Newton’s Laws of Motion

Science originates by observing nature and making inferences from them followed by devising and doing experiments to verify or refute theories. The three laws of motion discovered by Newton govern the motion of every object in nature all the time but due to the presence of friction and air resistance, they are a little difficult to see.

Newton’s first law is stated as:“In the absence of external forces, an object at rest remains at rest and an object in motion continues in motion with a constant velocity (that is, with a constant speed in a straight line)”.Though this is not what we observe everyday. A ball rolling on the floor eventually stops, faster on a sandy floor as compared to a marble floor. This is due to the force of friction present between the ball and the floor. An opposing force in the direction opposite to that of its velocity slows the ball down and eventually brings it to rest. If the ball were rolling on a frictionless floor(ideal case), it would never stop in the absence of external forces.

Newton’s second law states:The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.

∑F = ma

This law is a little easier to observe as compared to the first law. You can throw a lighter shot put farther than a heavier one even if you put all your energy (or force) in both the cases. This happens because the lighter one gets more acceleration as compared to the heavier one and it is able to cover more distance before falling down. Though there are a lot of other factors like angle of throwing, air drag, etc. which govern the distance covered by the shot put before landing but assuming those factors to be equivalent in both the throws, this should give you some insight that Newton’s second law holds.

Newton’s third law states:If two objects interact, the force F12 exerted by object 1 on object 2 is equal in magnitude to and opposite in direction to the force F21 exerted by object 2 on object 1:

F12 = -F21

This law can be understood by considering the following example. When you hit a football with a barefoot, the foot hurts less if you hit it softly and it hurts more if you hit it with greater strength. Thus, the football exerts as much force on your foot as you hit it with. Now, there can be a confusion regarding this rule when you think of all the bodies that the earth is attracting with its force of gravitation. What about the force they exert back on the earth? This is very true, anything that the earth attracts towards itself also attracts the earth towards themselves with a force equal to the magnitude of g (acceleration due to gravity), but the mass of earth is so large as compared to the magnitude of force applied that it effectively remains at rest whereas the body accelerates towards it and falls on the surface of the earth.

Free body diagramsWhile solving any problem on Newton’s laws of motion, we make use of free body diagrams. In these diagrams we represent all the external forces acting on the object and then apply newton’s second law to find its acceleration and other parameters. If the system to be analyzed involves more than one object, then their free body diagrams are drawn separately and then solved.

While drawing the free body diagram, you must include all the external forces on an object including any reactions which are not mentioned in the problem but are to be deduced using Newton’s third law of motion.

Friction

The force of friction is something you encounter all the time, so it’s the easiest to understand. The force of friction is very interesting in that its magnitude changes upto a maximum value depending on the external force applied on the object. Suppose you try to push a heavy box in order to slide it to another location. You start pushing it with a little force initially and you keep increasing the force until it starts to slide. Below that value of force, the box remains at rest whatever be the force applied on it by you. The free body diagram of the box looks likeFor the box to remain in equilibrium, the force of friction must always be equal to the force applied by you. Hence, it concludes that the force of friction (till the box doesn’t move) equals the applied force until it reaches a maximum after which it remains constant.

Thus, frictional force opposes (impending or actual) relative motion between two surfaces in contact. Static friction fs opposes impending relative motion; kinetic friction fk opposes actual relative motion. They are independent of the area of contact and satisfy the following approximate laws :

fs ≤ (fs)max = μsR

fk = μkRμs (co-efficient of static friction) and μk (co-efficient of kinetic friction) are constants characteristic of the pair of surfaces in contact. It is found experimentally that μk is slightly less than μs. It is found experimentally that the force of friction is independent of area of contact between the bodies as can be seen from the expressions for the force also.Static FrictionStatic friction is a self-adjusting force up to its limit μsN (fs ≤ μs N). You should not put fs= μsN without being sure that the maximum value of static friction is coming into play.Kinetic FrictionFrictional force that opposes relative motion between surfaces in contact is called kinetic or sliding friction.

Important Points:1) In all the problems on Newton’s Laws of motion, proceed by drawing the free body diagrams for each object in the system separately and then solving for the unknown.2) The physics (and probably the difficult part) in these problems is to recognize the constraints that bind the different parts of the system like the two objects have to move with the same acceleration or the object cannot lose contact with the surface of the incline, so the sum of forces on the object perpendicular to surface has to be zero. Rest is mathematics and comes easy after practicing a few problems. Once you master this ability, you can solve any problem on this topic.3) A string has same tension in it at all points. Each infinitesimal part of the string has the same tension trying to pull it apart in opposite directions.

