Here we will focus on solving numerical problems involving frictional force or friction and friction coefficient. We will take up a few sample problems as examples and solve those using the step-by-step process. We have already published a few posts on friction and if you want you can quickly browse through those posts for a quick revision before attempting the numerical problems.
Sample numerical problems based on friction and friction coefficient – with solution
Before targeting & solving the numerical problems, we should quickly go through some important points about friction as these will help you to understand and solve the problems easily. Here are those summary points as bullet points in the section below.
Summary Points to follow while solving numerical problems in physics from Friction chapter
• Friction is caused by bodies sliding over rough surfaces.
• The degree of surface roughness is indicated by the coefficient of friction, μ.
• The force of friction is calculated by multiplying the coefficient of friction by the normal force.
• The frictional force always opposes motion.
• Acceleration is caused by the net force which is found by subtracting the frictional force from the applied force.
Friction Problem set to solve now
Let’s solve some numerical problems using the concepts and formula of friction.
Example Problem 1 ] A box weighing 2000 N is sliding across a cement floor. The force pushing the box is 500 N and the coefficient of sliding friction between the box and the floor is 0.20. What is the acceleration of the box?
In this case, the normal force for the box is its weight.
Using the normal force and the coefficient of friction, we can find the frictional force.
We can also find the mass of the box from its weight since we know the acceleration due to gravity.
Then we can find the net force and the acceleration.
Friction = FF = μFN = (0.20)(2000 N) = 400 N
Mass of box = weight/g = (2000 / 9.8) Kg = 204 Kg
Fnet = Pushing force – frictional force = 500 N – 400 N = 100 N
Acceleration = a = Fnet / mass = (100 / 204 ) m/s2 = 0.49 m/s2
Example Problem 2 ] Two boxes are connected by a rope running over a pulley (see image). The coefficient of sliding friction between box A and the table is 0.20. (Ignore the masses of the rope and the pulley and any friction in the pulley.) The mass of box A is 5.0 kg and the mass of box B is 2.0 kg. The entire system (both boxes) will move together with the same acceleration and velocity.
Find the acceleration of the system.
The force tending to move the system is the weight of box B and the force resisting the movement is the force of friction between the table and box A.
The mass of the system would be the sum of the masses of both boxes. The acceleration of the system will be found by dividing the net force by the total mass.
FN(box A) = mg = (5.0 kg)(9.8 m/s^2) = 49 N
Friction = FF= μFN = (0.20)(49 N) = 9.8 N
Weight of box B = mg = (2.0 kg)(9.8 m/s^2) = 19.6 N
Fnet = 19.6 N – 9.8 N = 9.8 N
a = Fnet / (total mass of A and B) = 9.8 N/7.0 kg = 1.4 m/s^2
If you need quick guidance on ‘pulley based numerical problem‘ like the above one, then you can go through this post on pulley problem – how to address pulley numerical problems
Problem set 2 ( solution in the next page)
- A 52 N sled is pulled across a cement sidewalk at a constant speed. A horizontal force of 36 N is exerted. What
is the coefficient of sliding friction between the sidewalk and the metal runners of the sled?
- If the coefficient of sliding friction between a 25 kg crate and the floor is 0.45, how much force is required to
move the crate at a constant velocity across the floor?
- A smooth wooden 40.0 N block is placed on a smooth wooden table. A force of 14.0 N is required to keep
the block moving at constant velocity.
(a) What is the coefficient of sliding friction between the block and the tabletop?
(b) If a 20.0 N brick is placed on top of the wooden block, what force will be required to keep the block and
brick moving at constant velocity?
[The solution to the 3 problems is given in the next 3 pages. Use the pagination links below to navigate to those pages.]