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Problem 14-
A woman having a mass M stands in an elevator which has a downward acceleration a starting from rest. Determine the work done by her weight and the work of the normal force which the floor exerts on her when the elevator descends a distance s. Explain why the work of these forces is different.
Units Used: (^) kJ = 10 3 J
Given: (^) M = 70 kg g 9.81 m s^2
= a 4 m s^2
= s =6 m
Solution:
M g − Np= M a Np = M g −M a Np =406.7 N
UW = M g s UW =4.12 kJ
UNP = − sNp UNP =−2.44 kJ The difference accounts for a change in kinetic energy.
Problem 14-
The crate of weight W has a velocity vA when it is at A. Determine its velocity after it slides down the plane to s = s'. The coefficient of kinetic friction between the crate and the plane is μk.
Given:
W = 20 lb a = 3
v^ b^ =^4 A 12
ft s
s' =6 ft μk =0.
Solution:
θ atan a b
= NC = W cos( θ^ ) F = μk NC
Guess v' 1 m s
Given
W
g
vA^2 + W sin(^ θ)s'− F s'
W
g
= v'^2 v' = Find ( v') v' 17. ft s
Problem 14-
The crate of mass M is subjected to a force having a constant direction and a magnitude F, where s is measured in meters. When s = s 1 , the crate is moving to the right with a speed v 1. Determine its speed when s = s 2. The coefficient of kinetic friction between the crate and the ground is μk.
Given:
M = 20 kg F =100 N s 1 =4 m θ =30 deg
v 1 8 m s = a = 1
s 2 = 25 m b =1 m −^1 μk =0.
Solution:
Equation of motion: Since the crate slides, the friction force developed between the crate and its contact surface is F (^) f = μkN
N + F sin( θ^ )− M g= 0 N =M g −F sin( θ^ )
Principle of work and Energy: The horizontal component of force F which acts in the direction of displacement does positive work, whereas the friction force F (^) f= μk ( M g −F sin( θ ))does negative work since it acts in the opposite direction
to that of displacement. The normal reaction N, the vertical component of force F and the weight of the crate do not displace hence do no work.
F cos(^ θ)− μk N=M a F cos(^ θ)− μk (M g^ −F sin(^ θ))=M a
a
F cos( θ^ )−μk (^ M g −F sin( θ^ )) M = a 2. m s^2
v dv ds = a v 2 2
v 12 2 = +a s( 2 −s 1 )
v 2
v 12 2 +a s( 2 −s 1 )
= v 13. m s
The deformation is
T 1 + U 12 = T (^2)
M v^2 − F (^) avg x= 0
x
M
v^2 F (^) avg
= x =38.2 mm
Problem 14-
The crate of mass M is subjected to forces F 1 and F 2 , as shown. If it is originally at rest, determine the distance it slides in order to attain a speed v. The coefficient of kinetic friction between the crate and the surface is μk.
Units Used:
kN = 10 3 N
Given:
M = 100 kg v 6 m s
F 1 =800 N μ k =0. F 2 =1.5 kN g 9. m s^2
θ 1 =30 deg θ 2 =20 deg
Solution:
NC − F 1 sin( θ 1 )− M g+ F 2 sin( θ 2 )= 0 NC = F 1 sin( θ 1 ) +M g −F 2 sin( θ 2 ) NC =867.97 N T 1 + U 12 =T (^2)
F 1 cos( θ 1 )s− μk Nc s+ F 2 cos( θ 2 )s
= M v^2
s M v^2 2 (F 1 cos( θ 1 )− μk NC+F 2 cos( θ 2 ))
= s =0.933 m
Problem 14-
Design considerations for the bumper B on the train car of mass M require use of a nonlinear spring having the load-deflection characteristics shown in the graph. Select the proper value of k so that the maximum deflection of the spring is limited to a distance d when the car, traveling at speed v, strikes the rigid stop. Neglect the mass of the car wheels.