Problems:

1) A mass of 5 kg is suspended by a rope of length 2 m from the ceiling. A force of 45 N in the horizontal direction is applied at the midpoint R of the rope, as shown. What is the angle the rope makes with the vertical in equilibrium? (Take g = 10 ms-2). Neglect the mass of the rope.

Solution: We begin by drawing the free body diagram of the mass to find T2.As the mass is in equilibrium, the sum of all the external forces on it should be zero.Therefore, T2 = 50 N

Next, we draw the free body diagram of the point R.Since, point P is also in equilibrium, the sum of all the forces at this point must be zero.That is, T1Cosθ = 45 Nand       T1Sinθ = 50 N   Therefore, tanθ = (50/45), θ = 48o

2) A mass of 3 kg rests on a horizontal plane. The plane is gradually inclined until at an angle θ = 20° with the horizontal, the mass just begins to slide. What is the coefficient of static friction between the block and the surface?

Solution: Again we begin by drawing a figure containing all the forces acting on the mass. Now, instead of drawing another free body diagram, we should be able to see it in this figure itself.An important thing to keep in mind here is that we have resolved the force of gravity into its components and we must not consider mg during calculations if we are taking its components into account.Now, as θ increases, the self-adjusting frictional force fs increases until at θ = θmax, fs achieves its maximum value, (fs)max  = μsN.Therefore, tanθmax = μs or θmax = tan–1μs

When θ becomes just a little more than θmax, there is a small net force on the block and it begins to slide.Hence, for θmax = 20°,μs = tan 20° = 0.36

3) A small block B is placed is placed on another block A of mass 7 kg and length 15 cm. Initially the block B is near the right end of block A. 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.

Solution: As there is no friction between A and B, when the block A moves, block B remains at rest in its position.  

Now, acceleration of block A = 10/7 = 1.4 ms-2

As the block starts from rest, initial velocity, u = 0We know that s = ut + ½ at2

0.15 = 0 + 0.7t2

which gives t = 0.46 s

4) A ball of mass 5 kg and a block of mass 12 kg are attached by a lightweight cord that passes over a frictionless pulley of negligible mass as shown in the figure. The block lies on a frictionless incline of angle 30o. Find the magnitude of the acceleration of the two objects and the tension in the cord. Take g = 10 ms-2.

Solution.As the objects are connected by a cord (which we assume does not stretch), their accelerations have the same magnitude. Let us draw the free body diagrams for the

two objects.

Now, we know that the block never loses contact with the surface of the incline. Hence, in the direction perpendicular to the surface the sum of forces on the block must be zero.Therefore,  normal force, n = 120Cos30 = 60√3 = 104 NAlso, 120Sin30 – T = ma = 12aTherefore, a = (60-T)/12

Now, for the ball also, T – 50 = 5a (as both the objects have same acceleration)Putting the value of a from the above equation, we getT- 50 = 5(60-T)/12i.e. 17T = 900i.e. T = 52.94 NSubstituting the value of T in T – 50 = 5a, we geta = 0.59 ms-2 ≈ 0.6 ms-2

Note that here we chose the x axis for the incline along the surface of the incline and the y axis perpendicular to the surface of the incline. Such choices for the axes should be made which make the problem easier to solve. It would have been a tedious problem if we had stick to the conventional choice for the axes.

5) Two blocks of masses 5 kg and 7 kg are placed in contact with each other on a frictionless horizontal surface. A constant horizontal force 20 N is applied to the block of mass 7 kg. (a) Determine the magnitude of the acceleration of the two-block system.

(b) Determine the magnitude of the contact force between the two blocks.Solution: We know that the blocks remain in contact from our everyday experience.Now, given that the force of 20 N is the only force acting on the system in the horizontal direction and we know that the blocks move together as one, we can assume the system to be a mass of 12 kg in order to solve part (a).Therefore, acceleration, a = 20/12 = 1.67 ms-2.

To solve part (b) we draw the free body diagrams of both the blocks.                                        

 Applying Newton’s second law to the block of mass 5 kg gives∑Fx = P = 5a = 5 X 1.67 = 8.33 N

Now, to verify the value of P obtained apply Newton’s second law to the block of mass 7 kg,∑Fx = 20 – P = 11.67 Nwhich gives a = 11.67/7 = 1.67 ms-2.Hence, verified.

