EXERCISE 1. A block of mass 2 kg placed on a long frictionless horizontal table is pulled horizontally by a constant force F. It is found to move 10 m in the first two seconds. Find the magnitude of F. Sol: Given: mass of block, m = 2 kg; distance travelled, s = 10 m; time taken, t = 2 s; initial velocity, u = 0; Force, F =? Acceleration, a = F/m = F/2 We have, s = ut + ½ * a * t2 Or, 10 = 0 * 2 + ½ * (F/2) * 22
Or, F = 10 N. 2. A car moving at 40 km/h is to be stopped by applying brakes in the next 4.0 m. If the car weighs 2000 kg, what average force must be applied on it? Sol: Given: initial velocity, u = 40 km/h = 11.1 m/s; final velocity, v = 0; distance travelled, s = 4.0 m; mass of car, m = 2000 kg. We have, v2 – u2 = 2as Or, 02 – 11.12 = 2 * a * 4 Or, acceleration, a = – 15.4 m/s2 Average force must be applied = F = ma Or, F = 2000 * 15.4 = 30800 N Or, F = 3.08 *104 N ≈ 3.1 *104 N. 3. In a TV picture tube electrons are ejected from the cathode with negligible speed and reach a velocity of 5 x 106 m/s in travelling one centimeter. Assuming straight line motion, find the constant force exerted on the electron. The mass of the electron is 9.1 x 10-31 kg. Sol:
Given: mass of electron, m = 9.1 * 10-31 kg; initial velocity, u = 0; final velocity, v = 5 * 106 m/s; distance travelled, s = 1 cm = 0.01 m; F =? We have, v2 – u2 = 2as Or, (5 * 106)2 – 02 = 2 * a * 0.01 Or, a = 12.5 * 1014 Therefore, F = ma Or, F = (9.1 * 10-31) * (12.5 * 1014) Or, F = 1.1 * 10-15 N. 4. A block of mass 0.2 kg is suspended from the ceiling by a light string. A second block of mass 0.3 kg is suspended from the first block through another string. Find the tensions in the two strings. Take g = 10 m/s2. Sol:
T1 and T2 are the tensions in the two strings.
Taking the block 2 as the system. The forces acting on the system are (i) M2g downwards (ii) T2 upwards As the body is in equilibrium, these forces add to zero. T2 – M2g = 0 or, T2 = M2g = 0.3 * 10 = 3 N. Taking the block 1 as the system. The forces acting on the system are (i) M1g downwards (ii) T1 upwards (iii) T2 downwards As the body is in equilibrium, these forces add to zero. → T1 – T2 – M1g = 0 Or, T1 = M1g + T2 = 0.2 * 10 + 3 = 5 N. Therefore, tensions in the strings are 5 N and 3 N. 5. Two blocks of equal mass m are tied to each other through a light string. One of the blocks is pulled along the line joining them with a constant force F. Find the tension in the string joining the blocks. Sol:
Taking both the blocks as a system. Force acting on the system is F. Due to this force, let the system move with an acceleration, a. From Newton’s law of motion: We have, → Force = mass of system * acceleration Or, F = 2m * a Or, a = F/2m.
Now, taking first block as a system. Force acting on the system is T and acceleration of the system is F/2m. Mass of the system m. Again, from Newton’s law of motion: We have, → Force = mass of system * acceleration Or, T = m * a Or, T = m * (F/2m) = F/2. So, the tension in the string joining the blocks is F/2.
6. A particle of mass 50 g moves on a straight line. The variation of speed with time is shown in figure (5-E1). Find the force acting on the particle at t = 2, 4 and 6 seconds.
Sol: We know that, Slope of the curve in velocity v/s time graph = acceleration Therefore, a@ 2 s = (15/3) = 5 m/s2; a@ 4 s = 0 m/s2; a@ 6 s = – (15/3) = – 5 m/s2. We have, Force = mass * acceleration So, F@ 2s = (50/1000) * 5 Or, F@ 2s = 0.25 N along the motion. → F@ 4s = (50/1000) * 0 = 0. → F@ 6s = (50/1000) * (– 5) Or, F@ 6s = – 0.25 or 0.25 N opposite to the motion. 7. Two blocks A and B of mass mA and mB respectively are kept in contact on a frictionless table. The experimenter pushes the
block A from behind so that the blocks accelerate. If the block A exerts a force F on the block B, what is the force exerted by the experimenter on A? Sol: Given: mass of two blocks A and B are mA and mB respectively. Force exerted by block A on Block B is F. Therefore, the force exerted on block A by block B is also F. Let, the force exerted by the experimenter on A be P and acceleration be a.