Units Used:
Mg = 10 3 kg kN = 10 3 N MN = 10 3 kN Given:
M =5 Mg d =0.2 m v 4 m s
Solution:
1 2 M v^2 0
d k x^2 x
− d = 0
M v^2 k d^3 3 − = 0 k 3 M v^2 2 d^3
= k 15
MN
m 2
Determine the required height h of the roller coaster so that when it is essentially at rest at the crest of the hill it will reach a speed v when it comes to the bottom. Also, what should be the minimum radius of curvature ρ for the track at B so that the passengers do not experience a normal force greater than kmg? Neglect the size of the car and passengers.
Given:
v 100 km hr
k = 4
Solution:
T 1 + U 12 =T (^2)
m g h
= m v^2
h
v^2 g
h =39.3 m
k m g − m g m v^2 ρ
= ρ v^2 g k( − 1 ) = ρ =26.2 m
Guess x =10 mm
Given x+ δ
δ −k x x
d − M g r ( 1 −cos( θ^ ))
= M v^2 x = Find ( )x x =178.9 mm
Problem 14-
The force F , acting in a constant direction on the block of mass M, has a magnitude which varies with the position x of the block. Determine how far the block slides before its velocity becomes v 1. When x = 0, the block is moving to the right at speed v 0. The coefficient of kinetic friction between the block and surface is μk..
Given:
M = 20 kg c = 3
v 1 5 m s = d = 4
v 0 2 m s = k 50
N
m 2
μk = 0.3 g 9. m s^2
Solution:
NB − M g c c^2 +d^2
− k x^2 = 0 NB M g c c^2 +d^2
= + k x^2
Guess δ =2 m Given
1 2 M v 02 d c^2 +d^2
δ k x^2 x
0
δ x c c^2 +d^2
k x^2
− d
= M v 12
δ =Find( δ^ ) δ =3.413 m
The force F , acting in a constant direction on the block of mass M, has a magnitude which varies with position x of the block. Determine the speed of the block after it slides a distance d 1. When x = 0, the block is moving to the right at v0 .The coefficient of kinetic friction between the block and surface is μk.
Given:
M = 20 kg c = 3 d 1 = 3 m d = 4 v 0 2 m s = k 50
N
m 2
μk = 0.3 g 9. m s^2
Solution:
NB − M g c c^2 +d^2
− k x^2 = 0 NB M g c c^2 +d^2
= + k x^2
Guess v 1 2 m s
Given 1 2 M v 02 d c^2 +d^2
d (^1) k x^2 x
0
d (^1) x c c^2 +d^2
k x^2
− d
= M v 12
v 1 = Find (v 1 ) v 1 3. m s
Problem 14-
As indicated by the derivation, the principle of work and energy is valid for observers in any inertial reference frame. Show that this is so by considering the block of mass M which rests on the smooth surface and is subjected to horizontal force F. If observer A is in a fixed frame x, determine the final speed of the block if it has an initial speed of v 0 and travels a distance d, both directed to the right and measured from the fixed frame. Compare the result with that obtained by an observer B, attached to the x' axis and moving at a constant velocity of vB relative to A. Hint: The distance the block travels will first have to be computed for observer B before applying the principle of work and energy. Given:
M =10 kg F =6 N v 0 5 m s
d =10 m vB 2 m s
g 9. m s^2
Given 0 = 2 vA +vB
W (^) A d 2
sin( θ 1 ) −W B d sin( θ 2 ) μk W A cos( θ 1 )
d 2
+− μkW B cos( θ 2 )d
2 g
= (W^ A vA^2 +W B vB^2 )
vA vB
=Find (v A ,vB)
vB − 1. ft s = vA 0. ft s
Problem 14-
Block A has weight W (^) A and block B has weight W (^) B. Determine the speed of block A after it moves a distance d down the plane, starting from rest. Neglect friction and the mass of the cord and pulleys.