6) A block of mass 5 kg starts to slide down a frictionless plane having an inclination of 25.0° from rest at the top. The length of the incline is 2.00 m, find (a) the acceleration of the block and (b) its speed when it reaches the bottom of the incline. Take g = 10 ms -2.

Solution: We begin by drawing the free body diagram of the block.

(a) Applying Newton’s second law of motion along the surface of the incline, we get50 sin25 = 5aTherefore, a = 4.23 ms-2

(b) We know that v2 = u2 + 2asSince u = 0 (given that the block starts from rest)Therefore, v2 = 2 X 4.23 X 2which gives v = 4.1 m/s

7) A 75.0 kg man stands on a platform scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.20 m/s in 1.00 s. It travels with this constant speed for the next 10.00 s. The elevator then undergoes a uniform acceleration in the negative y direction for 1.70 s and comes to rest. What does the scale register (a) before the elevator starts to move?

(b) during the first 1.00 s? (c) while the elevator is traveling at constant speed? (d) during the time it is slowing down? Take g = 10 ms-2.Solution: The  scale registers the force applied by it on the man i.e. the normal force. We proceed by drawing the free body diagram for the person.

(a) Before the elevator starts to move, the man is in equilibrium. Hence the sum of all forces on him must be zero.Therefore, N = 750 NHence, the scale registers 750 N before the elevator starts to move.

(b) The man accelerates with the elevator at the same acceleration.Applying Newton’s second law in the vertical direction, we get∑Fy = 750 – N = 75a = 75 X 1.2which gives N = 660 NHence, the spring scale registers 660 N for the first 1 s.

(c) While the elevator is travelling at constant speed, the acceleration of the man is zero. Therefore, the forces applied on him must add up to zero.which implies N = 750 N

(d) The maximum speed of the elevator, v = u + atu = 0 (as the elevator starts from rest)Therefore, v = 1.2 X 1 = 1.2 m/sIt then comes to rest in 1.7 s after decelerating uniformly.i.e. 0 = 1.2 + a’t = 1.2 + a’ X 1.7which gives a’ = -0.7 ms-2

Now, applying Newton’s second law in the vertical plane, we get,∑Fy = 750 – N = 75a = 75 X (-0.7)which gives N = 802.5 N

8) A block of mass 2.50 kg is accelerated across a rough surface by a rope passing over a pulley, as shown in figure. The tension in the rope is 12.0 N, and the pulley is 10.0 cm above the top of the block. The coefficient of kinetic friction is 0.300. (a) Determine the acceleration of the block when x = 0.400 m. (b) Find the value of x at which the acceleration becomes zero. Take g = 10 ms-2.

Solution: To determine the acceleration of the block, we need to know all the forces

acting on it.

At x = 0.4 m, cosθ = 40/41.23 = 0.97 Sinθ = 0.24Since, T = 12 N, Tcosθ = 11.64 NTsinθ = 2.9 NApplying Newton’s second law in the vertical direction, we get∑Fy = 25 – N – 2.9 = 0 (as the block does not lose contact with the floor and hence its acceleration is zero in this direction)which gives N = 23.9 N

Therefore, Force of friction, fk = μk X 23.9 = .3 X 23.9 = 7.17 NNow, applying Newton’s second law in the horizontal direction, we get∑Fx = 11.64 – 7.17 = ma = 2.5awhich gives a = 1.8 ms-2.

(b) The acceleration becomes zero when Tcosθ equals frictional forcei.e. Tcosθ = μk X (25 – Tsinθ)12cosθ = 0.3 X (25 – 12sinθ)12cosθ + 3.6sinθ = 7.5

Sinαcosθ + cosαsinθ = 7.5/12.53Where sinα = 12/12.53 and cosα = 3.6/12.53So, sin(α + θ) = 0.6i.e. α + θ = 36.7o

Also, sin(α + θ) = 10x2+102 = 0.6

Solving, we get x = 13.33 m

9) A block is given an initial velocity of 5.00 m/s up a frictionless 20.0° incline. How far up the incline does the block slide before coming to rest? Take g = 10 ms-2.

