Laws of Motion, Friction & Dynamics of Circular Motion — HOTS Questions

A small block B of mass m = 0.5 kg is placed at one end of a block A of mass M = 10 kg and length l = 50 cm. A force F = 50 N is applied at the other end of the block as shown in Fig. Find the time elapsed before the block B falls down. Assume all the surfaces are frictionless and neglect the dimension of block B.

Answer

Given,

• Mass of block B, m = 0.5 kg
• Mass of block A, M = 10 kg
• Length of block A, l = 50 cm = 0.5 m
• Applied force, F = 50 N
• All the surfaces are frictionless

The time elapsed before block B falls down has to be calculated.

Since all the surfaces are frictionless, no horizontal force is transmitted from A to B. By Newton's second law, the acceleration of block A is

$\text a_A = \dfrac{\text F}{\text M} = \dfrac{50}{10} \\[1em] = 5\ \text{m s}^{-2}$

For block B there is no horizontal force at all, so

$\text a_B = 0$

Therefore, the acceleration of A relative to B is

$\text a_{AB} = \text a_A - \text a_B = 5 - 0 \\[1em] = 5\ \text{m s}^{-2}$

Block B falls off when A has slipped through its whole length relative to B, that is, when the relative displacement is l = 0.5 m. Both start from rest, so using $\text s = \text{ut} + \dfrac{1}{2}\text{at}^2$ with u = 0,

$\text l = \dfrac{1}{2}\text a_{AB}\text t^2 \\[1em] \text t = \sqrt{\dfrac{2\text l}{\text a_{AB}}}$

Substituting the values,

$\text t = \sqrt{\dfrac{2 \times 0.5}{5}} = \sqrt{0.2} \\[1em] = 0.45\ \text s$

Hence, block B falls down after 0.45 s.

Given fig. shows two blocks of masses m₁ = 5 kg and m₂ = 5 kg arranged by means of two massless inextensible strings and two light and smooth pulleys P₁ and P₂. The block m₂ is placed on a smooth platform. Find;

(a) the acceleration of two blocks.

(b) tension in the string connecting the block m₂ to pulley P₂.

(c) the force exerted by the clamp on the pulley P₁ that is pressure force, (not force per unit area) on axle of pulley P₁ (g = 10 ms^-2).

Answer

Given,

• Masses, m₁ = 5 kg and m₂ = 5 kg
• The platform, the pulleys and the strings are smooth and light
• Acceleration due to gravity, g = 10 m s^-2

The string carrying the tension T passes over the fixed pulley P₁ and then round the movable pulley P₂, so if a is the acceleration of the block m₁, the pulley P₂ and hence the block m₂ move with an acceleration $\dfrac{\text a}{2}$. Let T' be the tension in the string joining m₂ to the pulley P₂.

The platform is smooth, so the vertical forces on m₂ have no role to play. For the block m₂, by Newton's second law,

$\text T' = \text m_2\left(\dfrac{\text a}{2}\right) \quad \text{...............(1)}$

The pulley P₂ is light, so the net force on it must be zero,

$2\text T - \text T' = 0 \quad \Rightarrow \quad \text T' = 2\text T \quad \text{...............(2)}$

For the hanging block m₁,

$\text m_1\text g - \text T = \text m_1\text a \quad \text{...............(3)}$

(a) Multiplying equation (3) by 2,

$2\text m_1\text g - 2\text T = 2\text m_1\text a \quad \text{...............(4)}$

Adding equations (1), (2) and (4),

$2\text m_1\text g = \text m_2\left(\dfrac{\text a}{2}\right) + 2\text m_1\text a$

Here m₁ = m₂ = m = 5 kg, so

$2\text{mg} = \dfrac{\text{ma}}{2} + 2\text{ma} = \dfrac{5\text{ma}}{2} \\[1em] \text a = \dfrac{4\text g}{5}$

Substituting g = 10 m s^-2,

$\text a = \dfrac{4 \times 10}{5} \\[1em] = 8\ \text{m s}^{-2}$

Hence, the block m₁ moves with an acceleration of 8 m s^-2 and the block m₂ with an acceleration of 4 m s^-2.

(b) Substituting this value of a in equation (1),

$\text T' = 5 \times \dfrac{8}{2} \\[1em] = 20\ \text N$

Hence, the tension in the string connecting m₂ to the pulley P₂ is 20 N.

(c) From equation (2), the tension in the string passing over P₁ is

$\text T = \dfrac{\text T'}{2} = \dfrac{20}{2} \\[1em] = 10\ \text N$

The two segments of this string at the pulley P₁ are mutually perpendicular, one horizontal and the other vertical, and each pulls the pulley with a force T. Hence, the resultant force exerted on the axle of the pulley by the clamp is

$\text R = \sqrt{\text T^2 + \text T^2} = \text T\sqrt{2}$

Substituting the value of T,

$\text R = 10 \times 1.414 \\[1em] = 14.14\ \text N$

Hence, the force exerted by the clamp on the pulley P₁ is 14.14 N.

Figure shows three blocks of masses m₁ = 1 kg, m₂ = 2 kg and m₃ = 4 kg arranged by two light, inextensible strings and two light, frictionless pulleys P₁ and P₂ as shown in the figure. Find the acceleration of the blocks and tension in the strings.

