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Solved Examples    

1. A heavy particle of mass 0.50 kg is hanging from a string fixed with the roof. Find the force exerted by the string on the particle (as shown in figure). Take g = 9.8 m/s².

Sol. The force acting on the particle are

(a) pull of the earth, 0.50 kg × 9.8 m/s² = 4.9N, vertically downward .

(b) pull of the string , T vertically upward.

The particle is at rest with respect to the earth (which we assume to be an inertial frame). Hence , the sum of the forces should be zero. Therefore , T is 4.9N acting vertically upward.

2. A block of mass M is pulled on a smooth horizontal table by a string making an angle with the horizontal as shown in figure. If the acceleration of the block is `a' find the force applied by the string and by the table on the block.

Sol. Let us consider the block as the system.

The forces on the block are

(a) pull of the earth , Mg vertically downward.

(b) contact force by the table `N' vertically upward,

(c) pull of the string `T' along the string.

The free body digrams for the block is shown in figure

The accleartion of the block is horizontal and towards the right.Take this direction as the X-axis and vertically upward direction as the Y-axis. We have

component of Mg along the X-axis = 0

component of N along the X-axis = 0

component of T along the X-axis = Tcosθ

Hence , the total force along the X-axis = Tcosθ

Using Newton's law, Tcosθ = Ma ....(i)

component of Mg along the Y-axis = - Mg

component of N along the Y-axis = N

component of T along the Y-axis = T sinθ

Total force along the Y-axis = N + T sinθ - Mg

Using Newton's law, N + T sin θ - Mg = 0 ....(ii)

From equation (i) , T = Macosθ. Putting this in equation (ii) N= Mg -Ma tan θ.

4. The block shown in figure has a mass M and descends with an acceleration a. The mass of the string below the point A is m. Find the tension in the string at the point A and the lower end.

Sol. Consider "the block + part of the string below A" as the system. Let the tention at A be T. The forces acting on this system are

(a) (M+ m)g, downward, by the earth

(b) T, upward, by the upper part of the string.

The first is gravitional and the second is electromagnetic.We do not have to write the force by the string on the block. This electromagnetic force is by one part of the system on the other part. Only the forces acting on the system by the objects other than the system are to be included.

The system is desending with an accleration a ,Taking the downward direction as the X-axis is

(M + m )g-T = (M + m)a

or, T = (M + m)(g -a) .....(i)

We have omitted the free body digram. This you can do if you can draw the free body digram in your mind and write the equations correctly.

To get the tension T at the lower end we can put

m = 0 in (i)

Effectively , we take the point A at the lower end. Thus

we get T' = M (g -a).

5. A pendulum is hanging from the celling of a car having an acceleration a₀ with respect to the road. Find the angle made by the string with the vertical.

Sol. The situation is shwn in figure a. Suppose the mass of the bob is m and the string makes an angle θ with the vertical. We shall work from the car frame. This frame is noninertial as it has an acceleration  a⇀₀with respect to an inertial frame (the road). Hence , if we use Newton's second law we shall have to include a pseudo force.

Take the bob as the system.

The forces are :

(a) T along the string , by the string.

(b) mg downwards , by the earth

(c) ma₀ towards left (pseudo force)

The free body digram is shown in figure b. As the bob is at rest (remember we are discussing the motion with respect to the car) the forces in (a), (b), (c) should add to zero. Take X-axis along the forward horizontal direction and Y-axis along the upward vertical direction.

The components of the forces along the X-axis give

Tsinθ- ma₀ or , Tsinθ = ma₀ .......(i)

and the components Y-axis give

Tcosθ-mg = 0 or, Tcosθ = mg .......(ii)

Dividing (i) by (ii) tanθ = a₀ / g.

Thus the string makes an angle tan^_1(a₀ / g) with the vertical.

Questions for Short answer

1. The apparent weight of an object incresases in an elevator while accelerating upward . A moongphaliwala sells his moongphali using a beam balance in an elevator. Will he gain more if the elevator is acclerating up ?

2. A boy puts a heavy box of mass M on his head and jumps down from the top of a multistoried building to the ground. How much is the force exerted by the box on his head during his free fall ? Does the force greately increase during the period he balances himself after striking the ground ?