Solution: Initial velocity of the block, u = 5 m/s

The free body diagram of the block looks like Therefore, the force acting on the block in its direction of motion is mg sinθ = mg sin20Hence, a = -gsin20 taking positive x in the direction of motionWe know, that v2 = u2 + 2asTherefore, 0 = 25 + 2 X (-g sin20) X swhich gives s = 3.7 m Hence, the block comes to rest after travelling 3.7 m up the incline.

10) Two blocks of mass 5 kg and 9 kg are connected by a string of negligible mass that passes over a frictionless pulley. The inclines are frictionless. Find (a) the magnitude of the acceleration of each block and (b) the tension in the string. Take g = 10

ms-2.

Solution: Let the tension in the string be T and the acceleration of the blocks be ‘a’.Both the blocks being connected by the string move with same acceleration. The heavier block slides downwards ad the lighter block slides upwards the incline.Applying Newton’s second law to both the blocks,T – 50sin35 = 5a                                         (for 5 kg mass)90sin35 – T = 9a                                         (for 9 kg mass)Solving the two equations we get,a = 10sin35 = 5.7 NTherefore, T = 57 N

11) In figure blocks A and B have masses 40 kg and 20 kg respectively. (a) Determine the minimum weight of block C to keep A from sliding if the coefficient of friction, μs between A and the table is 0.18. Block C is suddenly lifted off A. (b) What is acceleration of block A if μk between A and the table is 0.15. Take g = 10 ms-2.

Solution: (a) Let the minimum weight of block C to keep A from sliding be m.If block A doesn’t slide, block B also remains in equilibrium, as they are connected by a string.Therefore, tension in the string, T = 200 N

Now, considering the free body diagram of system A and CApplying Newton’s second law in x direction, we havefs = T = μs X N i.e. 200 = 0.18 X (400 + mg)m = 71.1 kg

(b) If block is lifted off suddenly, the normal force becomes N = 400 NAnd as the block stars sliding, the kinetic friction, fk= μk X 400 = 0.15 X 400 = 60 NTherefore the acceleration of the block, a = (200-60)/40 = 3.5 ms-2.

12) Two masses 5 kg and 7 kg situated on a frictionless, horizontal surface are connected by a light string. A force of 50 N is exerted on one of the masses to the right. Determine the acceleration of the system and the tension T in the string.

Solution: The only external force acting on the system is 50 N in the x axis. As the blocks don’t lose contact with the floor the normal force is balanced by their weight for both the blocks.We could treat the system as one block of 12 kg as we can see that both the blocks have to have the same acceleration.Therefore, 50 = 12awhich gives a = 4.17 ms-2.

To find the tension T in the string, let us draw the free body diagrams for the two blocks

    Applying Newton’s second law on block of mass 5 kg, we getT = 5a = 5 X 4.17 = 20.8 NTo verify, apply Newton’s second law on block of mass 7 kg, which gives50 – T = 7 X 4.17T = 20.8 NHence verified.

13) A block of mass m1 = 5 kg on a frictionless horizontal table is connected to a block of mass m2 = 3 kg by means of a very light pulley P1 and a light fixed pulley P2 as shown in figure. If a1 and a2 are the accelerations of m1 and m2, respectively, (a) what is the relationship between these accelerations? Find (b) the tensions in the strings and (c) the accelerations a1 and a2. Take g = 10 ms-2.

                                                                                                       

Solution: Let the tensions in the strings of P1 and P2 be T1 and T2 respectively.For part (a), you need to visualize the problem carefully. Since the pulley P1 and mass m2 = 3 kg are connected by a string, their accelerations have to be the same as they have to move together. Now if the pulley goes forward a distance x, the mass m1 has to move forward by 2x in order for the string to not break, x for both the sides (upper and lower) of the string, which implies that the mass m1 will have to move with twice the acceleration as that of m2.Hence, a1 = 2a2.(b) Drawing the free body diagram of block m2, and applying Newton’s second law, we get

                                                                                                     

                                                                                                                                                          

30 – T2 = 3 X a2   …(1)Now, since the pulley P1 is massless, the sum of forces acting on it must be zero otherwise it will have infinite acceleration which cannot happen without destroying the system. Hence,2T1 = T2

Now for mass m1 we haveT1 = 5 X a1 = 10a2  or T2/2 = 10a2   …(2)Solving equations (1) and (2);a2 = 1.3 ms-2

Therefore, a1 = 2.6 ms-2

Using this value of a2, we getT1 = 13 Nand T2 = 26 N

14) A block A of mass 0.5 kg can slide on a frictionless incline of angle 30o and length 0.8 m kept inside an elevator going up with uniform velocity 2m/s. 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. Take g = 10 ms-2.