Taking block A as a system and forces acting on block A are P and F. Therefore, mA * a = P – F Or, P = mA * a + F ------------ (1) Taking block B as a system and forces acting on block A is F. Therefore, mB * a = F Or, a = F/ mB ------------- (2) Substituting the equation (2) in equation (1), we get P = mA * (F/ mB) + F = F (1 + mA/ mB).
8. Raindrops of radius 1 mm and mass 4 mg are falling with a speed of 30 m/s on the head of a bald person. The drops splash on the head and come to rest. Assuming equivalently that the drops cover a distance equal to their radii on the head, estimate the force exerted by each drop on the head. Sol: Given: radius of raindrop, r = 1 mm = 0.001 m; mass of raindrop, m = 4 mg = 4*10-6 kg; speed of raindrop = initial velocity = 30 m/s; distance covered = radius of raindrop, r = 1 mm; final velocity = 0. We have, v2 – u2 = 2 a s Or, 02 – 302 = 2 * a * 0.001 Or, a = – 450000 m/s2 Therefore, the force exerted is F = m*a Or, F = 4 * 10-6 * 450000 = 1.8 N. 9. A particle of mass 0.3 kg is subjected to a force F = – k x with k = 15 N/m. What will be its initial acceleration if it is released from a point x = 20 cm? Sol: Given: mass of particle, m = 0.3 kg; k = 15 N/m; x = 20 cm = 0.2 m. Force, F = – kx = – 15 * 0.2 = – 3.0 N
Therefore, its initial acceleration if it is released from a point x = 20 cm is = F/m = 3.0/0.3 = 10 m/s2. 10. Both the springs shown in figure (5-E2) are unstretched. If the block is displaced by a distance x and released, what will be the initial acceleration?
Sol:
Where, k1 and k2 are in parallel, so equivalent k = k1 + k2 Force, F = – kx = – (k1 + k2) * x So, acceleration, a = F/m = – (x/m)*(k1 + k2) The initial acceleration is (x/m)*(k1 + k2) opposite to the displacement. 11. A small block B is placed on another block A of mass 5 kg and length 20 cm. Initially the block B is near the right end of block A
(figure 5-E3). A constant horizontal force of 10 N is applied to the block A. All the surfaces are assumed frictionless. Find the time elapsed before the block B separates from A.
Sol: Given: mass of block A is, m = 5 kg and length, s = 20 cm = 0.2 m, force applied, F = 10 N. Acceleration of block A is a = F/m = 10/5 = 2 m/s2. As there is no friction between A & B, when the block A moves, Block B remains at rest in its position. Initial velocity, u = 0. We have, s = ut + ½ at2 Or, 0.2 = 0 * t + ½ * 2 * t2 Or, t = 4.5 s. So, the time elapsed before the block B separates from A is 0.45 s. 12. A man has fallen into a ditch of width d and two of his friends are slowly pulling him out using a light rope and two fixed pulleys as shown in figure (5-E4). Show that the force (assumed equal for both the friends) exerted by each friend on the rope
increases as the man moves up. Find the force when the man is at a depth h.
Sol:
Given: height = h; width = d; angle, tan θ = (d/2)/h or, θ = tan-1 (d/2)/h; cos θ = h/ [√ {(d/2)2 + (h)2}] = 2h/√ (d2 + 4h2). Force balance along y-axis we get, 2T cos θ = mg Or, T = mg/2cos θ Or, T = (mg/4h)*(√ (d2 + 4h2). 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?