3. A person drops a coin. Describe the path of the coin as seen by the person if he is in

(a) car moving at constant velocity and

(b) in a freely falling elevator.

4. Is it possible for a particle to describe a curved path if no force acts on it ? Does your answer depend on the frame of reference choosen to view the particle?

5. You are riding in a car. The driver suddenly applies the brakes and you are pushed forward. Who pushed you forward?

6. It is sometimes heard that inertial frame of reference is only an ideal concept and no such inertial frame actually exists. Comment

7. An object is placed far away from all the objects that can exert force on it. A frame of reference is constructed by taking the origin and axes fixed in this object. Will the frame be necessarily inertial?

8. Figure shows a light spring balance connected to two blocks of mass 20 kg each. The graduations in the balance measure the tension in the spring .

(a) What is the reading of the balance ?

(b) Will the reading change if the balance is heavy , say 2.0 kg.?

(c) What will happen if the spring is light but the blocks have unequal masses ?

9. The acceleration of a particle is zero as measured from an inertial frame of reference. Can we conclude that no force acts on the particle ?

10. Suppose you are running fast in a field when you suddenly find a snake in front of you. You stop quickly. Which force is responsible for your deceleration ?

11. If you jump barefooted on a hard surface , your legs get injured . But they are not injured if you jump on a soft surface like sand or pillow. Explain.

12. According to Newton's third law each team pulls the opposite team with equal force in a tug of war. Why then one team wins and the other looses ?

13. A spy jumps from an airplane with his parachute. The spy accelerates downward for some time when the parachute opens. The acceleration is suddenly checked and the spy slowly falls on the ground . Explain the action of parachute in checking the acceleration.

14. Consider a book lying on a table. The weight of the book and the normal force by the table on the book are equal in magnitude and opposite in direction. Is this an example of Newton's third law?

15. Two blocks of unequal masses are tied by a spring . The blocks are pulled streching the spring slightly and the system is released on a frictionless horizontal platform. Are the forces due to the spring on the two blocks equal and opposite ? If yes, is it an example of Newton's third law?

16. When a train starts , the head of a standing passenger seems to be pushed backward. Analysis the situation from the ground frame. Does it reallygo backward? Comming back to the train frame, how do you explain the backward movement of the head on the basis of Newton's laws?

17. A plumb bob is hung from the ceiling of a train compartment. If the train moves with an accleration `a' along a straight horizontal track, the string supporting the bob makes an angle angle tan^_1 a/g  with the normal to the ceiling. Suppose the train moves on an inclined straight track with uniform velocity. If the angle of incline is , tan^_1 a/g  the string again makes the same angle with the normal to the ceiling. Can a person sitting inside the compartment tell by looking at the plumb line whether the train is acclerated on a horizontal straight track or it is going on an incline ? If yes, how? if no, suggest a method to do so.

Objective - I

1. A body of weight w₁ is suspended from the ceiling of a room through a chain of weight w₂. The ceiling pulls the chain by a force

(a) w₁ (b) w₂ (c*) w₁ + w₂ (d) w₁+w₂ /2

2. When a horse pulls a cart , the force to move forward is the force exerted by

(a) the cart on the horse (b*) the ground on the horse

(c) the ground on the cart (d) the horse on the ground

Sol. Horce pushes the earth. Earth acts reaction force on the horse.

3. A car accelerates on a horizontal road due to the forse exerted by

(a) the engin of the car (b) the driver of the car (c) the earth (d*) the road

4. A block of mass 10 kg is suspended through two loght spring balance as shown in figure (5-Q2)

(a*) Both the sales will read 10 kg.

(b) Both the sales will read 5 kg.

(c) The upper sale will read 10 kg and the lower zero.

(d) The readings may be anything but their sum will be 10 kg.

5. A block of mass m is placed on a smooth inclined plane of inclination θ with the horizontal. The force exerted by the plane on the block has a magnitude

(a) mg (b) mg/cosθ (c*) mgcosθ (d) mgtanθ

6. A block of mass m is placed on a smooth wedge of inclinationθ . The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block has a magnitude.

