields:

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Engineering Mechanics - Dynamics Chapter 15

*e
v'
*

2*v' *cos 30 deg( )2
= *e
*

2 3

=

**Problem 15-90
**

Determine the angular momentum of particle
*A* of weight *W* about point *O*. Use a Cartesian
vector solution.

Given:

*W *2 lb= *a *3 ft=

*b *2 ft=*vA *12
ft
s

=

*c *2 ft=

*g *32.2
ft

s2
= *d *4 ft=

Solution:

**rOA
**

*c*−

*a b*+

*d
*

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

= **rv
**

*c
*

*b*−

*d*−

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

= **vAv ***vA
***rv
rv
**

=

**HO rOA ***W***vAv**( )×= **HO
**1.827−

0.000

0.914−

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

slug ft2

s ⋅=

**Problem 15-91
**

Determine the angular momentum **HO** of the
particle about point *O*.

Given:

*M *1.5 kg=

*v *6
m
s

=

*a *4 m=

*b *3 m=

*c *2 m=

*d *4 m=

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Engineering Mechanics - Dynamics Chapter 15

Solution:

**rOA
**

*c*−

*b*−

*d
*

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

= **rAB
**

*c
*

*a*−

*d*−

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

= **vA ***v
***rAB
rAB
**

=

**HO rOA ***M***vA**( )×= **HO
**42.0

0.0

21.0

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

kg m2⋅ s

=

***Problem 15-92
**

Determine the angular momentum **HO** of each of the particles about point *O*.

Given: θ 30 deg= φ 60 deg=

*mA *6 kg= *c *2 m=

*mB *4 kg= *d *5 m=

*mC *2 kg= *e *2 m=

*vA *4
m
s

= *f *1.5 m=

*vB *6
m
s

= *g *6 m=

*vC *2.6
m
s

= *h *2 m=

*a *8 m= *l *5=

*b *12 m= *n *12=

Solution:

**HAO ***a mA vA *sin φ( ) *b mA vA *cos φ( )−= **HAO **22.3
kg m2⋅

s =

**HBO ***f*− *mB vB *cos θ( ) *e mB vB *sin θ( )+= **HBO **7.18−
kg m2⋅

s =

**HCO ***h*− *mC
n
*

*l*2 *n*2+

⎛ ⎜ ⎝

⎞ ⎟ ⎠

*vC g mC
l
*

*l*2 *n*2+

⎛ ⎜ ⎝

⎞ ⎟ ⎠

*vC*−= **HCO **21.60−
kg m2⋅

s =

365

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Engineering Mechanics - Dynamics Chapter 15

**Problem 15-93
**

Determine the angular momentum **HP** of each of the particles about point *P*.

Given: θ 30 deg= φ 60 deg= *a *8 m= *f *1.5 m=

*mA *6 kg= *b *12 m= *g *6 m=*vA *4
m
s

=

*c *2 m= *h *2 m=
*mB *4 kg= *vB *6

m s

=
*d *5 m= *l *5=

*mC *2 kg= *vC *2.6
m
s

= *e *2 m= *n *12=

Solution:

**HAP ***mA vA *sin φ( ) *a d*−( ) *mA vA *cos φ( ) *b c*−( )−=

**HAP **57.6−
kg m2⋅

s =

**HBP ***mB vB *cos θ( ) *c f*−( ) *mB vB *sin θ( ) *d e*+( )+=

**HBP **94.4
kg m2⋅

s =

**HCP ***mC*−
*n
*

*l*2 *n*2+

⎛ ⎜ ⎝

⎞ ⎟ ⎠

*vC c h*+( ) *mC
l
*

*l*2 *n*2+

⎛ ⎜ ⎝

⎞ ⎟ ⎠

*vC d g*+( )−=

**HCP **41.2−
kg m2⋅

s =

**Problem 15-94
**

Determine the angular momentum **HO
**of the particle about point *O*.

Given:

*W *10 lb= *d *9 ft=

*v *14
ft
s

= *e *8 ft=

*a *5 ft= *f *4 ft=

*b *2 ft= *g *5 ft=

*c *3 ft= *h *6 ft=

366

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Engineering Mechanics - Dynamics Chapter 15

Solution:

**rOA
**

*f*−

*g
*

*h
*

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

= **rAB
**

*f e*+

*d g*−

*h*−

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

=

**vA ***v
***rAB
rAB
**

= **HO rOA ***W***vA**( )×= **HO
**16.78−

14.92

23.62−

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

slug ft2

s ⋅=

**Problem 15-95
**

Determine the angular momentum **HP** of the particle about point *P*.

Given:

*W *10 lb= *d *9 ft=

*v *14
ft
s

= *e *8 ft=

*a *5 ft= *f *4 ft=

*b *2 ft= *g *5 ft=

*c *3 ft= *h *6 ft=

Solution:

**rPA
**

*f*− *c*−

*b g*+

*h a*−

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

= **rAB
**

*f e*+

*d g*−

*h*−

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

=

**vA ***v
***rAB
rAB
**

= **HP rPA ***W***vA**( )×= **HP
**14.30−

9.32−

34.81−

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

slug ft2

s ⋅=

***Problem 15-96
**

Determine the total angular momentum **HO** for the system of three particles about point *O*. All the
particles are moving in the *x-y* plane.

Given:

*mA *1.5 kg= *a *900 mm=

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Engineering Mechanics - Dynamics Chapter 15

*vA *4
m
s

= *b *700 mm=

*mB *2.5 kg= *c *600 mm=

*vB *2
m
s

= *d *800 mm=

*mC *3 kg= *e *200 mm=

*vC *6
m
s

=

Solution:

**HO
**

*a
*

0

0

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

*mA
*

0

*vA*−

0

⎛⎜ ⎜ ⎜⎝

⎞⎟ ⎟ ⎟⎠

⎡⎢ ⎢ ⎢⎣

⎤⎥ ⎥ ⎥⎦

×

*c
*

*b
*

0

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

*mB
*

*vB*−

0

0

⎛⎜ ⎜ ⎜⎝

⎞⎟ ⎟ ⎟⎠

⎡⎢ ⎢ ⎢⎣

⎤⎥ ⎥ ⎥⎦

×+

*d*−

*e*−

0

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

*mC
*

0

*vC*−

0

⎛⎜ ⎜ ⎜⎝

⎞⎟ ⎟ ⎟⎠

⎡⎢ ⎢ ⎢⎣

⎤⎥ ⎥ ⎥⎦

×+=

**HO
**

0.00

0.00

12.50

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

kg m2⋅ s

=

**Problem 15-97
**

Determine the angular momentum **HO** of each of the two particles about point *O*. Use a scalar
solution.