Given:
W (^) A = 60 lb e = 3 W (^) B = 10 lb f = 4 d = 5 ft g 32. ft s^2
Solution:
L = 2 sA +sB 0 = 2 ΔsA+ ΔsB 0 = 2 vA +vB
0 W A
e e^2 +f 2
W A
g
vA^2
W B
g
= + ( 2 vA)^2
vA 2 g d W (^) A + 4 W (^) B
W A
e e^2 +f 2
⎛⎜ − 2 W B
= vA 7. ft s
The block A of weight WA rests on a surface for which the coefficient of kinetic friction is μk. Determine the distance the cylinder B of weight WB must descend so that A has a speed vA starting from rest.
Given:
W (^) A =3 lb W (^) B =8 lb μk =0.
vA 5 ft s
Solution:
L =sA + 2 sB Guesses d =1 ft Given
W (^) B d− μk W (^) A 2 d
2 g W (^) A vA^2 W (^) B
vA 2
2
d = Find ( )d d =0.313 ft
Problem 14-
The block of weight W slides down the inclined plane for which the coefficient of kinetic friction is μk.^ If it is moving at speed v when it reaches point A, determine the maximum deformation of the spring needed to momentarily arrest the motion.
Given:
W = 100 lb a =3 m
v 10 ft^ b^ =4 m s
d =10 ft k 200 lb ft = μk =0.
Solution:
N b a^2 +b^2
= W N =80 lb
Initial Guess
d (^) max =5 m
Solution:
NB − M g k a +b x
− sin(^ θ)= 0 NB M g k sin(^ θ) a +b x
U
x (^1)
x (^2) x k cos( θ^ ) a +b x
d μk x (^1)
x (^2) M g x k sin( θ^ ) a +b x
= − d U =173.177 N m⋅
M v 12 + U
= M v 22 v 2 v 12
2 U
M
= + v 2 15. m s
The motion of a truck is arrested using a bed of loose stones AB and a set of crash barrels BC. If experiments show that the stones provide a rolling resistance F (^) t per wheel and the crash barrels provide a resistance as shown in the graph, determine the distance x the truck of weight W penetrates the barrels if the truck is coasting at speed v 0 when it approaches A. Neglect the size of the truck.
Given:
F (^) t = 160 lb d =50 ft
W = 4500 lb k 1000 lb ft 3
v 0 60 ft s = g 32. ft s^2
Solution:
1 2
W
g
v 02 − 4 F (^) t d k x^4 4
x
2 W v 02 k g
16 F (^) t d k
1 4 = x =5.444 ft
Problem 14-
The crash cushion for a highway barrier consists of a nest of barrels filled with an impact-absorbing material. The barrier stopping force is measured versus the vehicle penetration into the barrier. Determine the distance a car having weight W will penetrate the barrier if it is originally traveling at speed v 0 when it strikes the first barrel.
Units Used:
kip = 103 lb
Given:
W =4000 lb
v 0 55 ft s
g 32. ft s^2
Solution:
1 2
W
g
v 02 − Area= 0
Area
W
g
= v 02 Area = 187.888 kip ft⋅ We must produce this much work with the barrels. Assume that 5 ft < x < 15 ft
Area =( 2 ft) 9 kip( ) + ( 3 ft) 18 kip( )+( x −5 ft) 27 kip( )
x Area −72 kip ft⋅ 27 kip = +5 ft x = 9.292 ft Check that the assumption is corrrect!
Problem 14-
The collar has a mass M and is supported on the rod having a coefficient of kinetic friction μk. The attached spring has an unstretched length l and a stiffness k. Determine the speed of the collar after the applied force F causes it to be displaced a distance s = s 1 from point A. When s = 0 the collar is held at rest.
Given:
M =30 kg μk =0. a =0.5 m θ =45 deg F = 200 N s 1 =1.5 m l =0.2 m g 9. m s^2
k 50
N
m
sB = 0.15 m k 100
N
m
Solution:
Δl = l^2 + b^2 − (l −sB)^2 +b^2
Guess F =1 N Given
F Δl − M g sB
− k sB^2
= M vB^2
F = Find ( F) F =43.9 N
Problem 14-
The collar has a mass M and is moving at speed v 1 when x = 0 and a force of F is applied to it. The direction θ of this force varies such that θ = ax, where θ is clockwise, measured in degrees. Determine the speed of the collar when x = x 1. The coefficient of kinetic friction between the collar and the rod is^ μk.