Sol: Taking block A as a system and forces acting on block A are
I) weight of the block A, mAg downward II) Reaction force on block A by Block B, N upward And acceleration of block, a = 2 m/s2 downward Thus, mAa = mAg – N Or, N = mA * (g – a) Or, N = 0.5*(10 – 2) = 4 N Force exerted by the block A on block B is equal and opposite to the force exerted by the block B on block A. Therefore, Force exerted by the block A on block B is = 4 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. Sol: Given: mass of bob, m = 50 g = 0.05 kg. (a) Goes up with acceleration, a = 1.2 m/s2 Thus, ma = T – mg Or, T = m (g + a) So, tension, T = 0.05 * (9.8 + 1.2) = 0.55 N. (b) Goes up with deceleration, a = – 1.2 m/s2 Thus, ma = T – mg Or, T = m (g + a) So, tension, T = 0.05*(9.8 – 1.2) = 0.43 N. (c) Goes up with uniform velocity, a = 0 Thus, ma = T – mg Or, T = m (g + a) So, tension, T = 0.05 * (9.8 – 0) = 0.49 N. (d) Goes down with acceleration, a = 1.2 m/s2 Thus, ma = mg – T or, T = m (g – a)
So, tension, T = 0.05*(9.8 – 1.2) = 0.43 N. (e) Goes down with deceleration, a = – 1.2 m/s2 Thus, ma = mg – T Or, T = m (g – a) So, tension, T = 0.05*(9.8 + 1.2) = 0.55 N. (f) Goes down with uniform velocity, a = 0 Thus, ma = mg – T Or, T = m (g – a) So, tension, T = 0.05*(9.8 – 0) = 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 the minimum weights recorded are 72 kg and 60 kg. Assuming that the magnitudes of the acceleration and the 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. Sol:
Given: maximum and the minimum weights recorded are 72 kg and 60 kg, g = 9.9 m/s2. Let the magnitudes of the acceleration and the deceleration be a. When elevator accelerated, weighing machine shows maximum weight (N = 72 * 9.9 = 712.8 N). So, ma = N – W Or, ma = 72g – mg ---------------- (1) When elevator decelerated, weighing machine shows minimum weight (N = 60 * 9.9 = 594 N). So, m * (– a) = N – W Or, – ma = 60g – mg ---------------- (2) From equation 1 and 2 we get, 2ma = 12g or, a = 6g/m Now, putting the value of a in equation (1) or (2) We get, m = 66 kg
(a) The true weight of the person is = 66 kg. (b) The magnitude of the acceleration, a = 6g/66 = 0.9 m/s2. 16. Find the reading of the spring balance shown in 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.
Sol:
The forces on this pulley are (i) T upwards by the upper string and (ii) 2T’ downwards by the lower strings.
As the mass of the pulley is negligible → T – 2T’ = 0 Or, T = 2T’ Motion of m = 1.5 kg: acceleration is (g/10 + a) in the upward direction. The forces on this mass are (i) 1.5*g downward by the earth and (ii) T’ upward by the string. Thus, 1.5 * (g/10 + a) = T’ – 1.5g -------- (1) Motion of m = 3.0 kg: acceleration is (g/10 – a) in the upward direction. The forces on this mass are (i) 3.0*g downward by the earth and (ii) T’ upward by the string. Thus, 3.0 * (g/10 – a) = T’ – 3.0g -------- (2) Solving equation (1) and (2) we get, T’ = 11g/5 N and a = – 11g/10 m/s2 So, T = 2T’ = 22g/5 N Therefore, the reading of the spring balance is = T/g = 22/5 = 4.4 kg. 17. A block of 2 kg is suspended from the ceiling through a massless 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? Sol: Given: mass of the block, m = 2 kg, spring constant, k = 100 N/m. Force, F = mg = 2*10 = 20 N We have, F = – kx (–ve sign shows that force is opposite to the displacement) Therefore, x = F/k = 0.2 m = 20 cm. After adding 1 kg of weight, the new force is F’ = 3*10 = 30 N. We have, F’ = – kx’ Therefore, x’ = 0.3 m = 30 cm. The elongation after adding 1 kg of weight is (0.3 – 0.2) = 0.1 m = 10 cm. 18. Suppose the ceiling in the previous problem is that of an elevator which is going up with an acceleration of 2.0 m/s2. Find the elongations. Sol:
Motion of the 2 kg block: the acceleration is a = 2 m/s2 upward. The forces on the block are (i) Force by spring is kx upward and (ii) 2g downward by the earth Thus, ma = kx – mg Or, x = m (a + g)/k = 0.24 m. Now, after adding of another 1 kg block, new mass, m’ = 3 kg. And elongation, x’ = m’ (a + g)/k Or, x’ = 0.36 m So, the elongation after adding 1 kg of weight is (0.36 – 0.24) = 0.12 m = 12 cm. 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 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? Sol:
Let, the air resistance force is F and Buoyant force is B. Given: Fa ∝ v, where v → velocity. → Fa = kv, where k → proportionality constant. When the balloon is moving downward, → B + kv = Mg Or, M = (B + kv)/g When the balloon is moving upward, → B = mg + kv Or, m = (B – kv)/g So, amount of mass that should be removed = M – m = (B + kv)/g – (B – kv)/g = 2 kv/g = 2 (Mg – B)/g = 2 (M – B/g). 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? Sol:
As the volume of the block in both case are same, so buoyancy force (B) in the both cases are same. Motion of the m kg block: the acceleration is a = g/6 m/s2 upward. The forces on the block are (i) Buoyancy force is B upward and (ii) mg downward by the earth Thus, m*g/6 = B – mg Or, B = 7mg/6 Motion of the m’ kg block: the acceleration is a = g/6 m/s2 downward. The forces on the block are (i) Buoyancy force is B upward and (ii) m’g downward by the earth Thus, m’*g/6 = m’g – B Or, B = 5m’g/6 Equating both the buoyancy we get the following relationship → m’ = (7/5)m
Therefore, sand should be put inside the box is = m’ – m = (2m/5). 21. A force F = v * A 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? Sol: For constant velocity, the magnitude of both the force must be equal and direction must opposite to each other. So, vA sin θ = mg Or, v = mg/A sin θ For the minimum speed, sin θ = 1 Therefore, the minimum speed, v = mg/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 = 300 g and m2 = 600 g. The system is released from rest, (a) Find the distance travelled 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.