Given:

*mA *2 kg= *c *1.5 m=

*mB *1.5 kg= *d *2 m=

*e *4 m=*vA *15
m
s

=

*f *1 m=

*vB *10
m
s

= θ 30 deg=

*a *5 m= *l *3=

*b *4 m= *n *4=

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Engineering Mechanics - Dynamics Chapter 15

Solution:

**HOA ***mA*−
*n
*

*n*2 *l*2+

⎛ ⎜ ⎝

⎞ ⎟ ⎠

*vA c mA
l
*

*n*2 *l*2+

⎛ ⎜ ⎝

⎞ ⎟ ⎠

*vA d*−= **HOA **72.0−
kg m2⋅

s =

**HOB ***mB*− *vB *cos θ( )*e mB vB *sin θ( ) *f*−= **HOB **59.5−
kg m2⋅

s =

**Problem 15-98
**

Determine the angular momentum **HP** of each of the two particles about point *P*. Use a scalar
solution.

Given:

*mA *2 kg= *c *1.5 m=

*d *2 m=*mB *1.5 kg=

*e *4 m=
*vA *15

m s

=
*f *1 m=

*vB *10
m
s

= θ 30 deg=

*a *5 m= *l *3=

*b *4 m= *n *4=

Solution:

**HPA ***mA
n
*

*n*2 *l*2+
*vA b c*−( ) *mA
*

*l
*

*n*2 *l*2+
*vA a d*+( )−= **HPA **66.0−

kg m2⋅ s

=

**HPB ***mB*− *vB *cos θ( ) *b e*+( ) *mB vB *sin θ( ) *a f*−( )+= **HPB **73.9−
kg m2⋅

s =

**Problem 15-99
**

The ball *B* has mass *M* and is attached to the end of a rod whose mass may be neglected. If the
rod is subjected to a torque *M *=* at*2+ *bt *+ *c*, determine the speed of the ball when *t* = *t1*. The ball
has a speed *v* = *v0* when *t* = 0.

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Engineering Mechanics - Dynamics Chapter 15

Given:

*M *10 kg=

*a *3
N m⋅

s2 =

*b *5
N m⋅

s =

*c *2 N m⋅=

*t1 *2 s=

*v0 *2
m
s

=

*L *1.5 m=

Solution: Principle of angular impulse momentum

*M v0 L
*0

*t1
ta t*2 *b t*+ *c*+

⌠ ⎮ ⌡

d+ *M v1 L*=

*v1 v0
*1

*M L *0

*t1
ta t*2 *b t*+ *c*+

⌠ ⎮ ⌡

d+= *v1 *3.47
m
s

=

***Problem 15-100
**

The two blocks *A *and *B *each have a mass *M0*. The blocks are fixed to the horizontal rods, and
their initial velocity is *v'* in the direction shown. If a couple moment of *M* is applied about shaft
*CD *of the frame, determine the speed of the blocks at time *t*. The mass of the frame is
negligible, and it is free to rotate about *CD*. Neglect the size of the blocks.

Given:

*M0 *0.4 kg=

*a *0.3 m=

*v' *2
m
s

=

*M *0.6 N m⋅=

*t *3 s=

Solution:

2*a M0 v' M t*+ 2*a M0 v*=

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Engineering Mechanics - Dynamics Chapter 15

*v v'
M t
*

2*a M0
*+= *v *9.50

m s

=

**Problem 15-101
**

The small cylinder* C* has mass *mC* and is attached to the end of a rod whose mass may be

neglected. If the frame is subjected to a couple *M* = *at2 + b*, and the cylinder is subjected to
force *F*, which is always directed as shown, determine the speed of the cylinder when *t* = *t1*.
The cylinder has a speed *v0* when *t* = 0.

Given:

*mC *10 kg= *t1 *2 s=

*a *8 N
m

s2
= *v0 *2

m s

=

*d *0.75 m=
*b *5 N m⋅=

*e *4=

*F *60 N= *f *3=

Solution:

*mC v0 d
*0

*t1
ta t*2 *b*+

⌠ ⎮ ⌡

d+
*f
*

*e*2 *f *2+

⎛ ⎜ ⎝

⎞ ⎟ ⎠

*F d t1*+ *mC v1 d*=

*v1 v0
*1

*mC d *0

*t1
ta t*2 *b*+

⌠ ⎮ ⌡

d
*f
*

*e*2 *f *2+

⎛ ⎜ ⎝

⎞ ⎟ ⎠

*F d t1*+
⎡
⎢
⎢
⎣

⎤ ⎥ ⎥ ⎦

+= *v1 *13.38
m
s

=

**Problem 15-102
**

A box having a weight *W* is moving around in a circle of radius *rA* with a speed *vA1* while
connected to the end of a rope. If the rope is pulled inward with a constant speed *vr*, determine
the speed of the box at the instant *r* = *rB*. How much work is done after pulling in the rope
from *A *to *B*? Neglect friction and the size of the box.

Given:

*W *8 lb=

*rA *2 ft=

371

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Engineering Mechanics - Dynamics Chapter 15

*vA1 *5
ft
s

=

*vr *4
ft
s

=

*rB *1 ft=

*g *32.21
ft

s2 =

Solution:

*W
g
*

⎛⎜ ⎝

⎞⎟ ⎠

*rA vA1
W
g
*

⎛⎜ ⎝

⎞⎟ ⎠

*rB vBtangent*=

*vBtangent rA
vA1
rB
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

= *vBtangent *10.00
ft
s

=

*vB vBtangent
*2 *vr
*

2+= *vB *10.8
ft
s

=

*UAB
*1
2

*W
g
*

⎛⎜ ⎝

⎞⎟ ⎠

*vB
*2 1

2
*W
g
*

⎛⎜ ⎝

⎞⎟ ⎠

*vA1
*2−= *UAB *11.3 ft lb⋅=

**Problem 15-103
**

An earth satellite of mass *M* is launched into a free-flight trajectory about the earth with initial
speed *vA* when the distance from the center of the earth is *rA*. If the launch angle at this position
is φ*A* determine the speed *vB *of the satellite and its closest distance *rB* from the center of the
earth. The earth has a mass *Me*.* Hint:* Under these conditions, the satellite is subjected only to the
earth’s gravitational force, **F**, Eq. 13-1. For part of the solution, use the conservation of energy.