Solution: The driving force on the block which causes it to move down the plane is F = mgsin30 = 2.5 NSo, acceleration = 2.5/0.5 = 5 ms-2

Initial velocity of block u = 0s = 0.8 mNow, we that s = ut + ½ at2

0.8 = 0 + 2.5t2

which gives t = 0.57 s

15) The elevator shown is descending with a constant acceleration of 1.5 ms-2. The mass of the block A is 1 kg. What force is exerted by block A on block B? Take g = 10 ms-2.

Solution: The acceleration of the two blocks is the same as that of the elevator as they do not lose contact with the floor of the elevator. The free body diagram of the block A is

Applying Newton’s second law of motion, we have,10 – N = 1.5which gives N = 6.67 NNow, from Newton’s third law of motion, the force exerted by block A on block B = N = 6.67 N

16) A block of mass 5.00 kg is pushed up against a wall by a force P that makes a 50.0° angle with the horizontal as shown in the figure. The coefficient of static friction between the block and the wall is 0.30. Determine the possible values for the magnitude of F that allow the block to remain stationary. Take g = 10 ms-2.

Solution: We draw the free body diagram of the block. In this problem the direction of friction can be upwards or downwards depending on the magnitude of F. Lets take friction downwards case first.

Applying Newton’s second law in x direction, we getF cos55 = N   …(1)Applying Newton’s second law in y direction, we getFsin55 = (μs X N) + mg = (0.3 X N) + 50   …(2)

Dividing equation (2) by (1), we getNtan55 = 0.3N + 501.13N = 50Normal force,N = 44.3 N

Using this value of N, we getF = 77.3 Nwhich means for any value of F more than 77.3 N, the block will start sliding upwards.

Now to find the minimum value of F, lets consider the friction upwards case.

Applying Newton’s second law in x direction, we getF cos55 = N   …(1)Applying Newton’s second law in y direction, we getFsin55 + (μs X N) = mg0.82F + (0.3 X N) = 50   …(2)

Putting N from equation (2) in (2), we get0.82F + 0.17F = 50 F = 50.44 Nwhich is the minimum value of F below which the block starts sliding downwards.

17) Consider the three connected objects shown in figure. If the inclined plane is frictionless and the system is in equilibrium, find (in terms of m, g, and θ (a) the mass M and (b) the tensions T1 and T2. If the value of M is double the value found in part (a), find (c) the acceleration of each object, and (d) the tensions T1 and T2.

Solution:  (a) For the ball to be in equilibrium,T2 = Mg   …(1)Similarly, for the mass m to be in equilibrium, we must haveT2 = T1 + mgsinθ   …(2)Again, for mass 2m to be in equilibrium, we must haveT1 = 2mgsinθ   …(3)

Putting the expression for T1 in equation (2)T2 = 3mgsinθ

Putting this in equation (1)M = 3msinθ

(b) Now, if the value of M is twice i.e. 6msinθFrom figure, we see that the magnitude of acceleration of each object has to be same (say a) as they are all connected by strings.Applying Newton’s second law on the ball, we have6mgsinθ – T2 = 6msinθ X a   …(4)Applying Newton’s second law on the mass m, we haveT2 – T1 – mgsinθ = ma   …(5)Applying Newton’s second law on the mass 2m, we haveT2 – 2mgsinθ = 2ma   …(6)Solving equations (4) and (6) simultaneously, we get4mgsinθ = 6masinθ + 2ma a = 2gsinθ/(3sinθ+1)   …(7)Putting (7) in (6), we get  T2 = 6mgsinθ ( 1+sinθ1+3sinθ )From equation (5), we get T1 = 3mgsinθ ( 1+sinθ1+3sinθ )

18) In the above problem, if the coefficient of static friction between m and 2m and the inclined plane is μs , and the system is in equilibrium, find (a) the minimum value of M and (b) the maximum value of M.Solution: (a) When the value of M is minimum, the masses have a tendency to slide down the incline if M were made infinitesimally smaller than its value.Hence, friction will be upwards at the minimum value of M.So, For the ball to be in equilibrium,T2 = Mg   …(1)Similarly, for the mass m to be in equilibrium, we haveT2 + μsmgcosθ= T1 + mgsinθ   …(2)Again, for mass 2m to be in equilibrium, we haveT1 + 2μsmgcosθ = 2mgsinθ   …(3)

