Newton's Laws of Motion - with Examples, Problems, Solutions and Visualizations





Target Audience: High School Students, College Freshmen and Sophomores, students preparing for the International Baccalaureate (IB), AP Physics B, AP Physics C, A Level, Singapore/GCE A-Level; 

Class 11/12 students in India preparing for ISC/CBSE and Entrance Examinations like the IIT-JEE/AIEEE Anyone else who needs this Tutorial as a reference!

This might also be helpful in studying topics required by Common Core Physics. 


A Quick Summary of what we'll study in this chapter and the kind of problems we'll solve

(after this intro, there is a comprehensive document with study material as well as solutions to problems.)

An Introduction to 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.


Topics which the tutorial will introduce and apply in various interesting problems :

Drawing free body diagrams and identifying forces acting on a body

While 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.

Friction : Static and dynamic 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 like
For 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 = μkR

μs (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 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. Frictional force that opposes relative motion between surfaces in contact is called kinetic or sliding friction


Important Points:

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.
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.
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.


Related Visualizations





Constrained Motion - 1 - A Visualization

Constrained Motion 1 - A Block and a Wedge



Here are some of the problems solved in this tutorial :


Q: 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.

Q: 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?

Q:

Q: o. Find the magnitude of the acceleration of the two objects and the tension in the cord. Take g = 10 ms-2.

Q:
(a) Determine the magnitude of the acceleration of the two-block system.
(b) Determine the magnitude of the contact force between the two blocks.

Q: -2.

Q: 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.

Q: x = 0.400 m. (b) Find the value of x at which the acceleration becomes zero. Take g = 10 ms-2.

Q: -2.

Q: -2.

Q: s between A and the table is 0.18. Block C is suddenly lifted off A. (b) What is the acceleration of block A if μk between A and the table is 0.15. Take g = 10 ms-2.

Q: T in the string.

Q: 1 = 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.        

Q: o 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.

Q: -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.

Q: 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.

Q: 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.

Q: 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.

Q:

Q: 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?

Q: 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.

Q:

Q: -2

Q: 2 = 2.5 kg and m3 = 3 kg. What should be the mass m1 so that it remains at rest? Take g = 10 ms-2.

Q: 2g/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.

Q: 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.

Q: s = 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.

Q:

Tutorial with solved problems and examples :

Mechanics-2-newtons-laws

MCQ Quiz #1

Newton's Laws: Some Basic Problems

Companion MCQ Quiz- test how much you know about the topic. Your score will be e-mailed to you at the address you provide.

MCQ Quiz: Newton's Laws (Basic Problems)

MCQ Quiz #2

Newton's Laws: Some more Problems Testing the Basics

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MCQ Quiz #3


Newton's Laws: Some Challenging Problems

Companion MCQ Quiz- test how much you know about the topic. Your score will be e-mailed to you at the address you provide.


Newton's Laws- Challenging Problems