Sol: Given: m1 = 300 g = 0.3 kg; m2 = 600 g = 0.6 kg. The acceleration of blocks m1 and m2 are shown in the figure.
Motion of the 0.3 kg block: the acceleration is a upward. The forces on the block are (i) Tension T upward and (ii) m1g downward by the earth Thus, m1a = T – m1g ------------- (1) Motion of the 0.6 kg block: the acceleration is a upward. The forces on the block are (i) Tension T upward and
(ii) m2g downward by the earth Thus, m2a = m2g – T ----------- (2) Solving equation 1 and 2 we get Acceleration, a = (m2 – m1) g/ (m2 + m1) g = 3.266 m/s2. And T = 3.9 N (a) t = 2 s; a = 3.266 m/s2; u = 0; So, distance travelled, s = ut + ½ at2 Or, s = 0 * 2 + ½ * 3.266 *22 = 6.5 m. (b) The tension in the string is = 3.9 N. (c) The force exerted by the clamp on the pulley is 2T = 2 * 3.9 = 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. Sol: In the previous problem we got the following value → a = 3.26 m/s2; T = 3.9 N. After 2 sec the velocity of mass (m2) V = u + at = 0 + 3.26 × 2 = 6.52 m/s upward. At this time m2 is moving 6.52 m/s downward.
At time 2 sec, m2 stops for a moment. But m1 is moving upward with velocity 6.52 m/s. It will continue to move until final velocity become zero. Here, v = 0; u = 6.52 m/s; a = – g = – 9.8 m/s2 [moving upward m1]. We know, V = u + at Or, 0 = 6.52 + (– 9.8) * t Or, t = 6.52/9.8 = 0.66 = 2/3 sec. During this period 2/3 sec, m2 mass also starts moving downward. So the string becomes tight again after a time of 2/3 sec. 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.
Sol: Mass of the rod per unit length is 3/30 = 0.1 kg/cm.
Mass of 10 cm rod is m1 = 0.1 * 10 = 1 kg. Mass of 20 cm rod is m2 = 0.1 * 20 = 2 kg. Both the pulley are smooth and massless. So, T1 = 20 N and T2 = 32 N Taking 30 cm rod as a system: → ma = T2 – T1 Or, 3a = 32 – 20 Or, a = 4 m/s. Now, taking 10 cm rod or 20 cm rod as a system: Motion of 10 cm rod: → m1a = R – T1 Or, R = 1 * 4 + 20 = 24 N. 25. Consider the situation shown in 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.
Sol:
→ (AC)2 = (AB)2 + (BC)2 Or, 52 = 32 + 42 → RHS = LHS So, ABC is a right angle triangle. Where, angle B = 900, angle A = sin-1 (4/5) = 53.130, angle A = sin1 (3/5) = 36.870. Motion of block- 1: acceleration, a shown in fig. → ma = mg sin A – T Or, 1a = 1g sin 53.130 – T ------------ (1) Motion of block- 2: acceleration, a shown in fig. → ma = T – mg sin C Or, 1a = T – 1g sin 36.870 ----------- (2)
Solving equation 1 and 2 we get, T = 7g/10 and a = g/10 Therefore, the magnitude of the acceleration of the two 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.