Units used: Mm 103 km=

Given:

φ*A *70 deg=*M *700 kg=

*vA *10
km
s

= *G *6.673 10 11−×
N m2⋅

kg2 =

*rA *15 Mm= *Me *5.976 10
24× kg=

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Engineering Mechanics - Dynamics Chapter 15

Solution: Guesses *vB *10
km
s

= *rB *10 Mm=

Given *M vA *sin φ*A*( )*rA M vB rB*=

1 2

*M vA
*2 *G Me M
*

*rA
*−

1 2

*M vB
*2 *G Me M
*

*rB
*−=

*vB
*

*rB
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

Find *vB rB*,( )= *vB *10.2
km
s

= *rB *13.8 Mm=

***Problem 15-104
**

The ball *B* has weight *W* and is originally rotating in a circle. As shown, the cord *AB* has a length of
*L* and passes through the hole *A*, which is a distance *h* above the plane of motion. If *L*/2 of the cord
is pulled through the hole, determine the speed of the ball when it moves in a circular path at *C*.

Given:

*W *5 lb=

*L *3 ft=

*h *2 ft=

*g *32.2
ft

s2 =

Solution: θ*B *acos
*h
L
*

⎛⎜ ⎝

⎞⎟ ⎠

= θ*B *48.19 deg=

Guesses *TB *1 lb= *TC *1 lb= *vB *1
ft
s

= *vC *1
ft
s

= θ*C *10 deg=

Given *TB *cos θ*B*( ) *W*− 0= *TB *sin θ*B*( ) *Wg
vB
*

2

*L *sin θ*B*( )
⎛⎜
⎜⎝

⎞⎟ ⎟⎠

=

*TC *cos θ*C*( ) *W*− 0= *TC *sin θ*C*( ) *Wg
vC
*

2

*L
*2

sin θ*C*( )

⎛⎜ ⎜ ⎜⎝

⎞⎟ ⎟ ⎟⎠

=

*W
g
*

⎛⎜ ⎝

⎞⎟ ⎠

*vB L *sin θ*B*( ) *Wg
*⎛⎜
⎝

⎞⎟ ⎠

*vC
L
*2

⎛⎜ ⎝

⎞⎟ ⎠

sin θ*C*( )=

373

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Engineering Mechanics - Dynamics Chapter 15

*TB
*

*TC
*

*vB
*

*vC
*

θ*C
*

⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

Find *TB TC*, *vB*, *vC*, θ*C*,( )=
*TB
*

*TC
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

7.50

20.85 ⎛ ⎜ ⎝

⎞ ⎟ ⎠

lb= θ*C *76.12 deg=

*vB *8.97
ft
s

= *vC *13.78
ft
s

=

**Problem 15-105
**

The block of weight *W* rests on a surface for which the kinetic coefficient of friction is μ*k*. It
is acted upon by a radial force **FR** and a horizontal force **FH**, always directed at angle θ from
the tangent to the path as shown. If the block is initially moving in a circular path with a speed
*v1* at the instant the forces are applied, determine the time required before the tension in cord
*AB *becomes *T*. Neglect the size of the block for the calculation.

Given:

*W *10 lb= μ*k *0.5=

*FR *2 lb= *T *20 lb=

*FH *7 lb= *r *4 ft=

*v1 *2
ft
s

= *g *32.2
ft

s2 =

θ 30 deg=

Solution:

Guesses *t *1 s= *v2 *1
ft
s

=

Given

*W
g
*

⎛⎜ ⎝

⎞⎟ ⎠

*v1 r FH*cos θ( )*r t*+ μ*k W rt*−
*W
g
*

⎛⎜ ⎝

⎞⎟ ⎠

*v2 r*=

*FR FH *sin θ( )+ *T*−
*W
g
*

−
*v2
*

2

*r
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

=

*t
*

*v2
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

Find *t v2*,( )= *v2 *13.67
ft
s

= *t *3.41 s=

374

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Engineering Mechanics - Dynamics Chapter 15

**Problem 15-106
**

The block of weight *W* is originally at rest on the smooth surface. It is acted upon by a radial
force **FR** and a horizontal force **FH**, always directed at θ from the tangent to the path as
shown. Determine the time required to break the cord, which requires a tension *T*. What is the
speed of the block when this occurs? Neglect the size of the block for the calculation.

Given:

*W *10 lb= θ 30 deg=

*FR *2 lb= *T *30 lb=

*FH *7 lb= *r *4 ft=

*v1 *0
ft
s

= *g *32.2
ft

s2 =

Solution:

Guesses *t *1 s= *v2 *1
ft
s

=

Given

*W
g
*

⎛⎜ ⎝

⎞⎟ ⎠

*v1 r FH*cos θ( )*r t*+
*W
g
*

⎛⎜ ⎝

⎞⎟ ⎠

*v2 r*=

*FR FH *sin θ( )+ *T*−
*W
g
*

−
*v2
*

2

*r
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

=

*t
*

*v2
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

Find *t v2*,( )= *v2 *17.76
ft
s

= *t *0.91 s=

**Problem 15-107
**

The roller-coaster car of weight *W* starts from
rest on the track having the shape of a
cylindrical helix. If the helix descends a
distance *h* for every one revolution, determine
the time required for the car to attain a speed *v*.
Neglect friction and the size of the car.

Given:

*W *800 lb=

*h *8 ft=

*v *60
ft
s

=

375

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Engineering Mechanics - Dynamics Chapter 15

*r *8 ft=

Solution:

θ atan
*h
*

2π*r
*⎛⎜
⎝

⎞⎟ ⎠

= θ 9.04 deg=

*FN W *cos θ( )− 0= *FN W *cos θ( )= *FN *790.06 lb=

*vt v *cos θ( )= *vt *59.25
ft
s

=

*HA tM
*⌠
⎮
⌡

d+ *H2*=
0

*t
tFN *sin θ( )*r
*

⌠ ⎮ ⌡

d
*W
g
*

⎛⎜ ⎝

⎞⎟ ⎠

*h vt*= *FN *sin θ( )*r t
W
g
*

⎛⎜ ⎝

⎞⎟ ⎠

*h vt*=

*t W
vt h
*

*FN *sin θ( )*g r
*⎛
⎜
⎝

⎞ ⎟ ⎠

= *t *11.9 s=

***Problem 15-108
**

A child having mass *M* holds her legs up as shown as she swings downward from rest at θ*1*. Her
center of mass is located at point *G1*. When she is at the bottom position θ = 0°, she *suddenly* lets her
legs come down, shifting her center of mass to position *G2*. Determine her speed in the upswing due
to this sudden movement and the angle θ*2* to which she swings before momentarily coming to rest.
Treat the child’s body as a particle.