Answer

Given,

• Masses, m₁ = 1 kg, m₂ = 2 kg and m₃ = 4 kg
• The strings are light and inextensible, and the pulleys are light and frictionless
• Acceleration due to gravity, g = 10 m s^-2

The block m₁ hangs from one side of the fixed pulley P₁, and the movable pulley P₂ hangs from the other side. The blocks m₂ and m₃ hang from the two sides of P₂.

Let a₁ be the absolute acceleration of m₁ (upwards), so that the pulley P₂ descends with the same acceleration a₁. Let a₂ be the acceleration of m₂ and m₃ relative to P₂. Then m₂ has an absolute upward acceleration (a₂ − a₁) and m₃ has an absolute downward acceleration (a₂ + a₁).

Let T₁ be the tension in the string over P₁ and T₂ that in the string over P₂. Applying Newton's second law,

$\text T_1 - \text m_1\text g = \text m_1\text a_1 \quad \text{...............(1)}$

$\text T_2 - \text m_2\text g = \text m_2(\text a_2 - \text a_1) \quad \text{...............(2)}$

$\text m_3\text g - \text T_2 = \text m_3(\text a_2 + \text a_1) \quad \text{...............(3)}$

Since the pulley P₂ is weightless,

$\text T_1 - 2\text T_2 = 0 \quad \Rightarrow \quad \text T_1 = 2\text T_2 \quad \text{...............(4)}$

Adding equations (2) and (3),

$\text m_3\text g - \text m_2\text g = \text m_2\text a_2 - \text m_2\text a_1 + \text m_3\text a_2 + \text m_3\text a_1$

Substituting m₂ = 2 kg and m₃ = 4 kg,

$4\text g - 2\text g = 2\text a_2 - 2\text a_1 + 4\text a_2 + 4\text a_1 \\[1em] 2\text g = 6\text a_2 + 2\text a_1 \\[1em] \text g = 3\text a_2 + \text a_1 \quad \text{...............(5)}$

Substituting T₁ from equation (4) in equation (1),

$2\text T_2 - \text m_1\text g = \text m_1\text a_1 \quad \text{...............(6)}$

Adding equation (6) and equation (3) multiplied by 2,

$2\text m_3\text g - \text m_1\text g = 2\text m_3\text a_2 + 2\text m_3\text a_1 + \text m_1\text a_1$

Substituting m₁ = 1 kg and m₃ = 4 kg,

$8\text g - \text g = 8\text a_2 + 8\text a_1 + \text a_1 \\[1em] 7\text g = 8\text a_2 + 9\text a_1 \quad \text{...............(7)}$

From equation (5), a₁ = g − 3a₂. Substituting this in equation (7),

$7\text g = 8\text a_2 + 9(\text g - 3\text a_2) \\[1em] 7\text g = 8\text a_2 + 9\text g - 27\text a_2 \\[1em] 19\text a_2 = 2\text g \quad \Rightarrow \quad \text a_2 = \dfrac{2\text g}{19}$

Substituting g = 10 m s^-2,

$\text a_2 = \dfrac{2 \times 10}{19} = \dfrac{20}{19} \\[1em] = 1.05\ \text{m s}^{-2}$

From equation (5),

$\text a_1 = \text g - 3\text a_2 = 10 - 3 \times 1.05 \\[1em] = 10 - 3.16 \\[1em] = 6.84\ \text{m s}^{-2}$

Therefore, the absolute accelerations of the blocks are

$\text{For } \text m_1 : \text a_1 = 6.84\ \text{m s}^{-2}\ (\text{upwards})$

$\text{For } \text m_2 : \text a_2 - \text a_1 = 1.05 - 6.84 = -5.79\ \text{m s}^{-2},\ \text{that is, } 5.79\ \text{m s}^{-2}\ (\text{downwards})$

$\text{For } \text m_3 : \text a_2 + \text a_1 = 1.05 + 6.84 = 7.89\ \text{m s}^{-2}\ (\text{downwards})$

From equation (1),

$\text T_1 = \text m_1(\text g + \text a_1) = 1 \times (10 + 6.84) \\[1em] = 16.84\ \text N$

From equation (4),

$\text T_2 = \dfrac{\text T_1}{2} = \dfrac{16.84}{2} \\[1em] = 8.42\ \text N$

Hence, the tensions in the two strings are 16.84 N and 8.42 N respectively.

Note: The question states m₃ = 4 kg, but the solution printed in the textbook substitutes m₃ = 3 kg and therefore obtains a₂ = 0.69 m s^-2, a₁ = 6.6 m s^-2, T₁ = 16.6 N and T₂ = 8.3 N. The above solution used the value of m₃ given in the question.

A person of mass 50 kg is standing inside a box of mass 25 kg by keeping a massless inextensible string. The other end of the strong string is passing over a light frictionless pulley P as shown in the diagram. The person manages to remain stationary in the box. What force is applied by him on the rope downwards?

Answer

Given,

• Mass of the person, m₁ = 50 kg
• Mass of the box, m₂ = 25 kg
• The string is massless and inextensible, and the pulley is light and frictionless
• The person and the box remain stationary

The force applied by the person on the rope has to be calculated.

Let T be the tension in the rope and R the normal reaction between the floor of the box and the person. By Newton's third law of motion, the person presses the floor with a force R downwards and the floor pushes the person up with the same force R.

Since the person and the box are both stationary, the net force on each is zero.