(a) mg (b*) mg/cosθ (c) mgcosθ (d) mgtanθ

7. Neglect the effect of rotation of the earth. Suppose the earth suddenly stops attracting objects placed near its surface. A person standing on the surface of the earth will

(a) fly up (b) slip along the surface

(c) fly along a tangent to the earth's surface (d*) remain standing

8. Three rigid rods are joined to form an equilateral triangle ABC of side 1m. Three particles carrying charges 20 μC each are attached to the vertices of the triangle. The whole system is at rest always in an inertial frame.The resultant force on the charged particle at A has the magnitude.

(A*) zero (B) 3.6 N (C) 3.63 N (D) 7.2 N

Sol. F=ma

a = acceleration of charge of particle at A = 0

 F_net = 0.

Since whole system is at rst then A is also at rest so resultant force on charge A is zero.

9. A force F₁ acts on a particle so as to accelerate it from rest to a velocity v. The force F₁ is then replaced by F₂ is then replaced by F₂ which decelerates it to rest.

(a) F₁ must be the equal to F₂ (b*) F₁ may be equal to F₂

(c) F₁ must be unequal to F₂ (d) None of these

10. Two objects A and B are thrown upward simultaneously with the same speed. The mass of A is greater than the mass of B .Suppose the air exerts a constant and equal force of resistance on the two bodies.

(a) The two bodies will reach the same height.

(b*) A will go higher than B.

(c) B will go higher than A.

(d) Any of the above three may happen depending on the speed with which the objects are thrown.

11. A smooth wedge A is fitted in a chamber hanging from a fixed ceiling near the earth's surface. A block B placed at the top of the wedge takes a time T to slidedown the length. If the block is placed at the top of the wedge and the cables supporting the chamber start accelerating it upward with an acceleration of `g', at the same instant, the block will. [HCV_Ch_5-obj-I_11]

(A) take a time loger than T to slide down the wedge

(B*) take a time shorter than T to slide down the wedge

(C) remain at the top of the wedge

(D) jump off the wedge

Sol.

When chamber starts moving up by acceleration `g', pseudo force mg acts downward on block.

Driving force is increased from mg sin q to 2 mg sin q hence acceleration is increased.

12. In an imaginary atmosphere, the air exerts a small force F on any particle in the direction of the particle's motion. A particle of mass m projected upward takes a time t₁ in reaching the maximum height and t₂ in the return journey to the original point. Then

(a) t₁ < t₂ (b*) t₁ > t₂  (c) t₁ = t₂ (d) the relation between t₁ and t₂ depends on the mass of the particle.

13. A person standing on the floor of an elevator drops a coin. The coin reaches the floor of the elevator is stationary and in time t₂ if it is moving uniformly. Then

(a*) t₁= t₂ (b)t₁ > t₂ (c) t₁ > t₂ (d) t₁ < t₂ or t₁ > t₂ depending

14. A free ²³⁸U nucleus kept in a train emits an alpha particle. When the train is stationary, a nucleus decays and a particle and the recoiling nucleus becomes x at time after the decay. If the decay takes place while the train is moving at a uniform velocity v, the distance between the alpha particle and the recoiling nucleus at a time t after the decay as ,measured by the passenger is

Objective - II

1. A reference frame attached to the earth

(a) is an inertial frame by definition

(b*) cannot be an inertial frame because the earth is revolving around the sun.

(c) is an inertial frame because Newton's laws are applicable in this frame.

(d*) cannot be an inertial frame because the earth is rotating about its angle.

2. A particle stays at rest as seen in a frame. We can conclude that

(a) the frame is inertial.

(b) resultant force on the particle is zero.

(c*) the frame may be inertial but the resultant force on the particle is zero.

(d*) the frame may be noninertial but there is a nonzero resultant force.

3. A particle is found to be at rest when seen from a frame S₁ and moving with a constant velocity when seen from another frame S₂. Markout the possible options.

(a*) Both the frames are inertial (b*) Both the frames are noninertial.

(c) S₁ is inertial and S₂ is noninertial. (d) S₁ is noninertial and S₂ is inertial.

4. Figure (5-Q3) shows the displacement of a particle going along the X-axis as a function of time. The force acting on the particle is zero in the region.

(a*) AB (b) BC (c*) CD (d) DE

5. Figure shows a heavy block kept on a frictionless surface and being pulled by two ropes of equal mass m. At t = 0, the force on the left rope is withdrawn but the force on the right and continues to act. Let F₁ and F₂ be the magnitudes of the forces by the right rope and the left rope on the block respectively.