Given:

*M *50 kg= *r1 *2.80 m= *g *9.81
m

s2 =

θ*1 *30 deg= *r2 *3 m=

Solution:

*v2b *2*g r1 *1 cos θ*1*( )−( )= *v2b *2.71 ms=

*r1 v2b r2 v2a*= *v2a
r1
r2
*

*v2b*= *v2a *2.53
m
s

=

θ*2 *acos 1
*v2a
*

2

2*g r2
*−

⎛⎜ ⎜⎝

⎞⎟ ⎟⎠

= θ*2 *27.0 deg=

376

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Engineering Mechanics - Dynamics Chapter 15

**Problem 15-109
**

A small particle having a mass *m *is placed inside the semicircular tube. The particle is
placed at the position shown and released. Apply the principle of angular momentum about
point *O* (Σ*M0* = *H0*), and show that the motion of the particle is governed by the differential
equation θ*''* + (*g* /* R*) sin θ = 0.

Solution:

Σ*M0
t
H0
*

d d

=

*R*− *m g *sin θ( )
*t
*

*m v R*( )d
d

=

*g *sin θ( )
*t
v*d

d
−= 2*t
*

*s*d

d

2 −=

But, *s R*θ=

Thus, *g *sin θ( ) *R*− θ*''*=

or, θ*''
g
R
*

⎛⎜ ⎝

⎞⎟ ⎠

sin θ( )+ 0=

**Problem 15-110
**

A toboggan and rider, having a
total mass *M*, enter horizontally
tangent to a circular curve (θ*1*)
with a velocity *vA*. If the track
is flat and banked at angle θ*2*,
determine the speed *vB* and the
angle θ of “descent”, measured
from the horizontal in a vertical
*x*–*z *plane, at which the
toboggan exists at *B*. Neglect
friction in the calculation.

Given:

*M *150 kg= θ*1 *90 deg= *vA *70
km
hr

= θ*2 *60 deg=

*rA *60 m= *rB *57 m= *r *55 m=

377

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Engineering Mechanics - Dynamics Chapter 15

Solution:

*h rA rB*−( )tan θ*2*( )=

Guesses *vB *10
m
s

= θ 1 deg=

Given 1 2

*M vA
*2 *M g h*+

1 2

*M vB
*2= *M vA rA M vB *cos θ( )*rB*=

*vB
*

θ

⎛ ⎜ ⎝

⎞ ⎟ ⎠

Find *vB *θ,( )= *vB *21.9 ms= θ 1.1− 10
3× deg=

**Problem 15-111
**

Water is discharged at speed *v* against the fixed cone diffuser. If the opening diameter of the
nozzle is *d*, determine the horizontal force exerted by the water on the diffuser.

Units Used:

Mg 103 kg=

Given:

*v *16
m
s

= θ 30 deg=

*d *40 mm= ρ*w *1
Mg

m3 =

Solution:

*Q
*π
4

*d*2*v*= *m' *ρ*w Q*=

*Fx m' v*− cos
θ
2

⎛⎜ ⎝

⎞⎟ ⎠

*v*+⎛⎜
⎝

⎞⎟ ⎠

=

*Fx *11.0 N=

***Problem 15-112
**

A jet of water having cross-sectional area *A* strikes the fixed blade with speed *v*. Determine the
horizontal and vertical components of force which the blade exerts on the water.

Given:

*A *4 in2=

378

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Engineering Mechanics - Dynamics Chapter 15

*v *25
ft
s

=

θ 130 deg=

γ*w *62.4
lb

ft3 =

Solution: *Q A v*= *Q *0.69
ft3

s =

*t
m*d

d
*m'*= ρ*Q*= *m' *γ*w Q*= *m' *1.3468

slug s

=

*vAx v*= *vAy *0
ft
s

= *vBx v *cos θ( )= *vBy v *sin θ( )=

*Fx
m'*−
*g
*

*vBx vAx*−( )= *Fx *55.3 lb=

*Fy
m'
g
*

*vBy vAy*−( )= *Fy *25.8 lb=

**Problem 15-113
**

Water is flowing from the fire hydrant opening of diameter *dB* with velocity *vB*. Determine the
horizontal and vertical components of force and the moment developed at the base joint *A*, if the
static (gauge) pressure at *A* is *PA*. The diameter of the fire hydrant at *A* is *dA*.

Units Used:

kPa 103 Pa=

Mg 103 kg=

kN 103 N=

Given:

*dB *150 mm= *h *500 mm=

*vB *15
m
s

= *dA *200 mm=

ρ*w *1
Mg

m3
=*PA *50 kPa=

Solution:

*AB *π
*dB
*2

⎛ ⎜ ⎝

⎞ ⎟ ⎠

2
= *AA *π

*dA
*2

⎛ ⎜ ⎝

⎞ ⎟ ⎠

2
= *m' *ρ*w vB*π

*dB
*2

⎛ ⎜ ⎝

⎞ ⎟ ⎠

2
= *vA
*

*m'
*ρ*w AA
*

=

379

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Engineering Mechanics - Dynamics Chapter 15

*Ax m' vB*= *Ax *3.98 kN=

*Ay*− 50π
*dA
*2

⎛ ⎜ ⎝

⎞ ⎟ ⎠

2
+ *m' *0 *vA*−( )= *Ay m' vA PA*π

*dA
*2

⎛ ⎜ ⎝

⎞ ⎟ ⎠

2
+= *Ay *3.81 kN=

*M m' h vB*= *M *1.99 kN m⋅=

**Problem 15-114
**

The chute is used to divert the flow of
water *Q*. If the water has a
cross-sectional area *A*, determine the
force components at the pin *A *and roller
*B *necessary for equilibrium. Neglect
both the weight of the chute and the
weight of the water on the chute.

Units Used:

Mg 103 kg= kN 103 N=

Given:

*Q *0.6
m3

s
= ρ*w *1

Mg

m3 =

*A *0.05 m2= *h *2 m=

*a *1.5 m= *b *0.12 m=

Solution:

*t
m*d

d
*m'*= *m' *ρ*w Q*=

*vA
Q
A
*

= *vB vA*=

Σ*Fx m' vAx vBx*−( )= *Bx Ax*− *m' vAx vBx*−( )=
Σ*Fy m' vAy vBy*−( )= *Ay m' *0 *vB*−( )−⎡⎣ ⎤⎦= *Ay *7.20 kN=

Σ*MA m' d0A vA d0B vB*−( )= *Bx
*1
*h
*

*m' b vA a b*−( )*vA*+⎡⎣ ⎤⎦= *Bx *5.40 kN=

*Ax Bx m' vA*−= *Ax *1.80− kN=

380

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Engineering Mechanics - Dynamics Chapter 15

**Problem 15-115
**

The fan draws air through a vent with speed *v*. If the cross-sectional area of the vent is *A*,
determine the horizontal thrust on the blade. The specific weight of the air is γ*a*.