For the box, the upward forces are the tension T and the reaction R exerted by the person on the floor acts downwards. Hence,

$\text T - \text R = \text m_2\text g \quad \text{...............(1)}$

For the person, the upward forces are the tension T (he pulls the rope downwards, so the rope pulls him upwards) and the normal reaction R of the floor. Hence,

$\text T + \text R = \text m_1\text g \quad \text{...............(2)}$

Adding equations (1) and (2),

$2\text T = (\text m_1 + \text m_2)\text g \\[1em] \text T = \dfrac{(\text m_1 + \text m_2)\text g}{2}$

Substituting the values with g = 9.8 m s^-2,

$\text T = \dfrac{(50 + 25) \times 9.8}{2} = \dfrac{75 \times 9.8}{2} \\[1em] = \dfrac{735}{2} \\[1em] = 367.5\ \text N$

By Newton's third law of motion, the force applied by the person on the rope downwards is equal in magnitude to the tension in the rope.

Hence, the person applies a force of 367.5 N on the rope in the downward direction.

A block A of mass m₁ = 100 kg rests on a block B of mass m₂ = 150 kg. A is tied with a horizontal string to a wall. Coefficient of friction between A and B is 0.25 and that between B and floor is 0.2. Draw the free body diagram of block A and B and calculate what horizontal force F is needed to move the block B? (Take g = 10 ms^-2).

Answer

Given,

• Mass of block A, m₁ = 100 kg
• Mass of block B, m₂ = 150 kg
• Coefficient of friction between A and B, μ₁ = 0.25
• Coefficient of friction between B and the floor, μ₂ = 0.2
• Acceleration due to gravity, g = 10 m s^-2

The horizontal force F needed to move block B has to be calculated.

Free body diagram of block A : The forces are the weight m₁g downwards, the normal reaction N₁ of B upwards, the tension T of the string and the frictional force f₁ exerted by B.

Free body diagram of block B : The forces are the weight m₂g downwards, the normal reaction N₁ of A pressing down on it, the normal reaction N₂ of the floor upwards, the applied force F, and the frictional forces f₁ (due to A) and f₂ (due to the floor).

For the vertical equilibrium of block A, there is no vertical motion, so

$\text N_1 = \text m_1\text g \quad \text{...............(1)}$

For the vertical equilibrium of block B,

$\text N_2 - \text N_1 - \text m_2\text g = 0 \\[1em] \text N_2 = \text N_1 + \text m_2\text g = (\text m_1 + \text m_2)\text g \quad \text{...............(2)}$

Block B is just made to move, so the frictional forces have their limiting values. For the horizontal equilibrium of block B,

$\text F = \text f_1 + \text f_2 = \mu_1\text N_1 + \mu_2\text N_2$

$\text F = \mu_1\text m_1\text g + \mu_2(\text m_1 + \text m_2)\text g$

Substituting the values,

$\text F = 0.25 \times 100 \times 10 + 0.2 \times (100 + 150) \times 10 \\[1em] = 250 + 0.2 \times 2500 \\[1em] = 250 + 500 \\[1em] = 750\ \text N$

Hence, a horizontal force of 750 N is needed to move the block B.

A heavy uniform ladder is in equilibrium with one end resting on the ground and the other end resting against a rough wall. If μ₁ and μ₂ be the coefficients of friction between the ladder and the ground and between the ladder and the wall respectively, show that the minimum angle of inclination of the ladder to the horizontal is

$\tan^{-1}\left(\dfrac{1 - \mu_1\mu_2}{2\mu_1}\right)$

Answer

Let the ladder AB of weight W and length l rest with its lower end A on the ground and its upper end B against the wall, making an angle θ with the horizontal. Since the ladder is uniform, its weight W acts vertically downwards at its mid-point, where its centre of gravity is situated.

Let N₁ and N₂ be the normal contact forces at A and B, and f₁ and f₂ the frictional forces at A and B respectively. At the minimum angle of inclination the ladder is just about to slip, so the frictional forces have their limiting values,

$\text f_1 = \mu_1\text N_1 \quad \text{and} \quad \text f_2 = \mu_2\text N_2$

Vertical equilibrium : The upward forces are N₁ and f₂, and the downward force is W,

$\text N_1 + \text f_2 = \text W \\[1em] \text N_1 + \mu_2\text N_2 = \text W \quad \text{...............(1)}$

Horizontal equilibrium : The wall pushes the ladder with N₂ and the friction at the ground opposes it,

$\text N_2 = \text f_1 = \mu_1\text N_1 \quad \text{...............(2)}$

Substituting N₂ from equation (2) in equation (1),

$\text N_1 + \mu_2\mu_1\text N_1 = \text W \\[1em] \text N_1(1 + \mu_1\mu_2) = \text W \\[1em] \text N_1 = \dfrac{\text W}{1 + \mu_1\mu_2} \quad \text{...............(3)}$

From equation (2),

$\text N_2 = \dfrac{\mu_1\text W}{1 + \mu_1\mu_2} \quad \text{...............(4)}$

Rotational equilibrium : Taking moments of all the forces about the lower end A. The forces N₁ and f₁ act at A itself, so their moments about A are zero. The weight W tends to rotate the ladder clockwise, while N₂ and f₂ tend to rotate it anticlockwise. The perpendicular distances are $\dfrac{ l}{2}\cos \theta$ for W, $l$ sin θ for N₂ and l cos θ for f₂. Hence, by the principle of moments,

$-\text W \times \dfrac{ l}{2}\cos \theta + \text N_2 \times l\sin \theta + \mu_2\text N_2 \times \text l\cos \theta = 0$

