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EXERCISES (continued)
13. The elevator shown in figure (5-E5) is descending with an acceleration of 2 m/s². The mass of the block A is 0.5 kg. What force is exerted by the block A on the block B?
The figure for problem number 13
Answer: If the elevator was not moving, the force applied by block A on block B would have been equal to its weight = mg = 0.5 kg x 9.8 m/s²= 4.9 N. But, in this case, the elevator is descending with an acceleration of a=2 m/s². So, in order to apply Newton's Laws of Motion with respect to the elevator we apply a pseudo force equal to ma on block A in the direction opposite to the acceleration ie upwards. The net force on the block is
= mg-ma= 4.9 - 0.5x2 = 4.9-1 = 3.9 N ≈ 4.0 N.
14. A pendulum bob of mass 50 g is suspended from the ceiling of an elevator. Find the tension in the string if the elevator (a) goes up with acceleration 1.2 m/s², (b) goes up with deceleration 1.2 m/s², (c) goes up with uniform velocity, (d) goes down with acceleration 1.2 m/s², (e) goes down with deceleration 1.2 m/s² and (f) goes down with uniform velocity.
Answer: Mass of the pendulum bob = 50 g = 0.05 kg.
Weight of the bob = 0.05 kg x 9.8 m/s² = 0.49 N (Downward)
It is also the tension in the string when the elevator is at rest.
When the elevator moves with an acceleration a, we have to apply a pseudo force equal to ma on the bob opposite to the direction of a. So let us take the different situations taking the downward direction as +ve:-
(a) Acceleration a = -1.2 m/s² (Upwards),
Pseudo force = ma = 0.05 kg x 1.2 m/s² = 0.06 N (Downwards)
Weight = 0.49 N (Downwards)
Total force on the bob = 0.49 + 0.06 N = 0.55 N (Downwards)
Tension in the string = 0.55 N
(b) Deceleration upwards = 1.2 m/s²
Acceleration a = 1.2 m/s² (Downward),
Pseudo force = ma = -0.05 kg x 1.2 m/s² = -0.06 N (Upward)
Weight = 0.49 N (Downward)
Total force on the bob = 0.49 - 0.06 N = 0.43 N (Downwards)
Tension in the string = 0.43 N
(c) The elevator goes up with uniform velocity,
Acceleration a = 0 m/s²,
Pseudo force = ma = 0.05 kg x 0 m/s² = 0
Weight = 0.49 N (Downward)
Total force on the bob = 0.49 N (Downwards)
Tension in the string = 0.49 N
(d) Acceleration a = 1.2 m/s² (Downwards),
Pseudo force = ma = -0.05 kg x 1.2 m/s² = -0.06 N (Upwards)
Weight = 0.49 N (Downwards)
Total force on the bob = 0.49 - 0.06 N = 0.43 N (Downwards)
Tension in the string = 0.43 N
(e) Deceleration downwards = 1.2 m/s²
Acceleration a = -1.2 m/s² (Upwnward),
Pseudo force = ma = 0.05 kg x 1.2 m/s² = 0.06 N (Downward)
Weight = 0.49 N (Downward)
Total force on the bob = 0.49 + 0.06 N = 0.55 N (Downwards)
Tension in the string = 0.55 N
(f) Elevator goes down with uniform velocity,
Acceleration a = 0 m/s²,
Pseudo force = ma = 0.05 kg x 0 m/s² = 0 N
Weight = 0.49 N (Downward)
Total force on the bob = 0.49 N (Downwards)
Tension in the string = 0.49 N
15. A person is standing on a weighing machine placed on the floor of an elevator. The elevator starts going up with some acceleration, moves with uniform velocity for a while and finally decelerates to stop. The maximum and minimum weights recorded are 72 kg and 60 kg. Assuming that the magnitude of acceleration and deceleration are the same, find (a) the true weight of the person and (b) the magnitude of the acceleration. Take g=9.9 m/s².
Answer: Let the true weight of the person be 'm' kg and the acceleration of the elevator = 'a' m/s². When the elevator starts going up with acceleration, the pseudo force acts downwards which adds to the apparent weight of the person. In this case apparent weight =mg+ma =m(g+a).
But the recorded apparent weight = 72 kg force, which gives the weight force = 72g N. These two should be equal.
So m(g+a) = 72g → mg + ma =72g ................(A)
During uniform velocity of the elevator, the weight does not change. When the elevator decelerates its acceleration is downwards and the pseudo force is upwards. So the apparent weight =mg-ma which is recorded as 60 kg. In terms of weight-force, these two can be equated as
mg - ma = 60g ...............................................(B)
Adding (A) and (B) we get
2 mg = 132g → m = 132/2 = 66 kg.