Given:

*v *12
ft
s

=

*A *2 ft2=

γ*a *0.076
lb

ft3 =

*g *32.20
ft

s2 =

Solution:

*m'
t
m*d

d
= *m' *γ*a v A*= *m' *0.05669

slug s

=

*T
m' v *0−( )

*g
*= *T *0.68 lb=

***Problem 15-116
**

The buckets on the *Pelton wheel* are
subjected to a jet of water of diameter *d*,
which has velocity *vw*. If each bucket is
traveling at speed *vb* when the water
strikes it, determine the power developed
by the wheel. The density of water is γ*w*.

Given:

*d *2 in= θ 20 deg=

*vw *150
ft
s

= *g *32.2
ft

s2 =

*vb *95
ft
s

=

γ*w *62.4
lbf

ft3 =

381

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Engineering Mechanics - Dynamics Chapter 15

Solution: *vA vw vb*−= *vA *55
ft
s

=

*vBx vA*− cos θ( ) *vb*+= *vBx *43.317
ft
s

=

Σ*Fx m' vBx vAx*−( )=

*Fx
*γ*w
g
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

π
*d*2

4

⎛ ⎜ ⎝

⎞ ⎟ ⎠

*vA vBx*− *vA*−( )−⎡⎣ ⎤⎦= *Fx *266.41
m

s2 lb⋅=

*P Fx vb*= *P *4.69 hp=

**Problem 15-117
**

The boat of mass *M* is powered by a fan *F *which develops a slipstream having a diameter *d*. If
the fan ejects air with a speed *v*, measured relative to the boat, determine the initial acceleration
of the boat if it is initially at rest. Assume that air has a constant density ρ*a* and that the entering
air is essentially at rest. Neglect the drag resistance of the water.

Given:

*M *200 kg=

*h *0.375 m=

*d *0.75 m=

*v *14
m
s

=

ρ*a *1.22
kg

m3 =

Solution:

*Q A v*= *Q
*π
4

*d*2*v*= *Q *6.1850
m3

s =

*t
m*d

d
*m'*= *m' *ρ*a Q*= *m' *7.5457

kg s

=

Σ*Fx m' vBx vAx*−( )=

*F *ρ*a Q v*= *F *105.64 N=

Σ*Fx M ax*= *F M a*=

382

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Engineering Mechanics - Dynamics Chapter 15

*a
F
M
*

= *a *0.53
m

s2 =

**Problem 15-118
**

The rocket car has a mass *MC* (empty) and carries fuel of mass *MF*. If the fuel is consumed
at a constant rate *c* and ejected from the car with a relative velocity *vDR*, determine the
maximum speed attained by the car starting from rest. The drag resistance due to the
atmosphere is *FD* = *kv*2 and the speed is measured in m/s.

Units Used:

Mg 103 kg=

Given:

*MC *3 Mg= *MF *150 kg=

*vDR *250
m
s

= *c *4
kg
s

=

*k *60 N
s2

m2 ⋅=

Solution:

*m0 MC MF*+= At time *t* the mass of the car is *m0 c t*−

Set *F k v*2= , then *k*− *v*2 *m0 c t*−( )
*t
v*d

d
*vDRc*−=

Maximum speed occurs at the instant the fuel runs out. *t
MF
c
*

= *t *37.50 s=

Thus, Initial Guess: *v *4
m
s

=

Given

0

*v
*

*v
*1

*c vDR k v
*2−

⌠⎮ ⎮ ⎮⌡

d

0

*t
*

*t
*1

*m0 c t*−

⌠ ⎮ ⎮ ⌡

d=

*v *Find *v*( )= *v *4.06
m
s

=

383

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Engineering Mechanics - Dynamics Chapter 15

**Problem 15-119
**

A power lawn mower hovers very close over the ground. This is done by drawing air in at speed *vA
*through an intake unit *A*, which has cross-sectional area *AA* and then discharging it at the ground,
*B*, where the cross-sectional area is *AB*. If air at *A* is subjected only to atmospheric pressure,
determine the air pressure which the lawn mower exerts on the ground when the weight of the
mower is freely supported and no load is placed on the handle. The mower has mass *M* with center
of mass at *G*. Assume that air has a constant density of ρ*a*.

Given:

*vA *6
m
s

=

*AA *0.25 m
2=

*AB *0.35 m
2=

*M *15 kg=

ρ*a *1.22
kg

m3 =

Solution: *m' *ρ*a AA vA*= *m' *1.83
kg
s

=

+↑ Σ*Fy m' vBy vAy*−( )= *P AB M g*− *m' *0 *vA*−( )−⎡⎣ ⎤⎦=

*P
*1

*AB
m' vA M g*+( )= *P *452 Pa=

***Problem 15-120
**

The elbow for a buried pipe of diameter *d* is
subjected to static pressure *P*. The speed of the
water passing through it is *v*. Assuming the pipe
connection at *A* and *B* do not offer any vertical
force resistance on the elbow, determine the
resultant vertical force **F** that the soil must then
exert on the elbow in order to hold it in
equilibrium. Neglect the weight of the elbow and
the water within it. The density of water is γ*w*.

Given:

*d *5 in= θ 45 deg=

384

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Engineering Mechanics - Dynamics Chapter 15

*P *10
lb

in2
= γ*w *62.4

lb

ft3 =

*v *8
ft
s

=

Solution:

*Q v
*π
4

*d*2⎛⎜
⎝

⎞⎟ ⎠

=

*m'
*γ*w
g
*

*Q*=

Also, the force induced by the water
pressure at *A* is

*A
*π
4

*d*2=

*F P A*= *F *196.35 lb=

2*F *cos θ( ) *F1*− *m' v*− cos θ( ) *v *cos θ( )−( )=

*F1 *2 *F *cos θ( ) *m' v *cos θ( )+( )=

*F1 *302 lb=

**Problem 15-121
**

The car is used to scoop up water that is lying in a trough at the tracks. Determine the force
needed to pull the car forward at constant velocity **v** for each of the three cases. The scoop has
a cross-sectional area *A* and the density of water is ρ*w*.