Substituting N₂ from equation (4) and dividing throughout by $l$,

$-\dfrac{\text W}{2}\cos \theta + \dfrac{\mu_1\text W}{1 + \mu_1\mu_2}\sin \theta + \dfrac{\mu_1\mu_2\text W}{1 + \mu_1\mu_2}\cos \theta = 0$

Dividing throughout by W and rearranging,

$\dfrac{1}{2}\cos \theta - \dfrac{\mu_1\mu_2}{1 + \mu_1\mu_2}\cos \theta = \dfrac{\mu_1}{1 + \mu_1\mu_2}\sin \theta$

$\dfrac{\sin \theta}{\cos \theta} = \dfrac{1 + \mu_1\mu_2}{2\mu_1} - \dfrac{\mu_1\mu_2}{\mu_1}$

$\tan \theta = \dfrac{1 + \mu_1\mu_2 - 2\mu_1\mu_2}{2\mu_1} = \dfrac{1 - \mu_1\mu_2}{2\mu_1}$

Therefore,

$\theta = \tan^{-1}\left(\dfrac{1 - \mu_1\mu_2}{2\mu_1}\right)$

Hence proved.

In given Fig. a uniform beam AB (= 5m long) of weight 20 kg is supported by three cables AC, BD and BE and are capable of holding a tension of 50 kg each. As the magnitude of W acting at 1 m length from A is slowly increased, which rope will break first and at what value of W?

Answer

Given,

• Length of the beam AB, l = 5 m
• Weight of the beam, W₁ = 20 kg, acting at its mid-point G (2.5 m from A)
• W acts at 1 m from A
• Each cable can hold a maximum tension of 50 kg
• The cable AC makes 37° and the cable BD makes 53° with the beam, BE being along the beam

Which cable breaks first, and the value of W at which it breaks, has to be found.

The components T₁ cos 37° of T₁, T₂ cos 53° of T₂ and the tension T₃ along BE are all parallel to the length of the beam, so they do not test the strength of the cables AC and BD. Only the vertical components T₁ sin 37° and T₂ sin 53° enter the moment equations.

Taking moments of the forces about A : The weights W and W₁ rotate the beam clockwise while T₂ sin 53° rotates it anticlockwise. Hence,

$-\text W \times 1 - 20 \times 2.5 + \text T_2\sin 53^\circ \times 5 = 0$

$\text T_2 = \dfrac{\text W + 50}{5\sin 53^\circ} \quad \text{...............(1)}$

Taking moments of the forces about B :

$-\text T_1\sin 37^\circ \times 5 + \text W \times 4 + 20 \times 2.5 = 0$

$\text T_1 = \dfrac{4\text W + 50}{5\sin 37^\circ} \quad \text{...............(2)}$

Comparing equations (1) and (2), the numerator of T₁ increases four times as fast with W as that of T₂, and sin 37° is smaller than sin 53°. Hence, as W is slowly increased, T₁ increases much more rapidly than T₂, and so the cable AC breaks first.

Putting T₁ = 50 kg in equation (2) and taking sin 37° = 0.6018,

$50 = \dfrac{4\text W + 50}{5 \times 0.6018} \\[1em] 4\text W + 50 = 50 \times 5 \times 0.6018 \\[1em] 4\text W + 50 = 150.45 \\[1em] \text W = \dfrac{100.45}{4} \\[1em] = 25\ \text{kg-wt}$

Hence, the cable AC will break first, and it will break when W becomes nearly 25 kg-wt.

A stream of water flowing horizontally with a speed of 15 m s^-1 gushes out of a tube of cross-sectional area 10^-2 m² and hits a nearby vertical wall. Find the force exerted on the wall by the impact of water. Assume that water does not rebound from the wall. Take density of water to be 1.0 × 10³ kg m^-3.

Answer

Given,

• Speed of the water, v = 15 m s^-1
• Area of cross-section of the tube, A = 10^-2 m²
• Density of water, ρ = 1.0 × 10³ kg m^-3
• The water does not rebound from the wall

The force exerted on the wall has to be calculated.

The volume of water striking the wall per second is Av, so the mass of water striking the wall per second is

$\text m = \text{Av}\rho$

Substituting the values,

$\text m = 10^{-2} \times 15 \times 1.0 \times 10^3 \\[1em] = 150\ \text{kg s}^{-1}$

The momentum carried to the wall per second is

$\text{mv} = 150 \times 15 \\[1em] = 2250\ \text{kg m s}^{-2}$

Since the water does not rebound from the wall, the momentum of the water after the impact is zero. Hence, the rate of change of momentum is 2250 kg m s^-2. By Newton's second law of motion, the force is equal to the rate of change of momentum,

$\text F = 2250\ \text N$

Hence, the force exerted on the wall by the impact of water is 2250 N.

Ten one-rupee coins are put on the top of one another on a table. Each coin has a mass of m kg. Give the magnitude and direction of (a) the force on the 7th coin (counted from the bottom) due to all the coins on its top, (b) the force on the 7th coin by the 8th coin, (c) reaction of the 6th coin on the 7th coin.

Answer

Given,

• Ten one-rupee coins are placed one above the other
• Mass of each coin = m kg

The magnitude and the direction of the force in each case has to be found.

(a) Above the 7th coin (counted from the bottom) there are three coins, namely the 8th, the 9th and the 10th. The force on the 7th coin due to all the coins on its top is the sum of their weights,

$\text F = 3\text{mg}\ \text N$

This force acts vertically downwards.