Sol:
Given: F = m2g/2 As the pulley is smooth and massless → T1 = T2 = T Motion of m1:
The acceleration is ‘a’ in horizontal direction. The forces on the m1 are (i) Applied force F (along –ve x-axis) (ii) Tension (T1) (along +ve x-axis) Thus, m1a = F – T1 Or, m1a = m2g/2 – T ----------- (1) Motion of m2: The acceleration is ‘a’ in vertical direction. The forces on the m2 are (i) m2g (along –ve Y-axis) (ii) Tension (T2) (along +ve Y-axis) Thus, m2a = T2 – m2g Or, m2a = T – m2g -------------- (2) Solving equation 1 and 2 we get Acceleration, a = – m2g/2(m1 + m2) Therefore, the acceleration of m1 is m2g/2(m1 + m2) towards right. 27. In figure (5-E11) m1 = 5 kg, 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.
Sol: Given: F = 1 N; m1 = 5 kg; m2 = 2 kg.
As the pulley is smooth and massless → T1 = T2 = T Motion of m1: The acceleration is ‘a’ in downward direction. The forces on the m1 are (i) Applied force F downward (ii) Tension (T1) upward (iii) m1g downward Thus, m1a = F – T1 + m1g Or, 5a = 1 – T + 5g ----------- (1)
Motion of m2: The acceleration is ‘a’ in upward direction. The forces on the m2 are (i) m2g downward (ii) Tension (T2) upward (iii) Applied force F downward Thus, m2a = T2 – m2g – F Or, 2a = T – 2g – 1 ------------ (2) Solving equation 1 and 2 we get Acceleration, a = (3/7) g = 4.3 m/s2. Therefore, the acceleration of either block is 4.3 m/s2. After the string breaks m1 move downward with force F acting downward. Therefore, m1a = F + m1g Or, 5a = 1 + 5g Or, a = 0.2 + g (downward direction). 28. Let m1 = 1 kg, m2 = 2 kg and m3 = 3 kg in figure (5-E12). Find the accelerations of m1, m2 and m3. The string from the upper pulley to ml is 20 cm when the system is released from rest. How long will it take before m1 strikes the pulley?
Sol: Given: m1 = 1 kg; m2 = 2 kg and m3 = 3 kg.
Let we assume the acceleration of m1 is a1 upward and a2 is the acceleration of the m2 and m3 with respect to pulley B. Motion of massless pulley B: the acceleration of pulley B is a1 downward. The forces on pulley B are (i) T upwards by the upper string and (ii) 2T’ downwards by the lower string. Thus, mpa1 = 2T’ – T = 0 [mp = 0] Or, T’ = T/2
Motion of m1: the acceleration of m1 is a1 downward. The forces on m1 are (i) T upwards by the upper string and (ii) m1g downwards by the earth. Thus, m1a1 = T – m1g Or, 1a1 = T – 1g Or, a1 = T – g ---------- (1) Equation of motion of m2: → m2 (a1 – a2) = m2g – T’ Or, 2 (a1 – a2) = 2g – T/2 ----------- (2) Equation of motion of m3: → m3 (a1 + a2) = m3g – T’ Or, 3 (a1 + a2) = 3g – T/2 ----------- (3) Solving equation 2 and 3 we get → a1 = g – 5T/24 ------------------- (4) Solving equation 1 and 4 we get → T = (48/29) g Substituting value of T = (48/29) g in equation 1. We get, a1 = (19/29) g (upward) Now, putting the value of a1 and T in (3)
We get, a2 = (2/29) g Acceleration of m2 = (a1 – a2) = (17/29) g downwards. Acceleration of m2 = (a1 + a2) = (21/29) g downwards. Distance of m1 from pulley, s = 20 cm = 0.2 m, a1 = 6.4 m/s2, u = 0. We have, s = ut + ½ at2 Or, 0.2 = 0 * t + ½ * 6.4 * t2 Or, t = 0.25 s It takes 0.25 s before m1 strikes the pulley. 29. In the previous problem, suppose m2 = 2.0 kg and m3 = 3.0 kg. What should be the mass m so that it remains at rest? Sol: According to the problem, acceleration of m1 is a1 = 0 (mass remains at rest), so acceleration of m2 = a2 upwards and acceleration of m3 = a2 downwards. Equation of motion of all the blocks: Block m1 → m1a1 = T – m1g Or, T = m1g --------- (1) Block m2 → m2a2 = T’ – m2g Or, 2 a2 = T/2 – 2g ----------- (2) Block m2 → m3a2 = m3g – T’
Or, 3 a2 = 3g – T/2 ------------ (3) Solving equation 2 and 3 we get, T = (24/5) g, and a2 = g/5 Putting the value of T in equation 1, we get The mass, m = m1 = 24/5 = 4.8 kg. 30. Calculate the tension in the string shown in figure (5-E13). The pulley and the string are light and all the surfaces are frictionless. Take g = 10 m/s2.