385

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Engineering Mechanics - Dynamics Chapter 15

Solution:

The system consists of the car and the scoop. In all cases

Σ*Fs m
t
v*d

d
*VDe
*

*t
me
*

d d

−=

*F *0 *V*ρ *A V*−= *F V*2ρ *A*=

**Problem 15-122
**

A rocket has an empty weight *W1* and carries fuel of weight *W2*. If the fuel is burned at the rate
*c* and ejected with a relative velocity *vDR*, determine the maximum speed attained by the rocket
starting from rest. Neglect the effect of gravitation on the rocket.

Given: *W1 *500 lb= *W2 *300 lb= *c *15
lb
s

= *vDR *4400
ft
s

= *g *32.2
ft

s2 =

Solution: *m0
W1 W2*+

*g
*=

The maximum speed occurs when all the fuel is consumed, that is, where *t
W2
c
*

= *t *20.00 s=

Σ*Fx m
t
v*d

d
*vDR
*

*t
me
*

d d

−=

At a time *t*, *M m0
c
g
*

*t*−= , where
*c
g t
*

*me
*d
d

= . In space the weight of the rocket is zero.

0 *m0 c t*−( )
*t
v*d

d
*vDRc*−=

Guess *vmax *1
ft
s

=

Given

0

*vmax
v*1

⌠ ⎮ ⌡

d

0

*t
*

*t
*

*c
g
*

*vDR
*

*m0
c
g
*

*t*−

⌠⎮ ⎮ ⎮ ⎮ ⎮⌡

d=

*vmax *Find *vmax*( )= *vmax *2068
ft
s

=

386

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Engineering Mechanics - Dynamics Chapter 15

**Problem 15-123
**

The boat has mass *M* and is traveling forward on a river with constant velocity *vb*, measured relative
to the river. The river is flowing in the opposite direction at speed *vR*. If a tube is placed in the water,
as shown, and it collects water of mass *Mw* in the boat in time *t*, determine the horizontal thrust *T* on
the tube that is required to overcome the resistance to the water collection.

Units Used:

Mg 103 kg=

Given:

*M *180 kg= *Mw *40 kg=

*vb *70
km
hr

= *t *80 s=

ρ*w *1
Mg

m3
=*vR *5

km hr

=

Solution: *m'
Mw
*

*t
*= *m' *0.50

kg s

=

*vdi vb*= *vdi *19.44
m
s

=

Σ*Fi m
t
v*d

d
*vdi m'*+=

*T vdi m'*= *T *9.72 N=

***Problem 15-124
**

The second stage of a two-stage rocket has weight *W2* and is launched from the first stage with
velocity *v*. The fuel in the second stage has weight *Wf*. If it is consumed at rate *r* and ejected with
relative velocity *vr*, determine the acceleration of the second stage just after the engine is fired.
What is the rocket’s acceleration just before all the fuel is consumed? Neglect the effect of
gravitation.

Given:

*W2 *2000 lb= *Wf *1000 lb= *r *50
lb
s

=

*v *3000
mi
hr

= *vr *8000
ft
s

= *g *32.2
ft

s2 =

387

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Engineering Mechanics - Dynamics Chapter 15

Solution:

Initially,

Σ*Fs m
t
v*d

d
*vdi
*

*t
me
*

d d

⎛ ⎜ ⎝

⎞ ⎟ ⎠

−=

0
*W2 Wf*+

*g
*⎛
⎜
⎝

⎞ ⎟ ⎠

*a vr
r
g
*

−= *a vr
r
*

*W2 Wf*+
⎛
⎜
⎝

⎞ ⎟ ⎠

= *a *133
ft

s2 =

Finally,

*a1 vr
r
*

*W2
*⎛
⎜
⎝

⎞ ⎟ ⎠

= *a1 *200
ft

s2 =0

*W2
g
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

*a1 vr
r
g
*

⎛⎜ ⎝

⎞⎟ ⎠

−=

**Problem 15-125
**

The earthmover initially carries volume *V* of sand having a density ρ. The sand is unloaded
horizontally through *A* dumping port *P *at a rate *m'* measured relative to the port. If the
earthmover maintains a constant resultant tractive force **F** at its front wheels to provide
forward motion, determine its acceleration when half the sand is dumped. When empty, the
earthmover has a mass *M*. Neglect any resistance to forward motion and the mass of the
wheels. The rear wheels are free to roll.
Units Used:

Mg 103 kg=

kN 103 N=

Given:

*A *2.5 m2= ρ 1520
kg

m3 =

*m' *900
kg
s

=
*V *10 m3=

*F *4 kN=

*M *30 Mg=

Solution:

When half the sand remains,

*M1 M
*1
2

*V*ρ+= *M1 *37600 kg=

388

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Engineering Mechanics - Dynamics Chapter 15

*t
m*d

d
*m'*= ρ*v A*= *v
*

*m'
*ρ *A
*

= *v *0.24
m
s

=

Σ *F m
t
v*d

d *t
m*d

d
*vDR*−= *F M1 a m' v*−=

*a
F m' v*+

*M1
*= *a *0.11

m

s2 =

*a *112
mm

s2 =

**Problem 15-126
**

The earthmover initially carries sand of volume *V* having density ρ. The sand is unloaded horizontally
through a dumping port *P* of area *A *at rate of *r* measured relative to the port. Determine the resultant
tractive force **F** at its front wheels if the acceleration of the earthmover is *a* when half the sand is
dumped. When empty, the earthmover has mass *M*. Neglect any resistance to forward motion and
the mass of the wheels. The rear wheels are free to roll.

Units Used:

kN 103 N=

Mg 1000 kg=

Given:

*V *10 m3= *r *900
kg
s

=

ρ 1520 kg

m3
= *a *0.1

m

s2 =

*A *2.5 m2= *M *30 Mg=

Solution:

When half the sand remains, *M1 M
*1
2

*V*ρ+= *M1 *37600 kg=

*t
m*d

d
*r*= *r *ρ*v A*= *v
*

*r
*ρ *A
*

= *v *0.237
m
s

=

*F m
t
v*d

d *t
m*d

d
*v*−= *F M1 a rv*−= *F *3.55 kN=

389

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Engineering Mechanics - Dynamics Chapter 15

**Problem 15-127
**

If the chain is lowered at a constant speed *v*, determine the normal reaction exerted
on the floor as a function of time. The chain has a weight *W* and a total length *l*.