So the true weight of the person is 66 kg.
Subtract (B) from (A)
2 ma = 72g - 60g = 12g
→ 2 x 66 a = 12 x 9.9
→ a = 6 x 9.9/66 = 0.9 m/s².
So the acceleration of the elevator = 0.9 m/s².
16. Find the reading of the spring balance shown in the figure (5-E6). The elevator is going up with an acceleration of g/10, the pulley and the string are light and the pulley is smooth.
The figure for problem number 16
Answer: Since the elevator goes up with acceleration g/10, so the pseudo forces due to this acceleration on both the blocks will be downwards and equal to mass times this acceleration.
Let the accelerations of the blocks be 'a' with respect to the elevator (Upwards for the left block and downwards for the right block). Let tension in the string be T. Considering the forces and acceleration of the left block, we have,
T-1.5g-1.5g/10 =1.5 a .............................(A)
Similarly for the right block.
3g+3g/10-T = 3a .....................................(B)
Adding the two equations we get,
1.5g+1.5g/10 = 4.5a
→a=g/3+g/30 =11g/30
Putting this value in (A),
T=1.5x11g/10 + 1.5x11g/30 =16.5x4g/30 = 66g/30 =11g/5 N
The pull on the spring balance = 2T = 2x11g/5 = 22g/5 =4.4g N
The scale on the spring balance reads in kg, so its reading will be
=4.4g/g =4.4 kg.
17. A block of 2 kg is suspended from the ceiling through a mass-less spring of spring constant k=100 N/m. What is the elongation of the spring? If another 1 kg is added to the block, what would be the further elongation?
Answer: Weight force of the block = 2g N =2 x 9.8 N = 19.6 N.
Elongation of the spring = 19.6/k = 19.6/100 m =0.196 m ≈ 0.2 m
When another 1 kg block is added, total weight = 3 kg
Now weight force = 3g = 3 x 9.8 N = 29.4 N
Total elongation = 29.4/100 =0.294 m ≈ 0.3 m
So further elongation = 0.3 - 0.2 m = 0.1 m
18. Suppose the ceiling in the previous problem is that of an elevator that is going up with an acceleration of 2.0 m/s². Find the elongations.
Answer: When the elevator goes up with an acceleration 'a', downwards pseudo force 'ma' acts on the block. So total force on the block = mg + ma = m(g+a) = 2 (9.8 + 2) = 2 x 11.8 = 23.6 N
Elongation of the spring = 23.6/k = 23.6/100 m = 0.236 m ≈ 0.24 m
When further 1 kg is added to the block, total mass = 3 kg,
Total force = 3(g + a) = 3 x 11.8 N = 35.4 N
Total elongation = 35.4/k = 35.4/100 m =0.354 m ≈ 0.36 m,
Hence further elongation = 0.36 - 0.24 = 0.12 m
19. The force of buoyancy exerted by the atmosphere on a balloon is B in the upward direction and remains constant. The force of air resistance on the balloon acts opposite to the direction of the velocity and is proportional to it. The balloon carries a mass M and is found to fall down near the earth's surface with a constant velocity v. How much mass should be removed from the balloon so that it may rise with a constant velocity v?
Answer: Let the force of air resistance for velocity v = F
When the balloon falls down with constant velocity v, the net force on it is = Mg - B - F (Taking the downwards direction as positive). Its value should be zero as there is no acceleration.
Mg - B - F = 0 ................................ (A)
→ F= Mg - B
When the balloon goes upwards with a constant velocity v (after removal of some mass, let remained mass be m), the net force on it is = mg - B + F
It should also be equal to zero, so
mg - B + F = 0 .................................(B)
Subtracting (B) from (A) we get
(M-m)g - 2F = 0
→ M-m = 2F/g, Putting the value of F in it, we get
→ M-m = 2 (Mg - B)/g = 2 (M-B/g)
It is the value of mass removed.
20. An empty plastic box of mass m is found to accelerate up at the rate of g/6 when placed deep inside water. How much sand should be put inside the box so that it may accelerate down at the rate of g/6.
Answer: Let the force of buoyancy on the box be B upwards. Weight of the box = mg, Net force on the box = B-mg (upwards) gives the box an acceleration = g/6. This gives the relation
B-mg = mg/6 ........................... (A)
Now let sand of mass m1 is added to the box to give it downward acceleration = g/6.