Sol:
Both the pulley A and B are frictionless and massless. Therefore, T1 = T2 = T3 = T.
Equation of motion of block A: → R – 1g = 0 Or, R = 10 N and mAa = T1 = T So, a = T Equation of motion of block B: → mBa = mBg – T3 Or, 1 * T = 1 * 10 – T Or, T = 5 N. So, the tension in the string is 5 N. 31. Consider the situation shown in figure (5-E14). Both the pulleys and the string are light and all the surfaces are frictionless. (a) Find the acceleration of the mass M. (b) Find the tension in the string, (c) Calculate the force exerted by the clamp on the pulley A in the figure.
Sol: Let the acceleration of 2M and M are a1 and a2 respectively.
Let the length of the string DABC be L (L = constant, because string is inextensible). So, L = BC + BA + AD Or, L = x1 + x1 + x2 = 2x1 + x2 Time rate of change of both the side of equation 𝑑2 𝐿
We get, 𝑑𝑡 2 =
𝑑2 (2𝑥1 ) 𝑑𝑡 2
+
𝑑2 (𝑥2 ) 𝑑𝑡 2
Or, 2a1 + a2 = 0 Therefore, 2a1 = a2 ---------- (1) Equation of motion of 2M mass: → T = 2Ma1 ------------------- (2) Equation of motion of massless pulley B: → T = 2T’ Equation of motion of M mass: Ma2 = Mg – T’ = Mg – T/2 --------- (3) Solving equation 1, 2 and 3 We get, a1 = g/3; a2 = 2g/3; T = 2Mg/3 and T’ = Mg/3.
(a) The acceleration of the mass M is 2g/3. (b) The tension in the string is Mg/3. (c)
The force exerted by the clamp on the pulley A is F = √ {(T’)2 + (T’)2} = (√2)T’ = √2Mg/3. And tan θ = T’/T’ = 1 Or, θ = 450 with the horizontal. 32. Find the acceleration of the block of mass M in the situation shown in figure (5-E15). All the surfaces are frictionless and the pulleys and the string are light.
Sol:
Let the acceleration of the block of mass M be a1 and the acceleration of the block of mass 2M be a2.
Length ABCD = L = AB + BC + CB + CB + DE = x1 + 2x2 + DE Time rate of change of L and DE is zero, because L and DE are constant (inextensible string). 𝑑𝐿
We get, 0 = 𝑑𝑡 =
𝑑(𝑥1 ) 𝑑𝑡
+
𝑑(2𝑥2 ) 𝑑𝑡
Again, time rate of change above equation We get, 0 =
𝑑2 (𝑥1 ) 𝑑𝑡 2
+
𝑑2 (2𝑥2 ) 𝑑𝑡 2
or, a1 + 2a2 = 0
Therefore, a1 = 2a2 ------------------ (1) Equation of motion of M mass: → T – Mg sin 30 = Ma1 ----------------- (2) Equation of motion of massless pulley C: → T’ = 2T Equation of motion of M mass: → 2Ma2 = 2Mg – T’
Or, 2Ma2 = 2Mg – 2T --------- (3) Solving equation 1, 2 and 3 We get, a1 = g/3 and a2 = g/6 So, the acceleration of the block of mass M is g/3 up the plane. 33. Find the mass M of the hanging block in figure (5-E16) which will prevent the smaller block from slipping over the triangular block. All the surfaces are frictionless and the strings and the pulleys are light.
Sol: Let the acceleration of blocks of mass M and M’ be a0. According to the problem, block of mass m is in rest on triangular block.
Equation of motion of block of mass M:
Ma0 = Mg – T (from F.B.D 1) ----------- (1) Equation of motion of block of mass M’: M’a0 = T – R1 sin θ (from F.B.D 2) ---------- (2) Equation of motion of block of mass m: a = 0 → ma = R1 sin θ – ma0 (from F.B.D 3) Or, ma0 = R1 sin θ ---------------- (3) And R1 cos θ = mg ------------------ (4) Solving equation (1), (2), (3) and (4) We get, R1 = mg/cos θ; a0 = g tan θ; T = (M’ + m) g/ tan θ and M = (M’ + m)/ (cot θ – 1). Therefore, the mass M of the hanging block is M = (M’ + m)/ (cot 𝛉 - 1). 34. Find the acceleration of the blocks A and B in the three situations shown in figure (5-E17).
