Download 1. SOLUTION: Because `B` is heavier and it sits on a steeper slope

1. SOLUTION:
Because ‘B’ is heavier and it sits on a steeper slope, we set up the force diagrams assuming that the crates
move rightward (i.e. ‘A’ moves upslope and ‘B’ downslope). The corresponding free-body diagrams
showing forces on each crate are shown. In these diagrams, we have denoted
TB
• Wa = MA g and Wb = Mb g: the weights of ‘A’ and ‘B’
Ff B
NA
• NA and NB : the normal forces on ‘A’ and ‘B’
TA
• Ff A = µk NA and Ff B = µk NB : the frictional forces
NB
A
B
• TA and TB : the forces of tension in the rope
60◦
30◦
Ff A
WA
Forces on A:
In the direction normal to the slope at ‘A’, the force balance gives
WB
NA −WA cos 30◦ = 0
In the along-slope direction, the forces result in an upslope acceleration aA such that
MA aA = TA − Ff A −Wa sin 30◦ .
Combining the results and solving for TA gives
TA = MA aA + µk (MA g cos 30◦ ) + MA g sin 30◦ .
(Explicitly, TA = 20aA + 0.3(20 × 9.81 cos 30◦ ) + 20 × 9.81 sin 30◦ = 20aA + 149)
(3 marks)
Forces on B:
In the direction normal to the slope at ‘B’, the force balance gives
NB −WB cos 60◦ = 0.
In the along-slope direction, the forces result in an downslope acceleration aB such that
MB aB = −TB − Ff B +WB sin 60◦ .
Combining the results and solving for TB gives
TB = −MB aB − µk (MB g cos 60◦ ) + MB g sin 60◦ .
(Explicitly TB = −30aB − 0.3(30 × 9.81 cos 60◦ ) + 30 × 9.81 sin 60◦ = −30aB + 210.)
(3 marks)
Connecting the results:
The rope remains tight, so TA = TB and aA = aB . Letting a ≡ aA = aB and equating the results for TA and
TB above gives
MA a + µk (MA g cos 30◦ ) + MA g sin 30◦ = −MB a − µk (MB g cos 60◦ ) + MB g sin 60◦ .
So the acceleration is
a=g
(2 marks)
(MB sin 60◦ − MA sin 30◦ ) − µk (MB cos 60◦ + MA cos 30◦ ) 210 − 149
≃
≃ 1.22 m/s2
MA + MB
20 + 30
2. SOLUTION:
Because the balloon ‘hovers’ before release the buoyancy force of the balloon, Fb , is in balance with the
weight of balloon and passenger and sandbag, Wt : Fb = Wt = Mt g (≃ 1000 × 9.81 ≃ 981 N)
(2 marks)
Motion of the sandbag
After release, the sandbag falls to the ground under the force of gravity starting from rest. Generally, its
position with respect to the balloon as a function of time is
zs = −gt 2 /2.
So it hits the ground when zs = −H = −100 m. This occurs at time
p
p
T = 2H/g (≃ 2 × 100/9.81 = 4.52 s).
(2 marks)
Motion of the balloon and passenger
Meanwhile, after release the hot air balloon still experiences a buoyancy force Fb ; but this acts on the
reduced weight Wb ≡ (Mt − Ms )g
(≃ (1000 − 20) × 9.81 ≃ 9614 N).
The unbalanced forces lead to a vertical acceleration a such that
(Mt − Ms )a = Fb −Wb = Mt g − (Mt − Ms )g = Ms g.
Explicitly,
(2 marks)
a=
Ms
Mt −Ms g
(≃
20
2
980 9.81 ≃ 0.200 m/s ).
Generally, the position of the balloon is given by
zb = H + at 2 /2.
So when the sandbag hits the ground the balloon is at height
1 2
Mt
1
Ms
2H
Ms
Z = H + aT = H +
=H
g
= H 1+
. ≃ 102 m .
2
2 Mt − Ms
g
Mt − Ms
Mt − Ms
(2 marks)
3. SOLUTION:
The yo-yo can only accelerate due to the combined force of the tension in the string and the weight of
the yo-yo. Because the string is vertical at the time under consideration, the acceleration of the yo-yo
must be vertical.
(2 marks)
At the instant the finger moves, the yo-yo’s acceleration must equal the component of acceleration normal
to the finger’s motion: v2 /L. Hence the acceleration of the yo-yo is
a=
v2 202
≃
≃ 8.00 cm/s2 upward
L
50
(2 marks)
[Alternately, by geometry one could denote the vertical
√ position of the yo-yo with respect to the finger
displaced by a small horizontal distance x as z = − L2 − x2 . This guarantees the string length is L and
that the yo-yo initially moves vertically.
√
So the yo-yo’s vertical velocity is dz/dt = (x/ L2 − x2 )dx/dt.
√
√
Hence a = d 2 z/dt 2 = (1/ L2 − x2 )(dx/dt)2 + (x2 /(L2 − x2 )3/2 )(dx/dt)2 + (x/ L2 − x2 )d 2 x/dt 2 .
At the time under consideration, x = 0. So a = (1/L)(dx/dt)2 = v2 /L.]