Given:

*W *5
lb
ft

=

*l *20 ft=

*v *4
ft
s

=

Solution:

At time *t*, the weight of the chain on the floor is *W M g v t*( )=

*t
v*d

d
0= *Mt M v t*( )=

*t
Mt
*

d d

*M v*=

Σ *Fs M
t
v*d

d
*vDt
*

*t
Mt
*

d d

+=

*R M g v t*( )− 0 *v M v*( )+=

*R M g v t v*2+( )= *R W
g
*

*g v t v*2+( )=

***Problem 15-128
**

The rocket has mass *M* including the fuel. Determine the constant rate at
which the fuel must be burned so that its thrust gives the rocket a speed
*v* in time *t* starting from rest. The fuel is expelled from the rocket at
relative speed *vr*. Neglect the effects of air resistance and assume that *g
*is constant.

Given:

*M *65000 lb= *vr *3000
ft
s

=

*v *200
ft
s

=

*g *32.2
ft

s2 =

*t *10 s=

390

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Engineering Mechanics - Dynamics Chapter 15

Solution:

*A System That Losses Mass*: Here,

*W m0
t
me
*

d d

*t*−⎛⎜
⎝

⎞ ⎟ ⎠

*g*=

Applying Eq. 15-29, we have

+↑Σ *Fz m tv
*d
d

*vDE
t
me
*

d d

−= integrating we find

*v vDE *ln
*mo
*

*m0
t
me
*

d d

*t*−

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

*g t*−=

with *m0 M*= *vDE vr*=

*v vr *ln
*m0
*

*m0
t
me
*

d d

*t*−

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

*g t*( )−=

*t
me
*

d d

*A*=
*m0*−

*e
*

*v g t*+
*vr
*

*m0*+
⎛⎜
⎜
⎜⎝

⎞⎟ ⎟ ⎟⎠

1
*t
*

= *A
m0*−

*e
*

*v g t*+
*vr
*

*m0*+
⎛⎜
⎜
⎜⎝

⎞⎟ ⎟ ⎟⎠

1
*t
*

= *A *43.3
slug

s =

**Problem 15-129
**

The rocket has an initial mass *m0*, including the fuel. For practical reasons
desired for the crew, it is required that it maintain a constant upward acceleration
*a0*. If the fuel is expelled from the rocket at a relative speed *ver*, determine the
rate at which the fuel should be consumed to maintain the motion. Neglect air
resistance, and assume that the gravitational acceleration is constant.

Solution:

*a0
t
v*d

d =

+↑ Σ*Fs* = *m
t
v*d

d
*ver
*

*t
me
*

d d

−

*mg*− *mao ver
t
m*d

d −=

*ver
*d*m
m
*

*a0 g*+( )d*t*=

391

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Engineering Mechanics - Dynamics Chapter 15

Since *ver* is constant, integrating, with *t* = 0 when *m* = *m0* yields

*ver *ln
*m
m0
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

*a0 g*+( )*t*=
*m
m0
*

*e
*

*a0 g*+

*ver
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

*t
*

=

The time rate fuel consumption is determined from Eq.[1]

*t
m*d

d
*m
*

*a0 g*+

*ver
*=

*t
m*d

d
*m0
*

*a0 g*+

*ver
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

*e
*

*a0 g*+

*ver
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

*t
*

=

Note : *ver * must be considered a negative quantity.

**Problem 15-130
**

The jet airplane of mass *M* has constant speed *vj* when it is flying along a horizontal straight line. Air
enters the intake scoops *S* at rate *r1*. If the engine burns fuel at the rate *r2* and the gas (air and fuel) is
exhausted relative to the plane with speed *ve*, determine the resultant drag force exerted on the plane
by air resistance. Assume that air has a constant density ρ. *Hint*: Since mass both enters and exits the
plane, Eqs. 15-29 and 15-30 must be combined.

Units Used:

Mg 1000 kg=

kN 103 N=

Given:

*M *12 Mg= *r2 *0.4
kg
s

=

*vj *950
km
hr

= *ve *450
m
s

=

*r1 *50
m3

s = ρ 1.22

kg

m3 =

Solution:

Σ*Fs m
t
v*d

d *t
me
*

d d

*vDE*( )−
*t
mi
*

d d

*vDi*( )+=

*t
v*d

d
0= *vDE Ve*= *vDi vj*=

*t
mi
*

d d

*r1*ρ=

392

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Engineering Mechanics - Dynamics Chapter 15

*A r1*ρ=
*t
me
*

d d

*r2 A*+= *B r2 A*+=

Forces *T* and *FD* are incorporated as the last two terms in the equation,

*FD ve B vj A*−= *FD *11.5 kN=

**Problem 15-131
**

The jet is traveling at speed *v*, angle θ with the horizontal. If the fuel is being spent at rate *r1* and the
engine takes in air at *r2* whereas the exhaust gas (air and fuel) has relative speed *ve*, determine the

acceleration of the plane at this instant. The drag resistance of the air is *FD* = *kv2* The jet has weight *W*.
*Hint:* See Prob. 15-130.

Given:

*v *500
mi
hr

= *ve *32800
ft
s

=

θ 30 deg= *k1 *0.7 lb
s2

ft2 =

*r1 *3
lb
s

= *W *15000 lb=

*r2 *400
lb
s

= *g *32.2
ft

s2 =

Solution:

*t
mi
*

d d

*r2
g1
*

= *A1 r2*=
*t
me
*

d d

*r1 r2*+

*g1
*= *B r1 r2*+= *v1 v*=

Σ*Fs m
t
v*d

d
*vDe
*

*t
me
*

d d

− *vDi
t
mi
*

d d

+=

*W*− sin θ( ) *k1 v1*2− *W a ve B*− *v1 A1*+=

*a
W*− sin θ( ) *k1 v1*2− *ve
*

*B
g
*

+ *v1
A1
g
*

− ⎛ ⎜ ⎝

⎞ ⎟ ⎠

*g
*

*W
*= *a *37.5

ft

s2 =

393

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Engineering Mechanics - Dynamics Chapter 15

***Problem 15-132
**

The rope has a mass *m'* per unit length. If the end length *y* = *h* is draped off the edge of the
table, and released, determine the velocity of its end *A *for any position *y*, as the rope uncoils
and begins to fall.

Solution:

*Fs m
t
v*d

d
*vDi
*

*t
mi
*

d d

+= At a time *t*, *m m' y*= and
*t
mi
*

d d

m'
*t
y*d

d
= *m' v*= .

Here, *vDi v*= ,
*t
v*d

d
*g*= .