(b) The 8th coin already supports the weights of the two coins above it (the 9th and the 10th) and has its own weight as well. Hence, the force exerted on the 7th coin by the 8th coin is the sum of these three weights,

$\text F = (2\text m + \text m)\text g = 3\text{mg}\ \text N$

This force also acts vertically downwards.

(c) The 6th coin experiences a downward force due to the weights of the four coins above it, that is, the 7th, 8th, 9th and 10th coins,

$\text F = 4\text{mg}\ \text N$

By Newton's third law of motion, the reaction of the 6th coin on the 7th coin is

$\text R = -\text F = -4\text{mg}\ \text N$

The negative sign shows that this reaction is directed vertically upwards.

A 25 kg block is raised by a 50 kg man in two different ways, as shown. What is the action on the floor by the man in the two cases? If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without yielding of the floor? (g = 9.8 ms^-2)

Answer

Given,

• Mass of the block, m = 25 kg
• Mass of the man, M = 50 kg
• Acceleration due to gravity, g = 9.8 m s^-2
• The floor yields to a normal force of 700 N

The action on the floor by the man in the two cases has to be calculated.

The force required to lift the block is its weight,

$\text F = \text{mg} = 25 \times 9.8 \\[1em] = 245\ \text N$

and the weight of the man himself is

$\text W = \text{Mg} = 50 \times 9.8 \\[1em] = 490\ \text N$

Case (a) : Here the man pulls the rope upwards, so he exerts the force in a direction opposite to his weight. He has to overcome his own weight and also exert the force to raise the block. Hence, the action on the floor is

$\text W + \text F = 490 + 245 \\[1em] = 735\ \text N$

Case (b) : Here the man pulls the rope in the direction of his weight, that is, his weight helps him in exerting the force. Hence, the action on the floor is

$\text W - \text F = 490 - 245 \\[1em] = 245\ \text N$

Since the floor yields to a normal force of 700 N, the action of 735 N in case (a) would make the floor yield, whereas the action of 245 N in case (b) would not.

Hence, the man should adopt mode (b) to lift the block without yielding of the floor.

A monkey of mass 40 kg climbs on a rope which can withstand a maximum tension of 600 N. In which case will the rope break? The monkey (a) climbs up with an acceleration of 6 ms^-2, (b) climbs down with an acceleration of 4 ms^-2, (c) climbs up with uniform speed of 5 ms^-1, (d) falls down the rope nearly freely under gravity? (Take g = 10 ms^-2, ignore mass of the rope.)

Answer

Given,

• Mass of the monkey, m = 40 kg
• Maximum tension the rope can withstand = 600 N
• Acceleration due to gravity, g = 10 m s^-2
• The mass of the rope is ignored

The case in which the rope breaks has to be found.

Let T be the tension in the rope, acting upwards, and mg the gravity force on the monkey, acting downwards.

(a) Climbing up with an acceleration of 6 m s^-2 : The net force on the monkey must be upwards,

$\text T - \text{mg} = \text{ma}_1 \quad \Rightarrow \quad \text T = \text m(\text g + \text a_1)$

$\text T = 40 \times (10 + 6) \\[1em] = 640\ \text N$

(b) Climbing down with an acceleration of 4 m s^-2 : The net force on the monkey must be downwards,

$\text{mg} - \text T = \text{ma}_2 \quad \Rightarrow \quad \text T = \text m(\text g - \text a_2)$

$\text T = 40 \times (10 - 4) \\[1em] = 240\ \text N$

(c) Climbing up with a uniform speed of 5 m s^-1 : The acceleration is zero, so

$\text T = \text{mg} = 40 \times 10 \\[1em] = 400\ \text N$

(d) Falling down nearly freely under gravity : Here a = g, so

$\text T = \text m(\text g - \text g) \\[1em] = 0$

The rope breaks only when the tension exceeds 600 N, and this happens only in case (a), where the tension is 640 N.

Hence, the rope will break in case (a), that is, when the monkey climbs up with an acceleration of 6 m s^-2.

A man of mass 65 kg is standing stationary with respect to a horizontal conveyor belt which is moving with an acceleration of 1.0 ms^-2, as shown. (a) Find the net force on the man. (b) If the coefficient of static friction between the shoes of the man and the belt is 0.2, up to what acceleration of the belt can the man continue to remain stationary relative to the belt? (g = 9.8 ms^-2)

Answer

Given,

• Mass of the man, m = 65 kg
• Acceleration of the belt, a = 1.0 m s^-2
• Coefficient of static friction, μ_s = 0.2
• Acceleration due to gravity, g = 9.8 m s^-2

(a) The man is stationary relative to the belt, so his acceleration is the same as that of the belt. The gravity force mg on him is balanced by the equal and opposite normal reaction R of the belt. Hence, the net force on the man is only due to his acceleration,

$\text F = \text{ma} = 65 \times 1.0 \\[1em] = 65\ \text N$

Hence, the net force on the man is 65 N, directed along the direction of motion of the belt. This force is provided by the force of friction between his shoes and the belt.