Net force on the box = (m+ m1)g-B. Now force-acceleration relation will be
(m+ m1)g-B = (m+ m1)g/6 ............(B)
Adding (A) and (B) we get,
(m+ m1)g - mg = mg/6 + (m+ m1)g/6
→ m1g= 2mg/6 + m1g/6
→ m1g - m1g/6 = mg/3
→ 5m1g/6 = mg/3
→ 5m1 = 2m → m1 = 2m/5
21. A force F=vxA is exerted on a particle in addition to the force of gravity, where v is the velocity of the particle and A is a constant vector in the horizontal direction. With what minimum speed a particle of mass m be projected so that it continues to move undeflected with a constant velocity?
Answer: The force vector F is a cross product of velocity vector v and constant vector A, so the direction of F will be perpendicular to both v and A. In order to move the particle undeflected, the net force on the particle should be zero. It will be only when the force of gravity mg is equal and opposite to F. Since gravity force always acts downwards, the direction of force F should be upwards. For it to happen the particle should be projected horizontally with an angle θ to the constant horizontal vector A.
Since the magnitude of F and gravity force are equal, we have
vA sinθ = mg
→ v = mg/Asinθ,
For the magnitude v to be minimum, sinθ should be maximum. The maximum value of sinθ is 1 for θ= 90°, so the minimum speed v of the particle will be equal to mg/A and projected horizontally at a right angle to the constant horizontal vector A.
22. In a simple Atwood machine, two unequal masses m1 and m2 are connected by a string going over a clamped light smooth pulley. In a typical arrangement (figure 5-E7) m1 = 300g and m2 =600g. The system is released from rest. (a) Find the distance traveled by the first block in the first two seconds. (b) Find the tension in the string. (c) Find the force exerted by the clamp on the pulley.
Figure for problem number 22
Answer: Let the tension in the string be T after the system is released from rest. Let the accelerations of the blocks be 'a'. For the first block, T-m1g =m1a
T = m1a + m1g
For the second block, m2g -T= m2 a,
→ m2g - m1a - m1g = m2 a (putting the value of T)
→ (m1 + m2 )a = (m2-m1)g
→ a = (m2-m1)g/(m1 + m2 )
(a) Let us calculate the distance 's' traveled after t=2 s of release. from the relation s=ut+½at², we have
s= 0+½ x (m2-m1)g/(m1 + m2 ).2² = 2(m2-m1)g/(m1 + m2 )
Putting the value m1 =0.3 kg and m2 = 0.6 kg we get,
s= 2 x0.3 x 9.8/0.9 m = 6.5 m
(b) Tension in the string T = m1a + m1g
= m1 (m2-m1)g/(m1 + m2 ) + m1g
= {m1 (m2-m1)+ m1 (m1 + m2 )} g/(m1 + m2 )
= 2 m1 m2 g/(m1 + m2 ) = 2 x 0.3 x 0.6 x 9.8/0.9 =3.9 N
(c) Force exerted by the clamp on the pulley
= 2T = 2 x 3.9 N =7.8 N.
23. Consider the Atwood machine of the previous problem. The larger mass is stopped for a moment 2.0 s after the system is set into motion. Find the time elapsed before the string is tight again.
Answer: Acceleration of the blocks, a = (m2-m1)g/(m1 + m2 )
= (0.6-0.3)x9.8/(0.6+0.3) = 0.3x9.8/0.9 =9.8/3 =3.26 m/s².
When the larger mass is stopped, the smaller mass moves up with a velocity v which can be calculated from the relation, v=u+at. Here u=0, t=2 s and a =3.26 m/s². We get,
v= 0+3.26x2 =6.52 m/s. From this instant, the string loses its tension and the smaller block moves up under the gravity force only with an acceleration -g. When the larger mass is released after the stop, it begins to gain velocity under the force of gravity only. The string gets tight again when both the blocks cover an equal distance in opposite directions. Let the time elapsed before the string is tight again be t'. During this time, the initial velocity of the smaller block = u' = 6.52 m/s (as calculated above). Distance covered in time t'
S =u't'-½gt'²
And the same distance traveled by the larger block
S =0+½gt'² =½gt'²
On equating,
½gt'² =u't' -½gt'²
→gt'² -u't' =0
→t'(gt' -u') =0
Either t' =0 or gt' -u' =0
At both of these times, the string is tight but we will not take t' =0 because it was the instant when the larger block was stopped.
Thus, gt' -u' =0
→t' =u'/g
→ t'=6.52/9.8 = 2/3 seconds.