Sol:
In all the three cases, acceleration of block B (aB) is half of the acceleration of block A (aA). This can be proved in the similar manner as shown in the problem no. 31 and 32. (a) We assume the acceleration of block A be ‘a’ downwards and the acceleration of block B is a/2 upwards.
Equation of motion of massless pulley A: mp = 0 → 2T – T’ = mpap = 0 Or, 2T = T’ Equation of motion of block A: → 4g – T = 4a -------------- (1) Equation of motion of block B: → T’ – 5g = 5a/2 Or, 4T – 10g = 5a ------------- (2) Solving equation (1) and (2) We get, a = (2/7) g
So, the acceleration of block A is (2/7) g downwards and the acceleration of block B is g/7 upwards. (b) We assume the acceleration of block A be ‘a’ towards right and the acceleration of block B is a/2 downwards.
Equation of motion of massless pulley A: mp = 0 → 2T – T’ = mpap = 0 Or, 2T = T’ Equation of motion of block A: T = 2a ------------------ (1) Equation of motion of block B: 5g – T’ = 5a/2 Or, 10g – 4T = 5a ---------- (2) Solving equation (1) and (2) We get, a = (10/13) g So, the acceleration of block A is (10/13) g towards right and the acceleration of block B is (5/13) g downwards.
(c) We assume the acceleration of block A be ‘a’ downwards and the acceleration of block B is a/2 upwards.
Equation of motion of massless pulley A: mp = 0 → 2T – T’ = mpap = 0 Or, 2T = T’ Equation of motion of block A: → 2g – T = 2a -------------- (1) Equation of motion of block B: → T’ – 1g = a/2 Or, 4T – 2g = a -------------- (2) Solving equation (1) and (2) We get, a = (2/3) g So, the acceleration of block A is (2/3) g downwards and the acceleration of block B is (1/3) g upwards. 35. Find the acceleration of the 500 g block in figure (5-E18).
Sol: All the pulleys are smooth and massless, so the tension on the both sides of the pulley are same. Strings are inextensible, so acceleration of all the blocks are same. Let the acceleration of all the blocks be a.
The equation of motion of mass 0.05 kg: → T1 – 0.05g = 0.05a -------------- (1) The equation of motion of mass 0.1 kg: → T2 – T1 – 0.1g sin 300 = 0.1a --------- (2) The equation of motion of mass 0.5 kg: → 0.5g – T2 = 0.5a ----------------- (3)
Solving equation (1), (2) and (3) We get, a = (8/13) g Therefore, the acceleration of the 500 g block is (8g/13) downward. 36. A monkey of mass 15 kg is climbing on a rope with one end fixed to the ceiling. If it wishes to go up with an acceleration of 1 m/s2, how much force should it apply to the rope? If the rope is 5 m long and the monkey starts from rest, how much time will it take to reach the ceiling? Sol:
Given: mass of monkey, m = 15 kg; a = 1 m/s2. The equation of motion of monkey: → T – mg = ma Or, T – 15g = 15 * 1 Therefore, the monkey should apply 165 N force to the rope. 2nd part:
Given: a = 1 m/s2, distance travelled, s = 5 m, initial velocity, u = 0. We have, s = ut + ½ at2 Or, 5 = 0 * t + ½ * 1 * t2 Or, t = √10 s Time required is √10 sec. 37. A monkey is climbing on a rope that goes over a smooth light pulley and supports a block of equal mass at the other end (figure 5-E19). Show that whatever force the monkey exerts on the rope, the monkey and the block move in the same direction with equal acceleration. If initially both were at rest, their separation will not change as time passes.
Sol:
Suppose the monkey accelerates upward with acceleration ’a’ & the block, accelerate downward with acceleration a1. Let Force exerted by monkey is equal to ‘T’. From the free body diagram of monkey ∴ T – Mg = Ma Or, T = Mg + ma --------------- (i) Again, from the FBD of the block, → Mg – T = Ma1. Or, Mg – Mg – Ma – Ma1 = 0 [From (i)] Or, ma = – ma1 Or, a = – a1. Acceleration ‘–a’ downward i.e. ‘a’ upward. Therefore, the block & the monkey move in the same direction with equal acceleration.
If initially they are rest (no force is exerted by monkey) no motion of monkey of block occurs as they have same weight (same mass). Their separation will not change as time passes. 38. The monkey B shown in figure (5-E20) is holding on 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 respectively. If A can tolerate a tension of 30 N in tail, what force should it apply on the rope in order carry the monkey B with it? Take g = 10 m/s2.