(b) The limiting frictional force between the man's shoes and the belt is

$\text f_s = \mu_s\text R = \mu_s\text{mg}$

$\text f_s = 0.2 \times 65 \times 9.8 \\[1em] = 127.4\ \text N$

The man can continue to remain stationary relative to the belt only so long as the force needed to accelerate him does not exceed this limiting friction. At the maximum acceleration a_max,

$\text{ma}_{max} = \text f_s \\[1em] \text a_{max} = \dfrac{\text f_s}{\text m} = \dfrac{127.4}{65} \\[1em] = 1.96\ \text{m s}^{-2}$

Hence, the man can remain stationary relative to the belt up to an acceleration of 1.96 m s^-2.

Two bodies A and B of respective masses 5 kg and 10 kg in contact with each other rest on a table against a rigid partition, as shown. The coefficient of friction between the bodies and the table is 0.15. A force F of 200 N is applied horizontally at A. Find (a) the reaction of the partition, (b) the action-reaction forces on A and B. What would happen if the partition be removed? Does the answer to (b) change, when the bodies are in motion? Ignore difference between μ_s and μ_k and take g = 9.8 ms^-2.

Answer

Given,

• Mass of body A, m₁ = 5 kg and mass of body B, m₂ = 10 kg
• Coefficient of friction between the bodies and the table, μ_s = 0.15
• Applied force, F = 200 N, horizontally at A
• Acceleration due to gravity, g = 9.8 m s^-2

(a) When the force F is applied at A towards the right, the limiting frictional force on the two bodies together, directed towards the left, is

$\text f_s =\text f_1+ \text f_2= \mu_s(\text m_1 + \text m_2)\text g = 0.15 \times (5 + 10) \times 9.8 \\[1em] = 0.15 \times 147 \\[1em] = 22\ \text N$

Hence, the net force exerted on the partition is

$\text F' = \text F - \text f_s = 200 - 22 \\[1em] = 178\ \text N,\ \text{towards the right}$

By Newton's third law of motion, the reaction of the partition is 178 N, directed towards the left.

(b) The force of limiting friction on body A alone is

$(\text f_1) = \mu_s\text m_1\text g = 0.15 \times 5 \times 9.8 \\[1em] = 7.35\ \text N$

Hence, the net force exerted by body A on body B is

$\text F'' = \text F - (\text f_1) = 200 - 7.35 \\[1em] = 192.65\ \text N,\ \text{towards the right}$

By Newton's third law of motion, the reaction of body B on body A is 192.65 N, directed towards the left.

If the partition is removed : The two bodies move together towards the right under the net force F' = 178 N, with an acceleration

$\text a = \dfrac{\text F'}{\text m_1 + \text m_2} = \dfrac{178}{15} \\[1em] = 11.9\ \text{m s}^{-2}$

The force producing this motion in A alone is

$\text m_1\text a = 5 \times 11.9 \\[1em] = 59.5\ \text N$

Hence, the net force now exerted by A on B is

$\text F'' - \text m_1\text a = 192.65 - 59.5 \\[1em] = 133.15\ \text N,\ \text{towards the right}$

and the reaction of B on A is 133.15 N towards the left.

Hence, the answer to part (b) does change when the bodies are in motion.

A block of mass 15 kg is placed on a long trolley. The coefficient of friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5 ms^-2 for 20 s and then moves with uniform velocity. Discuss the motion of the block as viewed by (a) a stationary observer on the ground, (b) an observer moving with the trolley (g = 9.8 ms^-2).

Answer

Given,

• Mass of the block, m = 15 kg
• Coefficient of friction between the block and the trolley, μ_s = 0.18
• Acceleration of the trolley, a = 0.5 m s^-2 for 20 s, and uniform velocity thereafter
• Acceleration due to gravity, g = 9.8 m s^-2

The force required to give the block the same acceleration as the trolley is

$\text F = \text{ma} = 15 \times 0.5 \\[1em] = 7.5\ \text N$

The weight mg of the block is balanced by the normal reaction R of the trolley. The limiting frictional force between the block and the trolley is

$\text f_s = \mu_s\text{mg} = 0.18 \times 15 \times 9.8 \\[1em] = 26.5\ \text N$

(a) As viewed by a stationary observer on the ground : As the trolley accelerates forward it exerts a reactionary force F' (= F = 7.5 N) on the block in the backward direction. As the block tends to move under this force, the force of friction comes into play opposite to F'. Since the limiting friction (26.5 N) is much larger than 7.5 N, the frictional force, being self-adjusting, simply adjusts itself to 7.5 N in the direction of motion of the trolley, and the block does not slide. Hence, during the accelerated motion of the trolley the block moves along with the trolley with the same acceleration of 0.5 m s^-2, appearing to be at rest relative to the trolley.

When the trolley later moves with a uniform velocity, no force F is needed, and the block shows no tendency to move relative to the trolley. Hence, the force of friction does not come into play at all, and the block moves along with the trolley with the same uniform velocity.

(b) As viewed by an observer moving with the trolley : During the first 20 s this observer has an accelerated motion, so his is a non-inertial frame in which the law of inertia is not valid. To him the block appears stationary, and he explains this by invoking a pseudo force of 7.5 N acting backwards on the block, which is balanced by the frictional force of 7.5 N acting forwards. Once the trolley moves with a uniform velocity, the observer becomes an inertial observer and the block still appears to him to be at rest, with no force acting on it.

The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end, as shown. The coefficient of friction between the box and the surface below it is 0.15. On a straight road, the truck starts from rest and accelerates with 2 m s^-2. At what distance from the starting point does the box fall off the truck? (Ignore the size of the box and take g = 9.8 ms^-2).