(a*) F₁ = F₂ = F for t < 0 (b) F₁ = F₂ = F + mg for t < 0

(c) F₁ = F, F₂ = F for t > 0 (d) F₁ < F, F₂ = F for t > 0

6. The force exerted by the floor of an elevator on the foot of a person standing there is more than the weight of the person if the elevator is

(A*) going up and speeding up (B) going up and speeding down

(C) going down and speeding up (D*) going down and speeding down

7. If the tension in the cable supporting an elevator is equal to the weight of elevator , the elevator may be

(a) going up with increasing speed. (b) going down with incresing speed.

(c*) going up with uniform speed. (d*) going down with uniform speed.

8. A particle is observed from frames two S₁ and S₂ . The frame S2 moves with respect to S₁ with an acceleration a.Let F₁ and F₂ be the pseudo forces on the particle when seen from S₁ and S₂ respectively. Which of the following are not possible ?

(a) F₁ = 0, F₂ ≠  0 (b) F₁≠0, F₂ = 0 (c) F₁≠ 0, F₂≠0 (d*) F₁ = 0, F₂ = 0

9. A person says that he measured the acceleration of a particle to be nonzero while no force was acting on the particle.

(a) He is a liar.

(b) His clock might have been longer than the standrad.

(c) His meter scale might have been longer than the standrad.

(d*) He might have used noninertial frame.

                                                 Worked Out Examples

1. A body of mass m is suspended by two strings making angles α and β with the horizontal. Find the tensions in the strings.

Sol. Take the body of mass m as the system. The forces acting on the system are

(i) mg downwards (by the earth)

(ii) T₁ along the first string (by the first string) and

(iii) T₂ along the second string (by the second string)

These forces are shown in figure. As the body is in equilibrium , these forces must add to zero. Taking horizontal components,

T₁cosα-T₂cosβ + mgcosπ/2 = 0

or, T₁cosα = T₂cosβ .......(i)

Taking vertical components,

T₁sinα + T₂sinβ- mg = 0 .......(ii)

Eliminating T₂ from (i) and (ii),

T₁sinα + T₁  cosα/cosβ ×sinβ = mg

or, T₁ =

   T₂ =

From (i), T₂ =

2. Two bodies of masses m₁ and m₂ are connected by a light string going over a smooth light pulley at the end of an incline. The mass m₁ lies on the incline and m₂ hangs vertically. The system is at rest. Find the angle of the incline and the force exerted by the incline on the body of mass m₁ .(figure)

Sol. Figure shows the situation with the forces on m₁ and m₂ shown. Take the body of mass m₂ as the system . The forces acting on it are

(i) m₂g vertically downward (by the earth)

(ii) T vertically upward (by the string)

As the system is at rest , these forces should add to zero.

This gives T = m₂g .....(i)
Next, consider the body of mass m₁ as the system. The forces acting on this system are

(i) m₁g vertically downward (by the earth)

(ii)T along the string up the incline (by the string),

(iii) N normal to the incline (by the incline).

As the string and the pulley are all light and smooth, the tension in the string is uniform everywhere. Hence same T is used for the equations of m₁ and m₂. As the system is in equilibrium , these forces should add to zero.

Taking components parallel to the incline,

T = m₁gcos(π/2-θ)= m₁gsinθ .......(ii)

Taking components along the normal to the incline,

N = m₁gcosθ .......(iii)

Eliminating T from (i) and (ii) ,

m₂g = m₁gsinθ

or, sinθ= m₂ / m₁

giving θ = sin^_1(m₂ / m₁)

From (iii) N = m₁g{1-(m1/m2)}²

3. A bullet moving at 250 m/s penetrates 5 cm into a tree limb before coming to rest. Assuming that the force

exerted by the tree limb is uniform, find its magnitude . Mass of the bullet is 10g.

Sol. The tree limb exerts a force on the bullet in the direction opposite to its velocity. This force causes decleration and hence the velocity decreases from 250 m/s to zero in 5 cm. We have to find the force exerted by the tree limb on the bullet. If a be the decleration of the bullet, we have,

u = 250 m/s, v = 0, x= 5 cm = 0.05m

giving, a = = 625000 m/s².