24. Figure (5-E8) shows a uniform rod of length 30 cm having a mass of 3.0 kg. The strings shown in the figure are pulled by constant forces of 20 N and 32 N. Find the force exerted by the 20 cm part of the rod on the 10 cm part. All the surfaces are smooth and the strings and the pulleys are light.
The figure for problem number 24
Answer: The length of the uniform rod is 30 cm and the mass is 3.0 kg, hence mass of 10 cm part = (10/30)x 3.0 kg =1 kg.
mass of 20 cm part = (20/30)x 3.0 kg = 2 kg.
Net force on the whole block = 32 N- 20 N =12 N (towards right)
Let the acceleration of the block be 'a'. From the relationship
Force = mass x acceleration
12 N = 3.0 kg x a → a = 4 m/s²
Let the force exerted by the 20 cm block on the 10 cm block be F (towards right). The resultant force on the 10 cm block = F- 20 N. Again from Force = mass x acceleration, we get,
F-20 = 1 x 4 → F = 20 + 4 = 24 N
25. Consider the situation shown in the figure (5-E9). All the surfaces are frictionless and the string and the pulley are light. Find the magnitude of the acceleration of the two blocks.
The figure for problem number 25
Answer: It is clear that the triangle is a right-angled triangle and the vertex angle at the pulley is 90°. See the diagram below.
Diagram showing forces on the blocks
Let angle ACB = θ. The component of weight mg along the 3 m side =mg.cosθ = 4mg/5. The component of weight mg along the 4 m side =mg.sinθ = 3mg/5.Let tension in the string be T and the acceleration of the blocks along planes be a.
Consider the forces and acceleration along the plane on the left side block,
4mg/5 - T = ma, →T=4mg/5-ma ........... (A)
Consider the forces and acceleration along the plane on the right side block,
T-3mg/5 =ma,
4mg/5-ma-3mg/5=ma, {Putting the value of T from (A)}
→ 2ma = mg/5
→ a=g/10.
So the acceleration of the blocks is g/10.
26. A constant force F=m2g/2 is applied on the block of mass m1 as shown in figure (5-E10). The string and the pulley are light and the surface of the table is smooth. Find the acceleration of m1.
The figure for problem number 26
Answer: As is clear from the figure, horizontal forces on the block m1 are tension in the string T (towards the right) and F= m2g/2 (towards the left). Let the acceleration of block m1 be 'a' towards the right. It gives the equation,
T- m2g/2= m1a → T=m2g/2+m1a
Block m2 will have the same acceleration 'a' downwards and forces on it will be tension T upwards and weight m2g downwards. It gives,
m2g - T = m2a,
→m2g - m2g/2- m1a = m2a, (Putting the value of T)
→ (m1 + m2)a = m2g/2
→ a = m2g/2(m1 + m2) . (for block m1 its direction will be towards the right as assumed initially)
27. In figure (5-E11) m1 =5 kg and m2 = 2 kg and F = 1 N. Find the acceleration of either block. Describe the motion of m1 if the string breaks but F continues to act.
The figure for problem number 27
Answer: Let the acceleration of blocks be 'a' and tension in the string T. Resultant downward force on block m1 is
=m1g+F-T =5g+1 -T, Downward acceleration is a. so we get,
5g+1 -T = 5a, → T= 5g+1-5a
Consider the block m2, Resultant upward forces = T- 2g -1, upward acceleration assumed 'a', we get,
T- 2g -1=2a (Put the value of T)
5g+1-5a -2g -1=2a → 7a = 3g
→a =3g/7 =3 x9.8/7
=3x1.4
=4.2 m/s².
If the string breaks, the upward pull of the string due to tension becomes zero and the resultant downward force on the block m1 is
=5g+1 N, Mass = 5 kg, hence Acceleration = Force/Mass
= (5g+1)/5 m/s² =g+0.2 m/s² (downwards)
===<<<O>>>===
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CHAPTER-7 - Circular Motion
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HC Verma's Concepts of Physics, Chapter-6, Friction,
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Click here for → OBJECTIVE-I
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Click here for → Exercises (11-20)
Click here for → EXERCISES (21-31)
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Click here for → Exercises (11-20)
Click here for → EXERCISES (21-31)
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Click here for→ Newton's laws of motion - Objective - I
Click here for → Newton's Laws of Motion - Objective -IIClick here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for→ Newton's Laws of Motion-Exercises(Q.No.28 to 42)
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HC Verma's Concepts of Physics, Chapter-4, The Forces
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Click here for "The Forces" - OBJECTIVE-I
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