Answer

Given,

• Mass of the box, m = 40 kg
• Distance of the box from the open end, s = 5 m
• Coefficient of friction, μ_s = 0.15
• Acceleration of the truck, a = 2 m s^-2, starting from rest
• Acceleration due to gravity, g = 9.8 m s^-2

The distance from the starting point at which the box falls off the truck has to be calculated.

The force on the box due to the accelerated motion of the truck is

$\text F = \text{ma} = 40 \times 2 \\[1em] = 80\ \text N$

As the truck accelerates in the forward direction, a reactionary force F' (= F = 80 N) acts on the box in the backward direction. As the box tends to move under this force, the force of friction comes into play in the forward direction. The limiting frictional force between the box and the surface below it is

$\text f_s = \mu_s\text{mg} = 0.15 \times 40 \times 9.8 \\[1em] = 58.8\ \text N$

Since F' is greater than f_s, the box slides backwards. The net force acting on the box in the backward direction is

$\text F' - \text f_s = 80 - 58.8 \\[1em] = 21.2\ \text N$

The acceleration produced in the box relative to the truck, in the backward direction, is

$\text a' = \dfrac{21.2}{40} \\[1em] = 0.53\ \text{m s}^{-2}$

If t is the time in which the box slides through 5 m relative to the truck and falls off, then using $\text s = \text{ut} + \dfrac{1}{2}\text a'\text t^2$ with u = 0,

$5 = \dfrac{1}{2} \times 0.53 \times \text t^2 \\[1em] \text t^2 = \dfrac{10}{0.53} = 18.87 \\[1em] \text t = 4.34\ \text s$

The distance covered by the truck from the starting point in this time is

$\text s = \text{ut} + \dfrac{1}{2}\text{at}^2 = 0 + \dfrac{1}{2} \times 2 \times (4.34)^2 \\[1em] = 18.8\ \text m$

Hence, the box falls off the truck at a distance of 18.8 m from the starting point.

Find the mass M of the hanging block in the figure 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. (θ = 30°)

Answer

Given,

• Mass of the smaller block, m = 2 kg
• Mass of the triangular block, M' = 5 kg
• Angle of the triangular block, θ = 30°
• All the surfaces are frictionless, and the strings and the pulleys are light

The mass M of the hanging block has to be calculated.

Let the hanging block M descend with an acceleration a. Since the string is light and inextensible and passes over a light pulley, the triangular block together with the smaller block placed on it moves horizontally with the same acceleration a, and the smaller block moves along with the triangular block without slipping.

Condition that the smaller block does not slip : In the frame of the accelerating triangular block, a pseudo force ma acts on the smaller block in the backward direction. Resolving the pseudo force and the gravity force along the inclined surface, the smaller block does not slip when these components balance,

$\text{ma}\cos \theta = \text{mg}\sin \theta$

Dividing throughout by m cos θ,

$\text a = \text g\tan \theta \quad \text{...............(1)}$

Substituting θ = 30°,

$\text a = \text g\tan 30^\circ = \dfrac{\text g}{\sqrt{3}}$

For the hanging block M : Let T be the tension in the string. The net downward force on M is Mg − T, so by Newton's second law,

$\text{Mg} - \text T = \text{Ma} \quad \text{...............(2)}$

For the triangular block together with the smaller block : This combination, of total mass (M' + m), is pulled horizontally by the tension T alone, the floor being frictionless. Hence,

$\text T = (\text M' + \text m)\text a \quad \text{...............(3)}$

Substituting the values,

$\text T = (5 + 2)\text a = 7\text a$

Adding equations (2) and (3),

$\text{Mg} = \text{Ma} + 7\text a = (\text M + 7)\text a$

$\text a = \dfrac{\text{Mg}}{\text M + 7} \quad \text{...............(4)}$

Comparing equations (1) and (4),

$\dfrac{\text{Mg}}{\text M + 7} = \text g\tan 30^\circ = \dfrac{\text g}{\sqrt{3}}$

Cancelling g and cross-multiplying,

$\sqrt{3}\text M = \text M + 7 \\[1em] \text M(\sqrt{3} - 1) = 7 \\[1em] \text M = \dfrac{7}{\sqrt{3} - 1}$

Rationalising the denominator,

$\text M = \dfrac{7}{\sqrt{3} - 1} \times \dfrac{\sqrt{3} + 1}{\sqrt{3} + 1} = \dfrac{7(\sqrt{3} + 1)}{3 - 1} \\[1em] = \dfrac{7(\sqrt{3} + 1)}{2}$

Substituting $\sqrt{3} = 1.732$,

$\text M = \dfrac{7 \times 2.732}{2} = \dfrac{19.124}{2} \\[1em] = 9.56\ \text{kg}$

Hence, the mass of the hanging block must be $\dfrac{7(\sqrt{3} + 1)}{2}$ kg, that is, nearly 9.56 kg.

A uniform sphere of weight W and radius 3 m is being held by a string of length 5 m attached to a frictionless wall as shown in the figure. Find tension in the string and normal reaction N of the wall in terms of weight W.

Answer

Given,

• Weight of the sphere = W
• Radius of the sphere, r = 3 m
• Length of the string, BO = 5 m
• The wall is frictionless

The tension in the string and the normal reaction of the wall have to be found in terms of W.

The forces acting on the sphere are its weight W, acting vertically downwards at the centre O, the tension T along the string, and the normal reaction N of the wall, acting horizontally. The wall is frictionless, so N has no vertical component.