The force on the bullet is F = ma = 6250 N.

4. The force on a particle of mass 10 g is (i10 + j5). If it starts from rest , what would be its position at time t = 5s?

Sol. We have F_x = 10N giving

a_x = F_x/m = 10N/0.01KG = 1000 m/s².

As this is a case of constant acceleration in x_direction,

x = u_xt + 1/2a_x t² = 1/2 × 1000 m/s² × (5s)²= 12500 m.

Similarly , a_y = F_Y/m = 5N/0.01kg = 500 m/s² and y = 6250 m

Thus the position of the particle at t = 5s is,

= (i12500 + j6250) m .

5. With what acceleration 'a' should the box of figure descend so that the block of mass M exerts a force Mg/4 on the floor of the box?

Sol. The block is at rest with respect to the box which is accelerated with respect to the ground. Hence the acceleration of the block with respect to the ground is `a' downward. The forces on the block are

(i) Mg downward (by the earth) and

(ii) N upward (by the floor)

The equation of motion of the block is , therefore

Mg - N = Ma.

If N = Mg / 4 , the above equations gives a = 3g / 4. The block and hence the box should descend with an acceleration

3g / 4 .

6. A block `A' of mass is tied to a fixed point C on a horizontal table through a string passing round a massless smooth pulley B (figure). A force F is applied by the experimenter to the pulley. Show that if the pulley is displaced by a distance x and the block will be displaced by 2x. Find the acceleration of the block and the pulley.

Sol. Suppose the pulley is displayed to B and the block to A' (Figure). The length of the string is CB + BA and is also equal to CB + BB' + B'B + BA'. Hence CB + BA' + AA' = CB + BB' + B'B + BA' or, A'A = 2BB'.

The displacement of A is , therefore , twice the displacement of B in any given time interval. Differentiating twice, we find that the acceleration of A is twice the accelaration of B.

To find the acceleration of the block we will need the tension in the string. That can be obtained by considering the pulley as the system.

The forces acting on the pulley are

(i) F towards right by the experimenter ,

(ii) T towards left by the portion BC of the string and

(iii) T towards left by the portion BA of the string .

The vertical forces , if any add to zero as there is no vertical motion.

As the mass of the pulley is zero, the equation of motion is

F -2T = 0 giving T = F / 2

Now consider the block as the system. The only horizontal force acting on the block is the tension T towards right. The acceleration of the block is , therefore a = T / M = F/2m. The acceleration of the pulley is a / 2 = F/4m.

7. A smooth ring A of mass m can slide on a fixed horizontal rod.A string tied to the ring passes over a fixed pulley B and carries a block C of mass M (= 2 m) as shown in figure. At an instant the string between the ring and the pulley makes an angle θ with the rod. (a) Show that, if the ring slides with a speed v, the block descends with speed vcosθ.(b) With what acceleration will the ring start moving if the system is relesed from rest with θ = 30º ?

Sol. (a) Suppose in a small time interval Dt the ring is displaced from A to A' (figure) and the block from C to C'. Drop a perpendicular A'P from A' to AB. For small displacement A'BPB. Since the length of the string is constant, we have

AB + BC = A'B + BC'

or, AP + PB + BC = A'B + BC'

or, AP = BC' -BC = CC' (as A'B ≈ PB)

or, AA' cosθ = CC'

or, =

or, (velocity of the ring) cosθ = (velocity of the block).

(b) If the initial accelaration of the ring is a, that of the block will be a cosθ. Let T be the tension in the string at this instant. Consider the block as the system. The forces acting on the block are

(i) Mg downwards due to the earth, and

(ii) T upwards due to the string.

The equation of motion of the block is

Mg -T = Macosθ .....(i)

Now consider the ring as the system. The forces on the ring are

(i) Mg downwards due to gravity

(ii)N upward due to the rod.

(iii) T along the string due to the string.

Taking components along the rod, the equation of motion of the ring is

T cosθ = ma ....(ii)

From (i) and (ii),

Mg -ma/cosθ = M a cosθ 

 a= 

Putting q = 30°, M = 2 m and g = 9.8 m/s² ; a = 6.78 m/s².