Sol: Given: mass of monkey A = mA = 5 kg; mass of monkey A = mB = 2 kg; tension in the tail, T2 = 30 N and g = 10 m/s2. Let monkey move upward with acceleration ‘a’.
The equation of motion of monkey B: → T2 – mBg = mBa Or, 30 – 2*10 = 2a Or, a = 5 m/s2. The equation of motion of monkey A: → T1 – mAg – T2 = mAa Or, T1 = mAa + mAg + T2 Or, T1 = 5 * 5 + 5 * 10 + 30 Or, T1 = 105 N. So, A can apply a maximum force of 105 N in the rope to carry the monkey B with it. For minimum force, acceleration a = 0; The equation of motion of monkey B: → T2 – mBg = mBa = 0 Or, T2 = 2 * 10 = 20 N The equation of motion of monkey A: → T1 – mAg – T2 = mAa =0 Or, T1 = mAg + T2 Or, T1 = 5 * 10 + 20 = 70 N So, A can apply a minimum force of 70 N in the rope to carry the monkey B with it.
Therefore, the monkey A should apply force between 70 N and 105 N to carry the monkey B with it. 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?
Sol: Given: mass of man is M = 60 kg; mass of box is m = 30 kg; box and man at rest; weight shown by the machine, R =?
Taking man of mass 60 kg as the system. The force acting on the system are (i) Normal reaction force by the box (upwards) (ii) Mg = 60g by the earth (downwards) (iii) Tension (T) by string (upwards) As the system at rest, these force should add to zero. This gives, T + R = 60g ------------ (1) Taking box of mass 30 kg as the system. The force acting on the system are (i) Normal reaction force by the man (downwards) (ii) mg = 30g by the earth (downwards) (iii) Tension (T) by string (upwards) As the system at rest, these force should add to zero. This gives, T – R = 30g ------------ (2) Solving equation (1) and (2) We get, T = 45g and R = 15g. Therefore, the weight shown by the machine is 15 kg. 2nd part: To get his correct weight suppose the applied force is ‘T’ and so, accelerates upward with ‘a’.
In this case, given that correct weight = R = 60g, where g = acceleration due to gravity.
The equation of motion of man: → 60a = T + R – 60g = T + 60g - 60g Or, T = 60a --------------- (1) The equation of motion of box: → 30a = T – R – 30g = T – 60g – 30g Or, T = 30a + 90g -------------- (2) Solving equation (1) and (2) We get, a = 3g = 30 m/s2 and T = 1800 N So, he should exert 1800 N force on the rope to get correct reading. 40. A block A can slide on a frictionless 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.
Sol: The force acting on the block is F = mg sin θ along the plane. So, acceleration = g sin θ Initial velocity of block u = 0. s = ℓ, a = g sin θ Now, S = ut + ½ at2 Or, ℓ = 0 * t + ½ * g sin θ * t2 Or, t = √ (2ℓ/g sin θ) So, the time taken by the block to slide down the length of the incline if it is released from the top of the incline is √ (2ℓ/g sin θ). 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. Sol: (i) Let the mass of the ball is ‘m’ and makes an angle of θ with the vertical.
→ T cos θ – mg = 0 Or, T cos θ = mg ------- (1) And, ma – T sin θ = 0 Or, ma = T sin θ --------- (2) From equation (1) and (2) We get, tan θ = a/g Or, θ = tan-1 (a/g). The angle is tan–1(a/g) with vertical. (ii) The mass of block is ‘m’ Suppose the angle of incline is ‘θ’
From the diagram → ma cos θ – mg sin θ = 0 Or, ma cos θ = mg sin θ Or, tan θ = a/g Or, θ = tan–1(a/g). Therefore, the angle of the incline with the horizontal is tan– 1 (a/g). 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. Sol: Because, the elevator is moving downward with an acceleration 12 m/s2 (>g), the block gets separated from elevator. So, the block moves with acceleration g = 10 m/s2 [freely falling body] and the elevator move with acceleration 12 m/s2. Now, the block has acceleration = g = 10 m/s2.
Given: u = 0, t = 0.2 sec So, the distance travelled by the block is given by → s = ut + ½ at2 Or, s = 0 + ½ * 10 * 0.22 Or, s = 5 * 0.04 = 0.2 m = 20 cm. The displacement of body is 20 cm during first 0.2 sec.
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