The sphere is uniform and rests in equilibrium under these three forces, so their lines of action must be concurrent. Hence, the line of the string passes through the centre O of the sphere.

Let A be the point at which the sphere touches the wall and B the point on the wall at which the string is attached. Since the radius drawn to the point of contact is perpendicular to the wall, AO is horizontal and

$\text{AO} = \text r = 3\ \text m$

while AB lies along the wall and is therefore vertical. Thus, the triangle ABO is right-angled at A, and

$\text{AB} = \sqrt{(\text{BO})^2 - (\text{AO})^2} = \sqrt{(5)^2 - (3)^2} \\[1em] = \sqrt{25 - 9} = \sqrt{16} \\[1em] = 4\ \text m$

Let θ be the angle which the string makes with the horizontal at O. From the triangle ABO,

$\sin \theta = \dfrac{\text{AB}}{\text{BO}} = \dfrac{4}{5} \quad \text{and} \quad \cos \theta = \dfrac{\text{AO}}{\text{BO}} = \dfrac{3}{5}$

Resolving the tension into its vertical and horizontal components, for the equilibrium of the sphere,

$\text T\sin \theta = \text W \quad \text{...............(1)}$

$\text T\cos \theta = \text N \quad \text{...............(2)}$

From equation (1),

$\text T = \dfrac{\text W}{\sin \theta} = \dfrac{\text W}{4/5} \\[1em] = \dfrac{5\text W}{4} = 1.25\ \text W$

Substituting this value in equation (2),

$\text N = \text T\cos \theta = \dfrac{5\text W}{4} \times \dfrac{3}{5} \\[1em] = \dfrac{3\text W}{4} = 0.75\ \text W$

Hence, the tension in the string is $\dfrac{5\text W}{4}$ and the normal reaction of the wall is $\dfrac{3\text W}{4}$.

A block of mass 15 kg hangs from three light chords A, B and C and is in equilibrium. What are the tensions in the chord A and B? Give your answer in kg-wt.

Answer

Given,

• Mass of the block, m = 15 kg, so its weight W = 15 kg-wt
• Length of the chord A = 4 m and length of the chord B = 3 m
• The angle between the chords A and B at their meeting point is 90°

The tensions in the chords A and B have to be calculated.

The chord C carries the whole weight of the block, so the tension in it is 15 kg-wt, acting vertically downwards at the point where the three chords meet.

Let P and Q be the points at which the chords A and B are fixed to the horizontal ceiling, and let C be the point at which the three chords meet. Then

$\text{PC} = 4\ \text m, \quad \text{QC} = 3\ \text m \quad \text{and} \quad \angle \text{PCQ} = 90^\circ$

Since the triangle PQC is right-angled at C,

$\text{PQ} = \sqrt{(\text{PC})^2 + (\text{QC})^2} = \sqrt{(4)^2 + (3)^2} \\[1em] = \sqrt{16 + 9} = \sqrt{25} \\[1em] = 5\ \text m$

PQ lies along the ceiling and is therefore horizontal. Let θ = ∠QPC be the angle which the chord A makes with the horizontal, and α = ∠PQC the angle which the chord B makes with the horizontal. From the right-angled triangle PQC,

$\sin \theta = \dfrac{\text{QC}}{\text{PQ}} = \dfrac{3}{5} \quad \text{and} \quad \cos \theta = \dfrac{\text{PC}}{\text{PQ}} = \dfrac{4}{5}$

$\sin \alpha = \dfrac{\text{PC}}{\text{PQ}} = \dfrac{4}{5} \quad \text{and} \quad \cos \alpha = \dfrac{\text{QC}}{\text{PQ}} = \dfrac{3}{5}$

The point C is in equilibrium under the three tensions. Resolving the tensions T_A and T_B into horizontal and vertical components,

Horizontal components :

$\text T_A\cos \theta = \text T_B\cos \alpha \quad \text{...............(1)}$

Vertical components :

$\text T_A\sin \theta + \text T_B\sin \alpha = \text W \quad \text{...............(2)}$

Substituting the values in equation (1),

$\text T_A \times \dfrac{4}{5} = \text T_B \times \dfrac{3}{5} \\[1em] 4\text T_A = 3\text T_B \quad \Rightarrow \quad \text T_B = \dfrac{4}{3}\text T_A$

Substituting this in equation (2),

$\text T_A \times \dfrac{3}{5} + \dfrac{4}{3}\text T_A \times \dfrac{4}{5} = 15 \\[1em] \dfrac{3\text T_A}{5} + \dfrac{16\text T_A}{15} = 15 \\[1em] \dfrac{9\text T_A + 16\text T_A}{15} = 15 \\[1em] \dfrac{25\text T_A}{15} = 15 \\[1em] \text T_A = \dfrac{15 \times 15}{25} \\[1em] = 9\ \text{kg-wt}$

Therefore,

$\text T_B = \dfrac{4}{3} \times 9 \\[1em] = 12\ \text{kg-wt}$

Hence, the tension in the chord A is 9 kg-wt and that in the chord B is 12 kg-wt.

The result may be checked by noting that the chords A and B are at right angles to each other, so the resultant of the two tensions is

$\sqrt{(\text T_A)^2 + (\text T_B)^2} = \sqrt{(9)^2 + (12)^2} = \sqrt{81 + 144} \\[1em] = \sqrt{225} = 15\ \text{kg-wt}$

which is exactly the weight of the block, as required for equilibrium.