8. A light rope fixed at one end to a wooden clamp on the ground passes over a tree branch and hangs on the other side. It makes an angle of 30° with the ground. A man weighing (60 kg) wants to climb up the rope. The wooden clamp can come out of the ground if on upward force greater than 360N is applied to it. Find the maximum acceleration in the upward direction with which the man can climb safely. Neglect friction at the tree branch. Take g = 10 m/s².

Sol. Let T be the tenstion in the rope. The upward force on the clamp is T sin 30° = T/2. The maximum tension that will not detach the clamp from the ground is, therefore, given by

T/2 = 360 N   or T = 720 N

If the acceleration of the man in the upward direction is a, the equation of motion of the man is

T -600 N = (60 kg) a

the maximum acceleration of the man for safe climbing is, therefore,

a =  (720N-600N)/60Kg= 2 m/s².

9. Three blocks of masses m₁, m₂ and m₃ are connected as shown in the figure. All the surfaces are frictionless and the string and the pulleys are light. Find the acceleration of m₁.

Sol. Suppose the accelerationof m₁ is a₀ towards right. That will also be the downward acceleration of the pulley B because the string connecting m₁ and B is constant in length. Also the string connecting m₂ and m₃ has a constant length. This implies that the decreases in the separation between m₂ and B equals the increase in the separation between m₃ and B. So, the upward acceleration of m₂ with respect to B equals to the downward acceleation of m₃ with respect to B. Let this acceleration be a.

The acceleration of m₂ with respect to the ground = a₀ -a (downward) and the acceleration of m₃ with respect to the ground = a₀ + a (downward).

These accleration will be used in Netwon's laws. Let the tension be T in the upper string and T' in the the lower string. Consider the motion of the pulley B.

The forces on this light puelly are

(a) T upwards by the upper string and

(b) 2T' downwards by the lower string.

As the mass of the pulley is negligible,

2T' = T = 0

given T' = T/2 ................ (i)

Motion of m₁ :

The acceleration is a₀ in the horizontal direction. The force on m₁ are

(a) T by the string (horizontal)

(b) m₁g by the earth (vertically downwards) and

(c) N by the table (vertically upwards)

In the horizontal direction, the equation is

T = m₁a₀ ................. (ii)

Motion of m₂ : acceleration is a₀ -a

(a) m₂g downward by the earth and

(b) T' = T/2 upward by the string

Thus, m₂g -T/2 = m₂ (a₀ -a) .................. (iii)

Motion of m₃ : The accelaration is (a₀ + a) downward. The forces on m₃ are

(a) m₃g downward by the earth

(b) T' = T / 2upward by the string. Thus,

m₃ g -T / 2= m₃ (a₀ + a) .....(iv)

We want to calculate a₀, so we shall eliminate T and from (ii) , (iii) and (iv)

Putting T from (ii) and (iii) and (iv),

a₀ -a = = g -

and a₀ + a = = g -

Adding 2a₀ = 2g _

or, a₀ = g -

or, a₀ = g

or, a₀ =

10. A particle slides down a smooth inclined plane of elevation θ fixed in an elevator going up with an acceleration a₀(figure). The base of the incline has a length L. Find the time taken by the particle to reach the bottom.

Sol. Let us work in the elevator frame. Apseudo force ma₀ in the frame . A pseudo force ma₀ in the downward direction is to be applied on the particle of mass m together with the real forces.

Thus , the forces on m are (figure)

(i) N normal force,

(ii) mg downward (by the earth),

(iii) ma₀ downward (pseudo)

Let a be the acclearation of the particles with respect to the incline. Taking components of the forces parallel to the incline and applying Newton's law,

mgsinθ + ma₀sinθ = ma

or, a = (g + a₀)sinθ

This is the acceleration with respect to the elevator. In this frame , the distance travelled by the particle is L / cos. Hence,

L /cosθ = 1/2(g + a₀) sinθ.t²

or, t =

11. All the surface shown in figure are assumed to be frictionless. The block of mass m slides on the prism which in turn slides backward on the horizontal surface. Find the acceleration of the smaller block with respect to the prism.

Sol. Let the acceleration of the prism be a₀ in the backward direction. Consider the motion of the smller block from the frame of the prism.