Putting the expression for T1 from (3) in equation (2);T2 = 3mgsinθ - 3μsmgcosθPutting this in equation (1);  Mmin = 3gsinθ - 3μsgcosθ

(b) Again, when the value of M is maximum, the masses have a tendency to slide upwards if M were made infinitesimally smaller than its value.Hence, friction will be downwards at the minimum value of M.Applying Newton’s second law to each mass, we haveT2 = Mg   …(4)                                                      (for the ball)T2 = T1 + mgsinθ + μsmgcosθ  …(5)                   (for mass m)T1 = 2mgsinθ+ 2μsmgcosθ    …(6)                     (for mass 2m)Putting the expression for T1 from equation (6) in equation (5);T2 = 3mgsinθ + 3μsmgcosθPutting this in equation (1);Mmax = 3gsinθ + 3μsgcosθ

19) A mass M = 16 kg is held in place by an applied force F and a pulley system as shown in figure. The pulleys are massless and frictionless. Find (a) the tension in each section of rope, T1, T2, T3, T4, and T5 and (b) the magnitude of F. Take g = 10 ms-2.

Solution: The free body diagram of the mass looks likeFor the mass to be in equilibrium, T5 = 160 NNow, the sum of forces across all pulleys should also be zero for them to be in equilibrium.Therefore,  T2 + T3 = T5 = 160 N   …(1)Again, for the fixed pulley, we haveT1 + T2  + T3= T4   …(2)Now, the string being pulled by force F has a tension of F in it everywhere for it to not break, i.e. T1 = F   …(3)

Also from the figure, it can be seen that T1, T2 and T3 are tensions at different parts of the same string. One part of a string cannot accelerate w.r.t. another part of the same string

because it will break and the system will collapse. Therefore, T1 = T2 = T3   ...(4)From equation (1), we getT2 = T3 = 80 NFrom (4); T1 = 80 NFrom (2); T4 = 240 NHence, F = 80 N from equation (3).

20) What horizontal force must be applied to the cart shown in figure so that the blocks remain stationary relative to the cart? Assume all surfaces, wheels, and pulley are frictionless.

   Solution: Let the system have an acceleration ‘a’ when the masses are in equilibrium w.r.t. the cart.Hence, we can assume the system be one object of mass M+m1+m2 moving with an acceleration of ‘a’.Therefore, F = (M+m1+m2)a   …(1)Now, considering, the free body diagrams of the two blocks,

The only force acting on mass m2 that accelerates it in the x direction is the push from the cart denoted by P,Therefore, P = m2a   …(2)Also, the tension in the string, T = m2g   …(3) as the blocks are stationary w.r.t. the cart i.e. they have no acceleration in the vertical direction.The mass m1 is given its acceleration by the tension in the string;i.e. T = m1a   …(4)Solving (3) and (4);a = m2g/m1   …(5)Now, applying Newton’s second law on the cart, we haveF – P = MaUsing equations (2) and (5), we getF = ( M+m2m1 )m2g

21) In figure blocks of mass m1 = 3 kg and m2 = 4 kg are connected by a string of negligible mass and are initially held in place. The heavier block is on a frictionless surface tilted at 30o. The coefficient of kinetic friction between block of mass 3 kg and the horizontal surface is 0.3. The pulley has negligible mass and friction. Once they are

released, the blocks move. What then is the tension in the string? Take g = 10 ms -2.

Solution: Note that since no static friction coefficient is mentioned, we assume fs is not relevant to this computation i.e. the blocks are already moving with acceleration ‘a’.  Let the tension in the string be T.Applying Newton's second law to each block's x axis, which for m1 is positive rightward and for m2 is positive downhill:T – fk = m1a                                                    (for mass 3 kg)   …(1)m2g sinθ – T = m2a                                       ( for mass 4 kg)   …(2)Adding the equations, we obtain the acceleration:a = m2gsinθ- fkm1+m2

Now, fk = μkN = μk m1g = 0.3 X 3 X 10 = 9 NTherefore, a = (40sin30 – 9)/7 = 1.57 ms-2.Putting this value in equation (1), we get T = 13.7 N.