*m' g y m' y
t
v*d

d
*v m' v*( )+=

*g y y
t
v*d

d
*v*2+= Since *v
*

*t
y*d

d
= , then d*t
*

d*y
v
*

=

*g y v y
y
v*d

d
*v*2+=

Multiply both sides by 2*y*d*y
*

2*g y*2 d*y *2*v y*2 d*v *2*y v*2 d*y*+=

*y*2*g y*2
⌠⎮
⎮⌡

d *v*2 *y*21
⌠
⎮
⌡

d= 2 3

*g y*3 *C*+ *v*2 *y*2=

*C
*2−

3
*g h*3=*v *0= at *y h*=

2 3

*g h*3 *C*+ 0=

2 3

*g y*3
2
3

*g h*3− *v*2 *y*2= *v
*2
3

*g
y*3 *h*3−

*y*2

⎛⎜ ⎜⎝

⎞⎟ ⎟⎠

=

**Problem 15-133
**

The car has a mass *m0* and is used to tow the smooth chain having a total length *l* and a mass per
unit of length *m'.* If the chain is originally piled up, determine the tractive force **F** that must be
supplied by the rear wheels of the car, necessary to maintain a constant speed *v* while the chain is
being drawn out.

394

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Engineering Mechanics - Dynamics Chapter 15

Solution:

+⎯⎯→ Σ*Fs* = *m
t
v*d

d
*vDi
*

*t
mi
*

d d

+

At a time *t*, *m m0 c t*+=

Where, *c
t
mi
*

d d

= *m'
t
x*d

d
= *m'v*=

Here, *vDi v*=
*t
v*d

d 0=

*F m0 m'v t*−( ) 0( ) *v m' v*( )+= *m'v*2= *F m'v*2=

**Problem 15-134
**

Determine the magnitude of force **F** as a function of time, which must be applied to the end of the
cord at *A* to raise the hook *H* with a constant speed *v*. Initially the chain is at rest on the ground.
Neglect the mass of the cord and the hook. The chain has a mass density ρ.

Given:

*v *0.4
m
s

= ρ 2 kg m

= *g *9.81
m

s2 =

Solution:

*t
v*d

d
0= *y v t*=

*mi m y*= *m v t*=

*t
mi
*

d d

*m v*=

+↑ Σ*Fs m tv
*d
d

*vDi
t
mi
*

d d

⎛ ⎜ ⎝

⎞ ⎟ ⎠

+=

*F m g v t*− 0 *v m v*+= *F m g v t v m v*+=

*F *ρ*g v t v*2+=

*f1 *ρ*g v*= *f1 *7.85
N
s

= *f2 *ρ*v
*2= *f2 *0.320 N=

*F f1 t f2*+=

395

docsity.com

Engineering Mechanics - Dynamics Chapter 16

**Problem 16-1
**

A wheel has an initial clockwise angular velocity ω and a constant angular acceleration α. Determine
the number of revolutions it must undergo to acquire a clockwise angular velocity ω*f*. What time is
required?

Units Used: rev 2π rad=

Given: ω 10 rad s

= α 3 rad

s2
= ω*f *15

rad s

=

Solution: ω*f
*2 ω2 2α θ+= θ

ω*f
*2 ω2−

2α = θ 3.32 rev=

ω*f *ω α *t*+= *t
*ω*f *ω−

α
= *t *1.67 s=

**Problem 16-2
**

A flywheel has its angular speed increased uniformly from ω*1* to ω*2* in time *t*. If the diameter of the
wheel is *D*, determine the magnitudes of the normal and tangential components of acceleration of a
point on the rim of the wheel at time *t*, and the total distance the point travels during the time period.

Given: ω*1 *15
rad
s

= ω*2 *60
rad
s

= *t *80 s= *D *2 ft=

Solution: *r
D
*2

=

ω*2 *ω*1 *α *t*+= α
ω*2 *ω*1*−

*t
*= α 0.56

rad

s2 =

*at *α*r*= *at *0.563
ft

s2 =

*an *ω*2
*2*r*= *an *3600

ft

s2 =

θ
ω*2
*

2 ω*1
*2

−

2α = θ 3000 rad=

*d *θ*r*= *d *3000 ft=

**Problem 16-3
**

The angular velocity of the disk is defined by ω = *at*2 + *b*. Determine the magnitudes of the velocity
and acceleration of point *A* on the disk when *t *= *t1*.

396

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Engineering Mechanics - Dynamics Chapter 16

Given:

*a *5
rad

s3 =

*b *2
rad
s

=

*r *0.8 m=

*t1 *0.5 s=

Solution: *t t1*=

ω *a t*2 *b*+= ω 3.25
rad
s

=

α 2*a t*= α 5.00
rad

s2 =

*v *ω*r*= *v *2.60
m
s

=

*a *α*r*( )2 ω2*r*( )2+= *a *9.35 m
s2

=

***Problem 16-4
**

The figure shows the internal gearing of a “spinner” used for drilling wells. With constant
angular acceleration, the motor *M *rotates the shaft *S *to angular velocity ω*M* in time *t* starting
from rest. Determine the angular acceleration of the drill-pipe connection *D *and the number of
revolutions it makes during the start up at *t*.

Units Used: rev 2π=

Given:

ω*M *100
rev
min

= *rM *60 mm=

*rD *150 mm= *t *2 s=

Solution:

ω*M *α*M t*=

α*M
*ω*M
*

*t
*= α*M *5.24

rad

s2 =

α*M rM *α*DrD*=

397

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Engineering Mechanics - Dynamics Chapter 16

α*D *α*M
rM
rD
*

⎛ ⎜ ⎝

⎞ ⎟ ⎠

= α*D *2.09
rad

s2 =

θ 1 2

α*Dt
*2= θ 0.67 rev=

**Problem 16-5
**

If gear *A *starts from rest and has a constant angular acceleration α*A*, determine the time needed for
gear *B *to attain an angular velocity ω*B*.

Given:

α*A *2
rad

s2
= *rB *0.5 ft=

*rA *0.2 ft=ω*B *50
rad
s

=

Solution:

The point in contact with both gears has a speed of

*vp *ω*B rB*= *vp *25.00
ft
s

=

Thus,

ω*A
vp
rA
*

= ω*A *125.00
rad
s

=

So that ω α*c t*= *t
*ω*A
*α*A
*

= *t *62.50 s=

**Problem 16-6
**

If the armature *A *of the electric motor in the drill
has a constant angular acceleration α*A*, determine
its angular velocity and angular displacement at
time *t*. The motor starts from rest.

Given:

α*A *20
rad

s2
= *t *3 s=

Solution:

ω α*c t*= ω α*A t*= ω 60.00
rad
s

=

398

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