Q#38
The monkey B shown in figure (5-E20) is holding on to the tail of the monkey A which is climbing up a rope. The masses of the monkeys A and B are 5 kg and 2 kg respectively. If A can tolerate a tension of 30 N in the tail, what force should it apply on the rope in order to carry the monkey B with it? Take g = 10 m/s².
The force applied by the monkey A on the rope is equal to the tension in the rope. If the monkey A just holds the rope or goes up with a uniform velocity carrying monkey B the tension in the rope is equal to the total weight of both monkeys.
T = 5g + 2g = 5 x 10 + 2 x 10 = 50 + 20 = 70 N
If it goes up with some acceleration the force applied by A on the rope (equal to tension T in the rope) will increase proportionately. Let us calculate maximum T. Maximum force allowed in the tail of A = 30 N. Let the acceleration of A for this tension be 'a'. Same will be the acceleration of B as he is holding the tail of A. Mass of B = 2 kg. Weight of B = 2g = 2 x 10 = 20 N. Net upward force on B = 30 – 20 = 10 N.
So, acceleration of B = a = 10 N/2 kg = 5 m/s² = acceleration of A.
Consider the jointly A and B. Total mass m = 5 + 2 = 7 kg. Net upward force = T – 7g. It gives,
T – 7g = 7a
T = 7 x 10 + 7 x 5 = 70 + 35 = 105 N.
So to carry the monkey B with it while going up, the monkey can apply a force between 70 N to 105 N on the rope.
Q#39
Figure (5-E21) shows a man of mass 60 kg standing on a light weighing machine kept in a box of mass 30 kg. The box is hanging from a pulley fixed to the ceiling through a light rope, the other end of which is held by the man himself. If the man manages to keep the box at rest, what is the weight shown by the machine ? What force should he exert on the rope to get his correct weight on the machine?
Answer:
Let the weight of the man shown on the weighing machine be F Newtons and the tension in the rope be T Newtons throughout. Consider the box. Total downward force on the rope by the box is weight of the box plus weight of the man shown in the weighing machine (F). it gives,
T= 30g + F
Consider the man. His actual weight 60g Newtons is reduced by the apparent F because F is also the force applied by the box on the man. Som in this case tension in the rope T is balanced by 60g – F, ie.
T = 60g – F
Equating these two, we get,
30g + F = 60g – F
2F = 30g
F = 15g N. In terms of kg the weighing machine will show the weight of the man = 15 kg.
For the weighing machine to show actual weight 60 kg of the man, let the force applied by the man on the rope be P Newtons. So new tension in the rope will be P. Due to this force let the acceleration gained by both the man and the box be 'a'. Consider the case of the man. Upward force on him is by the box the weight shown in the weighing machine equal to 60g N and the tension in the rope P while downward force is his weight 60g N. Net upward Force is
= P – 60g – 60g = P, it gives,
P = ma = 60 a
a = P/60
Consider the case of the box now, Net upward force = P – 30g – 60g
P – 30g – 60g = 30a
P = 90g + 30(P/60) (Replace a)
60P – 30P = 90 g x 60
P = 5400g/30 = 1800 N.
Q#40
A block A can slide on a friction-less incline of angle θ and length l, kept inside an elevator going up with uniform velocity v (figure 5-E22). Find the time taken by the block to slide down the length of the incline if it is released from the top of the incline.
Answer:
Let the mass of the block be M kg. Since the elevator is going up with uniform velocity no additional forces will be added to the bock and it will behave as if the elevator is at rest. The acceleration of the block will be the component of the acceleration due to gravity g along the incline = gsinθ. Let the time taken by the block to slide down the length 'l' of incline be 't' s. From the relation s = ut + ½at². for u = 0 we get,
l = 0 + ½ gsinθ.t²
gt²sinθ = 2l
t² = 2l/gsinθ
t= √(2l/gsinθ)
Q#41
A car is speeding up on a horizontal road with an acceleration 'a'. Consider the following situations in the car. (i) A ball is suspended from the ceiling through a string and is maintaining a constant angle with the vertical. Find this angle. (ii) A block is kept on a smooth incline and does not slip on the incline. Find the angle of the incline with the horizontal.
Answer:
Case (i)
Let the constant angle made by the string with the vertical be θ, mass of the ball be 'm' and tension in the string be T. We want to apply Newton's Laws of Motion with respect to the accelerating car which is a non-inertial frame. So additional pseudo force equal to 'ma' will have to be considered acting on the ball opposite to the direction of acceleration. Now forces on the ball are T, 'ma' and weight of the ball 'mg' as shown in the diagram below:
Tcosθ = mg (1)
Equating the forces in horizontal direction we get,
Tsinθ = ma (2)
Dividing (2) by (1),
tanθ = a/g
θ = tan-1(a/g).
Case (ii)
Let the angle of the incline with the horizontal be θ, mass of the block "m" and the normal force on the block by the incline N. We want to apply Newton's Laws of Motion with respect to the accelerating car which is a non-inertial frame. So additional pseudo force equal to 'ma' will have to be considered acting on the block opposite to the direction of acceleration. Now forces on the block are weight mg downward, pseudo force ma opposite to 'a' and the normal force N perpendicular to incline and upward as shown in the diagram below:
Since the block can move along the incline let us calculate the component of all the forces along the incline. Taking direction down the plane as positive, total force along the incline
= mgsinθ – macosθ (Component of N is zero as Tcos90°= 0)
For the block not to slip on the incline this force should be zero. Thus,
mgsinθ – macosθ = 0
gsinθ = acosθ,
tanθ = a/g,
θ = tan-1(a/g).
Q#42
A block is kept on the floor of an elevator at rest. The elevator starts descending with an acceleration of 12 m/s². Find the displacement of the block during the first 0.2 s after the start. Take g= 10 m/s².
Answer:
Since the elevator descends with an acceleration of 12 m/s² which is greater than the acceleration due to gravity 'g', the block kept on the floor of the elevator will separate and fall under the acceleration due to gravity. Let us calculate the displacement of the block during the first 0.2 s. Here,
u = 0 m/s, t = 0.2 s, displacement 'h' = ?
from the relation,
h = ut + ½gt²,
h = 0 + ½ x 10 x (0.2)²
h = 5 x 0.04 = 0.2 m = 20 cm.
= mgsinθ – macosθ (Component of N is zero as Tcos90°= 0)
For the block not to slip on the incline this force should be zero. Thus,
mgsinθ – macosθ = 0
gsinθ = acosθ,
tanθ = a/g,
θ = tan-1(a/g).
Q#42
A block is kept on the floor of an elevator at rest. The elevator starts descending with an acceleration of 12 m/s². Find the displacement of the block during the first 0.2 s after the start. Take g= 10 m/s².
Answer:
Since the elevator descends with an acceleration of 12 m/s² which is greater than the acceleration due to gravity 'g', the block kept on the floor of the elevator will separate and fall under the acceleration due to gravity. Let us calculate the displacement of the block during the first 0.2 s. Here,
u = 0 m/s, t = 0.2 s, displacement 'h' = ?
from the relation,
h = ut + ½gt²,
h = 0 + ½ x 10 x (0.2)²
h = 5 x 0.04 = 0.2 m = 20 cm.
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