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Friction, circular dynamics, and Work-Kinetic Energy.
If the block were to remain at rest relative to the truck, the fric- tion force would need to cause an acceleration of 2.20 m/s^2 , ;How- ever, the maximum acceleration possible to to static friction is (0.19)(9.80)=1.86 m/s^2 , so the block will accelerate relative to the truck. Acceleration a with respect to ground is (from Fx = +μsmg = mabox): a = μk g = (0.15)(9.80) = 1. 47 m/s^2. (1)
The acceleration a�^ with respect to truck is:
a = abox − atruck = 1. 47 − 2. 20 = − 0. 73 (2)
The box will move 1.80 m relative to truck with a�^ acceleration.
2∆x 2(− 1 .80) t = = = 2.22s (3) abox − atruck − 0. 73
In this time, the truck moves the distance D:
1 1 D = atruck t^2 = (2.20)(2.22)^2 = 5.43m. (4) 2 2
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Please refer to figure (1).
Let’s denote the common magnitude of acceleration as a. If block B is to remain at rest on A , the sum of forces acting on B should � (^) −→ (^) −→ produce the acceleration a. Setting up the FB = m a :
B: �
FBy : NAB − mB g = 0 ⇒ NAB = mB g (5)
FBx : fs = mB a ≤ μsNAB (6)
NOTE: Here is a case in which friction causes positive acceleration.
A:
FAy : NA − NAB − mAg = 0 ⇒ NA = (mA + mB )g (7)
FAx : T −μk NA −fs = mAa ⇒ T = μk g(mA +mB )+(mA +mB )a (8)
∴ T = (mA + mB )(a + μk g) (9)
NOTE: The friction force between block A and B is LESS THAN or EQUAL to μsNAB. We don’t know it yet and must assign it an unknown variable fs.
C: �
FCy = T − mC g = mC aCy = −mC a (10)
Replacing T from above into equation (10) and solving for a:
mC − μk (mA + mB ) a = g (11) mA + mB + mC
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a)Adding these two equations and solving for a:
(mS + mL)g[cos θ(μSk + μLk ) − sin θ] a = (20) mS + mL
a = g[cos θ(μSk + μLk ) − sin θ] (21)
substitution gives a = − 2. 21 m/s^2. (22)
b)Substituting the above acceleration in either of (17) or (19) gives T=2.27 N.
c)The upper block will slide down with more acceleration until they collide! You could solve this section with the above formalism and you will get negative T which means that the string must be a stiff rod ,which supports compressive forces, in order to prevent the collision.
Please refer to figure (2)
Make sure that you have understood Example 5.22 p.184.
a)
Fy = 0 ⇒ cos β TU − cos β TL − mg = 0 (23)
mg TL = TU −. (24) cos β
The net force inward is
Frad = sin β TU + sin β TL = sin β (TU + TL) = mar. (25)
b) Solving v^2 4 π^2 R Frad = marad = m = m (26) R τ 2
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(where τ is the period of oscillations) for period τ :
mR τ = 2π (27) Frad
Using hypotenuse theorem the radius of oscillation is
R = (1.25)^2 − 1 = 0.75m (28)
and cos β = 4 ; sin β = 3 5 5. If you plug in the above numbers into equations we derived you’ll get:
TL = 31. 0 N. (29)
τ = 1. 334 s (30)
so the system makes 1/0.02223 ≈ 45 rev/min.
c) When the lower string becomes slack, the system is the same as the conical pendulum considered in Example 5.22 with cos β = 0.8, the period is
(1.25)(0.80) τslack = 2π = 2.007s. (31)
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which equivalent to 30 rev/min.
d)For oscillation with less revolutions the Tension in lower string is still Zero and the problem is again the conical pendulum problem; the block will drop to a smaller angle.
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(x 1 − x 0 ) = vavetstop (42)
∴ tstop =
2(x 1 − x 0 ) (43) v 0
d)Using (32) Here Wtot = Wperson + Wf using (35) and noting that s = x 2 − x 1 we have Wf = −Ff (x 2 − x 1 ). Using the result of part b we get
Wf = −
mv 02
x 2 − x 1 (44) 2 x 1
Collecting all the information we got we have
1 2 1 Wtot = Wperson + Wf = Wperson − mv 0
x 2 − x 1 = + mv 12 (45) 2 x 1 2
1 2 x 2 − x (^1 1 ) ∴ Wperson = mv 0 + mv 1 (46) 2 x 1 2
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6) U-Control Model Airplane
Please refer to figure (3)
We choose the x axis along the acceleration so we have 0 accelera- tion along the y axis! If you choose the x axis along the wire you have have T along and F ⊥. However you should project −→a and elimination of two equations becomes difficult.
� (^) −→ (^) −→ F = m a (47)
Fy = may = 0 ⇒ F cos θ − T sin θ − mg = 0 (48)
� (^) mv 2 mv 2 mv^2 Fx = max = = ⇒ F sin θ + T cos θ = (49) R L cos θ L cos θ
Eliminating F from (9) and (10) we get:
mv^2 T − + mg sin θ = 0 (50) L
The physical condition T≥ 0 will set a restriction on θ for a fixed v.
v^2 sin θ ≤ (51) Lg
So (^2) v θcrit = arcsin (52) Lg
Clearly for v^2 = Lg even 90 o will be fine. So there exists a finite vsaf e:
vsaf e = Lg (53)
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Figure 2: 5.
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Figure 3: U-Control Model Airplane
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