Given:
M = 5 kg v 1 8 m s
F =60 N
g 9. m s^2
μk =0.
x 1 = 3 m a 10 deg m
Solution:
N =F sin(^ θ)^ +M g
Guess v 5 m s
Given (^1) 2 M v 12 0
x (^1) F cos ( a x) x
x (^1) F sin ( a x) +M g x
− d
= M v^2
v = Find ( )v v 10. m s
Problem 14-
Cylinder A has weight W (^) A and block B has weight W (^) B. Determine the distance A must descend from rest before it obtains speed vA. Also, what is the tension in the cord supporting block A? Neglect the mass of the cord and pulleys. Given:
W (^) A = 60 lb vA 8 ft s
W (^) B = 10 lb g 32. ft s^2
Solution:
L = 2 sA +sB 0 = 2 vA +vB System
0 + W (^) A d− W (^) B 2 d
W A
g
vA^2
W B
g
= + ( 2 vA)^2
d
W A + 4 W B
2 g
vA^2 W (^) A − 2 W (^) B = d =2.484 ft
Block A alone
0 + W (^) A d− T d
W A
g
= vA^2 T W (^) A
W (^) A vA^2 2 g d = − T =36 lb
Problem 14-
The conveyor belt delivers crate each of mass M to the ramp at A such that the crate’s velocity is vA, directed down along the ramp. If the coefficient of kinetic friction between each crate and the ramp is μk, determine the speed at which each crate slides off the ramp at B. Assume that no tipping occurs.
Given:
M =12 kg vA 2. m s
μk =0.
g 9. m s^2
θ =30 deg a =3 m
vB F (^) N v'
= Find (v (^) B , F (^) N,v') vB 42. ft s = F (^) N = 50.6 lb v' 26. ft s^2
Problem 14-
When the block A of weight W 1 is released from rest it lifts the two weights B and C each of weight W 2. Determine the maximum distance A will fall before its motion is momentarily stopped. Neglect the weight of the cord and the size of the pulleys.
Given:
W 1 =12 lb W 2 =15 lb a =4 ft
Solution:
Guess y =10 ft
Given W 1 y − 2 W 2 (^ a^2 + y^2 −a)= 0 y = Find ( )y y =3.81 ft
Problem 14-
The catapulting mechanism is used to propel slider A of mass M to the right along the smooth track. The propelling action is obtained by drawing the pulley attached to rod BC rapidly to the left by means of a piston P. If the piston applies constant force F^ to rod BC such that it moves it a distance d, determine the speed attained by the slider if it was originally at rest. Neglect the mass of the pulleys, cable, piston, and rod BC.
Units Used:
kN = 10 3 N
Given:
M = 10 kg F = 20 kN d =0.2 m
Solution:
0 + F d
= M v^2 v 2 F d M = v 28. m s
Problem 14-
The collar has mass M and slides along the smooth rod. Two springs are attached to it and the ends of the rod as shown. If each spring has an uncompressed length L and the collar has speed v 0 when s = 0, determine the maximum compression of each spring due to the back-and-forth (oscillating) motion of the collar.
Given:
M = 20 kg a =0.25 m
L = 1 m kA 50
N
m
v 0 2 m s = kB 100
N
m
Solution:
1 2 M v 02
− (k (^) A +kB) d^2 = 0 d
M
kA + kB = v 0 d =0.73 m
The cyclist travels to point A, pedaling until he reaches speed vA. He then coasts freely up the curved surface. Determine the normal force he exerts on the surface when he reaches point B. The total mass of the bike and man is M. Neglect friction, the mass of the wheels, and the size of the bicycle.
Units Used:
kN = 10 3 N
Given:
vA 8 m s
M =75 kg a =4 m
Solution:
When y = x 2 y = a y a 4 = y =1 m
1 2 M vA^2 − M g y
= M vB^2 vB = vA^2 − 2 g y vB 6. m s
Now find the radius of curvature (^) x + y= a^1 2 x
dx
2 y