22) A block slides with constant velocity down an inclined plane that has slope angle θ. The block is then projected up the same plane with an initial speed u. (a) How far up the plane will it move before coming to rest? (b) After the block comes to rest, will it slide

down the plane again?

Solution: The free-body diagram of the block when it is sliding downhill with zero

acceleration looks likeApplying Newton’s second law gives N  = mg cosθ   …(1)fk = mg sinθ = μk X N   …(2)Dividing equation (2) by (1);μk  = tanθ

Now when the block is projected uphill the friction direction is reversed. Let its

acceleration be ‘a’. Again applying Newton’s second law givesmgsinθ + fk =  mgsinθ + μkN =  ma   …(3)mgcosθ – N = 0   (as the block doesn’t lose contact with the surface of the incline ever)

We can see that the acceleration is downhill, and therefore, the speed of the block will decrease as it moves up the incline.

(a) Putting μk  = tanθ and N = mgcosθ in equation (3);a = 2gsinθ downwards the incline.The distance travelled by the block before coming to a stop can be obtained by using the equation v2 = u2 + 2as0 = u2 + 4sgsinθwhich gives s = u2/4gsinθ

(b) We usually expect μk to be less than μs. The “angle of repose” (the minimum angle necessary for a stationary block to start sliding downhill) is N  = mg cosθrepose   …(1)fs = mg sinθrepose = μs X N   …(2)Dividing equation (2) by (1);μs  = tanθrepose

but in this problem we found that μk  = tanθ. Therefore, θrepose should be greater than θ as μs  > μk. Consequently, when the block comes to rest, the incline is not steep enough to cause it to start slipping down the incline again.

23) A block of mass m1 = 4.0 kg is put on top of a block of mass m2 = 5.0 kg. To cause the top block to slip on the bottom one while the bottom one is held fixed, a horizontal force of at least 12 N must be applied to the top block. The assembly of blocks is now placed on a horizontal, frictionless table. Find the magnitudes of (a) the maximum horizontal force that can be applied to the lower block so that the blocks will move together and (b) the resulting acceleration of the blocks. Take g = 10 ms-2.

Solution: When the force applied is at a maximum, the frictional force between the two blocks must also be a maximum. Since F = 12 N of force has to be applied to the top block for slipping to take place, usingF = fs = μs X N = μs X mg = 12, we haveμs = 12/40 =  0.3, the coefficient of static friction between the two blocks.

Now, we’re interested in the maximum force that can be applied on the bottom block such that the two will move together. If we pull the bottom block too hard, the top block will slip on the bottom one. The free-body diagrams for the two blocks are:

Now, applying Newton’s second law to both the blocks, we have;fs = m1a = 4a                            (for mass m1)F – fs = m2a = 5a                      (for mass m2)When F is maximum, fs will also be maximum i.e. equal to 12 NTherefore, a = 3 ms-2

and F = 27 N

24) Find the acceleration of block of mass in the figure shown. All the surfaces are frictionless and the pulleys and the string are light.

Solution: Let the tension in the string be T. From figure, we can see that the acceleration of block of mass M has to be 2a.The free body diagrams of the blocks look like

                                                        Applying Newton’s second law on the blocks, we have,T - Mg sin30 = 2Ma                                     (for mass m)

2Mg-2T = 2Ma                                       (for mass 2M)Solving the two equations, we getT = 5Mg/6and hence a = g/6Therefore, acceleration of mass M is g/6 up the plane.

25) Find the acceleration of the 5 kg block in figure. Take g = 10 ms-2

Solution: Let the tensions in the strings be denoted by T1 and T2 for strings carrying m1 and m3 respectively. We begin by drawing the free body diagrams for each block,

                                                                                             

Applying Newton’s second law to each block, we have,50 – T1 = 5a                           …(1)                       (for mass m1)  T1 – T2 – 10 sin30 = 1a         …(2)                      (for mass m2)T2 - 25 = 2.5a                         …(3)                      (for mass m3)From equations (2) and (3);T1 = 30 + 3.5aPutting in (1), we get,a = 20/8.5 = 2.35 ms-2.

26) In the given figure, suppose m2 = 2.5 kg and m3 = 3 kg. What should be the mass m1 so that it remains at rest? Take g = 10 ms-2.