The forces on the block are (figure 5-W15a)

(i) N normal force,

(ii) mg downward (gravity)

(iii) ma₀ forward (pseudo)

The block slides down the plane . Components of the forces parallel to the incline give

ma₀cosθ + mgsinθ = ma

or, a = a₀cosθ + gsinθ ...(i)

Components of the force perpendicular to the incline give

N + ma₀sinθ = mgcosθ ...(ii)

Now consider the motion of the prism from the lab frame. No pseudo force is needed as the frame used in interial. The forces are (Figure 5-w15b)

(i) Mg downward,

(ii) N normal to the incline (by the block),

(iii) N' upward (by the horizontal surface)

Horizontal components give

N sinθ = Ma₀ or N = Ma₀ / sinθ .....(iii)

Putting in (ii)

 ma₀/sinθ + ma₀sinθ = mgcosθ

or, a₀ =

From (i) , a = + gsinθ =

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 10m in the first two seconds. Find the magnitude of F.

[Ans : 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 weights 2000 kg, what average force must be applied on it ? [Ans : 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 × 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 × 10^_31 kg.

[Ans : 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 anotherstring. Find the tensions in the two string joining the blocks.

(g = 10 eh-@ls-²)

[Ans : 5 N & 3N]

5. Two blocks of equal mass m are tied to each other through a line 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.

Ans : F / 2

6. A particle of mass 50 g moves on a straight line. The variation of speed with time is shown in figure. Find the force acting on the particle at t = 2, 4 and 6 seconds.

[Ans : 0.25 N along the motion , zero and 0.25 N opoosite to the motion]

7. Two blocks A and B of mass m_A and m_B 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 exert a force F on the block B, what is the force exerted by the experimenter on A ?

Ans : F{1+m_1/ m₂}

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.

Ans : 1.8 N

9. A particle of mass 0.3 kg is subjected to a force F = -kx with k = 150 N.m. What will be its intial acceleration if it is released from a point x = 20 cm ?

Ans : 10 m / s² ]

10. Both the springs shown in figure are unstretched. If the block is displaced by a distance x and released, what will be intial acceleration.

[Ans : (k₁ + k₂)x/m opposite to the displacement ]

11. A small block B is placed on another block A of mass 5 kg and length 20 cm. Intially the block B is near the right end of block A (figure). 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 . HCV_Ch-5_Ex._11

Ans. 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. 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.

Ans :

13. The elevator shown in figure 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 ? Solve the problem taking (a) ground as the frame (b) lift as the frame. [HCV_Q. 13_chapter_5]

[4]

Sol.

m_A = 0.5 kg

(a) in ground frame

mg -N = m × 2

∴ N = 0.5 (10 -2)

N = 4 Newton

(b) in lift frame

net force zero

 N + mg = ma

∴ N = 4 newtons

[Ans : 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.

[Ans : (a) 0.55 N (b)0.43 N (c) 0.49 N (d) 0.43 N (e) 0.55 N (f) 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.

[Ans : 66 kg and 0.9 m/s|]

16. Find the reading of the spring balance as 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.

[Ans : 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 ?

Ans : 0.2 m and 0.1 m

18. Suppose the celing in the previous problem is that of an elevator which is going up with an acceleration of 2.0 m/s².Find the elongations.

Ans : 0.24 m, 0.12 m

19. The force of buoyancy exerted by the atmosphere on a ballon is B in the upward direction and remains constant.The force of air resistance on the ballon acts opposite to the direction of velocity and is propotional to it. The ballon 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 ballon so that it may rise with a constant velocity v ?

Ans : 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 placed inside the box so that it may accelerate down at the rate of g / 6.

Ans : 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.

Ans : 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 ) m₁ = 300 g and m₂ = 600g. 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. [HCV_Ex.22]

Ans : (a) 6.5 m (b) 3.9 N (c) 7.8 N]

23. Consider the Atwood machine of the previous problem. The larger mass is stopped for a moment at 2.0 s after the system is set into motion. Find the time elapsed before the string is tight again.

Ans : 2 / 3 s ]

24. Figure 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. HCV_Ch-5_Ex._24

[Ans : 24 N ]

25. Consider the situation shown in figure. All the surface are frictionless and the string and the pulley are light. Find the magnitude of the acceleration of the two blocks.