Solution: Let the tensions in the strings be T1 and T2 carrying masses m1 and m2 respectively. The free body diagrams of the masses are    

 

                                                                         For the mass m1 to be at rest, T1 = m1g   …(1)For the sum of forces across the pulley 2 carrying masses m2 and m3 to be zero, we have2T2 = T1   …(2)Let the acceleration of mass m2 be ‘a’. Applying Newton’s second law, we have

T2 – 25 = 2.5a   …(3)Applying Newton’s second law on mass m3, we have30 - T2 = 3a   …(4)Solving equations (3) and (4), we geta = 0.91 ms-2

T2 = 27.3 NPutting in (2), we get T1 = 54.6 NPutting in (1), we getm1 = 5.5 kg

27) A constant force F = m2g/3 is applied on the mass m1 as shown in figure. The pulley and the string are light and the surface of the table is smooth. Find the acceleration of m1.

Solution: Let the acceleration of mass m1 be ‘a’. The acceleration of mass m2 is also ‘a’.Their free body diagrams are          

                                                                                  Applying Newton’s second law on both the blocks,T – m2g/3 = m1a                                       (for mass m1)m2g – T = m2a                                           (for mass m2)Solving the two equations, we getm2g – m2g/3 - m1a = m2aa(m1 + m2) = 2m2g/3Therefore, a = 2m2g/3(m1 + m2)

28) Block 1 of mass m1 = 2.0 kg and block 2 of mass m2 = 1.0 kg are connected by a string of negligible mass. Block 2 is pushed by force of magnitude 25 N and angle θ = 35°.The coefficient of kinetic friction between each block and the horizontal surface is

0.25. What is the tension in the string? Take g = 10 ms-2.

Solution: Let the acceleration of the two blocks be ‘a’ and the tension in the string be ‘T’. Applying Newton’s second law on mass m2, we getFcosθ – μk X N – T = m2a                                 x-axism2g + Fsinθ = N                                                y-axisSolving the two equations, we get  m2a = Fcosθ - μk(m2g + Fsinθ) – T   a = 0.82F – 2.5 – 0.14F - T            ….(1)

Similarly, applying Newton’s second law on mass m1, we getT - μk X N = m1a                                                 x-axisN = m1g                                                               y-axisSolving the two equations, we get  m1a = T - μk m1g   2a = T – 5                                          ….(2)

Solving (1) and (2), we get;T = 11.33 Na =3.17 ms-2

Q:29 The two blocks shown in figure are not attached to each other. The coefficient of static friction between the blocks is μs = 0.35, but the surface beneath the larger block is frictionless. What is the minimum magnitude of the horizontal force required to keep the smaller block from slipping down the larger block? Take g = 10 ms-2.

Solution: Let the contact force between the two blocks be ‘P’. The free body diagrams

of the two blocks are                                                                    

 

  

Let the acceleration of the system be ‘a’ in the +ve x-direction. Applying Newton’s second law on the two blocks as a system, we haveF = (15 + 85)a or a = F/100   …(1)Now, analyzing the blocks individually, for the small block, we haveF – P = 15a   …(2)fs = μs X P = 150 Nwhich gives P = 428.6 NSimilarly, for the large block,N + fs = 850N = 700 N       (using fs = 150 N)P = 85a   …(3)428.6 = 85awhich gives a ≈ 5.04 ms-2.Putting in (1);F = 504 N

Q:30 Find the mass of the hanging block in figure which will prevent the smaller block from slipping over the triangular block. All the surfaces are frictionless and the strings and the pulley are light.

Solution: Let the acceleration of the triangular block be ‘a’ towards the right. Then block of mass ‘M’ has acceleration ‘a’ downwards and the mass ‘m’ has acceleration ‘a’ towards the right as no slipping occurs.We begin by drawing the free body diagram for each block,

From the free body diagram of mass ‘M’, we have

Mg - T = Ma   …(1)From the free body diagram of mass ‘m’, we haveNsin(90-θ) = mgor Ncosθ = mg   …(2)and Ncos(90-θ) = Nsinθ = ma   or gtanθ = a …(3)                [using (2)]Similarly, from the free body diagram of triangular bock, we haveT - Ncos(90-θ) = M’aor T - Nsinθ = M’a   …(4)Putting (1), (2) and (3) in (4);Mg - Mgtanθ - mgtanθ = M’gtanθor M = (M’ + m)tanθ/(1- tanθ)or M = (M’ + m)/(cotθ - 1)