[Ans : g / 10 ]

26. A constant force F = m₂g / 2 is applied on the block of mass m₁ as shown in figure. The string and the pulley are light and the surface of the table is smooth. Find the acceleration of m₁.

Sol. a = net driving force /Total mass

a =  a = toward right. Ans

27. In figure m₁ = 5 kg, m₂ = 2 kg and F = 1N. Find the acceleration of either block. Describe the motion of m₁, if the string breaks but F continues to act.

[Ans : 4.3 m/s², moves downward with acceleration g + 0.2 m/s² ]

Sol. a = = 3/7 × 10 = 4.3 m/s²

when string breakes

a =  ( m₁g + F/m₁ )= g + 0.2 = 10.2 m/s² downward.

28. Let m₁ = 1 kg, m₂ = 2 kg and m₃ = 3 kg in figure shown. Find the acceleration of m₁, m₂ and m₃.The string from the upper pulley to m₁ is 20 cm when the system is released from rest. How long will it take before m₁ strikes the pulley ? 

[Ans : 19/29g (up),11/29g (down),  21/29 g (down), 0.25 s ]

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 ?

Ans : 4.8 kg

30. Calculate the tension in the string shown in figure.The pulley and the string are light and all the surfaces are frictionless. Take g = 10 m/s².

[Ans : 5N ]

31. Calculate the situation shown in figure. 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.

[Ans : (a) 2g /3 (b) Mg / 3 (c) 2^1/2 Mg/ 3 at an angle of 45º with the horizontal ]

32. Find the acceleration of the block of mass M in the situation shown in figure. All the surfaces are frictionless and the pulleys and the string are light. HCV_Ch-5_Ex._32

[Ans : g /3 up the plane ]

33. Find the mass M of the hanging block in figure which will prevent the smaller block from slipping over the triangular block. All the surface are frictionless and the string and the pulleys are light.

HCV_Ch-5_Ex._33

Ans :M'+m/cotθ-1

34. Find the acceleration of the blocks A and B in the three situations shown in figure.

[Ans : (a) 2/7g downward , g/7 upward

(b) 10/13 g forward , 5/13g downward

(c) 2/3g downward, g/3 upward ]

Sol. (a) For A :

4g -T = 4a ..............(i)

For B :

2T -5g = 5(a/2) .............(ii)

Solving (i) & (ii)

Acceleration of A = a = 2/7g downward

Acceleration of B = a/2 = 1/7g upwards.

(b) For A :

T = 2a ................(i)

For B :

5g -2T = 5(a/2) .............(ii)

Solving (i) & (ii)

Acceleration of A = a = (10/13)g Rightwards

Acceleration of B = a/2 = 5/13g downward.

(c) For A :

2g -T = 2a .............(i)

For B :

2T -1 g = (a/2) ...........(ii)

Solving (1) & (2)

Acceleration of A Þ a = (2/3)g  (downward)

Acceleration of B = g/3­ (upward) .

35. Find the acceleartion of the 500 g block in figures.

[Ans : (8/13)g 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 5m long and the monkey starts from rest, how much time will it take to reach the ceiling ?

[Ans : 165 N, 10^1/2s ]

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). Show that whatever force the monkey exerts on the rope, the monkey and the block move in the same direction with equal acceleration. If intially both were at rest, their separation will not change as time passes.

38. The monkey B shown in figure is holding on to the tail of the monkey A which is climbing on 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 its tail, what force should it apply on the rope in order to carry the monkey B with it ? Take g = 10 m/s².

[HCV_Ch-5_Ex._38]

[Ans : Between 70 N and 105 N]

39. Figure 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 ar rest, what is the weightshown by the machine ? What force should be exert on the rope to get his correct weight on the machine ?

[Ans : 15 kg , 1800 N]

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). 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.

Ans : (2L/gsinθ)^1/2

41. A car is speeding up on a horizontal roadwith a constant acceleration a. Calculate in the following situations in the car. (i) A ball is suspended from the ceiling. Find the angle made by the string if the block & string remain static w.r. to car. (ii) A block is kept on a smooth fixed incline and does not slip on the fixed incline with the horizontal. HCV_Ch-5_Ex_41

[Ans : tan^_1(a/g) in each case]

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 0.2s after the start. Take g = 10 m/s².

[Ans : 20 cm]