Force of the static and the kinetic friction – problems and solutions

Solved problems in Newton’s laws of motion – Force of the static and the kinetic friction

1. An object rests on a horizontal floor. The coefficient static friction is 0.4 and acceleration of gravity is 9.8 m/s². Determine (a) The maximum force of the static friction (b) The minimum force of F 

Solution

Known :

Mass (m) = 1 kg

The coefficient static friction (μ_s) = 0.4

The acceleration of gravity (g) = 9.8 m/s²

Weight (w) = m g = (1 kg)(10 m/s²) = 10 kg m/s² = 10 Newton

Normal force (N) = w = 10 Newton

Wanted :

(a) The maximum force of the static friction (b) The minimum force of F

Solution :

(a) The maximum force of the static friction

f_s = μ_s N

f_s = (0.4)(9.8 N) = 3.92 Newton

(b) The minimum force of F

If the force F is exerted on the object but the object isn’t moved, so there must be the force of static friction exerted by the floor on the object. If the object will start to move, the force of the static friction is exceeded, there must be the force of the kinetic friction. Object start moves if F is greater than the maximum force of the static friction.

So the minimum force of F = maximum force of the static friction = 3.92 Newton.

2. 1 kg box is pulled along a horizontal surface by a force F, so the box is moving at a constant velocity. If the coefficient kinetic friction is 0.1, determine the magnitude of the force F! (g = 9.8 m/s²)

Known :

The coefficient kinetic friction (μ_k) = 0.1

Box’s mass (m) = 1 kg

Acceleration of gravity (g) = 9.8 m/s²

Weight (w) = m g = (1 kg)(9.8 m/s²) = 9.8 kg m/s² = 9.8 Newton

Normal force (N) = w = 9.8 Newton

Wanted : F

Solution :

Newton’s first law states that if no net force acts on an object, every object continues in it’s state of rest, or constant velocity in a straight line.

So if the object moves at a constant velocity, there must no net force (ΣF = 0). Force F is exerted on the object in the right direction so that the force of the kinetic friction is exerted on the object to the left direction.

∑F = 0

F – f_k = 0

F = f_k

The force of the kinetic friction :

f_k = μ_k N = (0.1)(9.8 N) = 0.98 Newton

object moves with constant velocity, F = f_k = 0.98 Newton

3. An object slides down an inclined plane with constant velocity. Determine coefficient kinetic friction (μ_k). g = 9.8 m/s²

Solution

w = weight, w_x = horizontal component of weight, points along the incline, w_y = vertical component of weight, perpendicular to the inclined plane, N = normal force, f_k = the force of the kinetic friction.

Known :

Mass (m) = 1 kg

Acceleration of gravity (g) = 9.8 m/s²

weight (w) = m g = (1 kg)(9.8 m/s²) = 9.8 kg m/s² = 9.8 Newton

w_x = w sin 30^o = (9.8 N)(0.5) = 4.9 Newton

w_y = w cos 30^o = (9.8 N)(0.5)√3 = 4.9√3 Newton

Normal force (N) = w_y = 4.9√3 Newton

Wanted : coefficient kinetic friction (μ_k)

Solution :

Object slides down an inclined plane with constant velocity so that the net force = 0.

∑F = 0

w_x – f_k = 0

w_x = f_k

w_x = μ_k N

5 = μ_k (5√3)

μ_k = 5 / 5√3

μ_k = 1 /√3

μ_k = 0.58

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1. Mass and weight
2. Normal force
3. Newton’s second law of motion
4. Friction force
5. Motion on horizontal surface without friction force
6. Motion of two bodies with the same acceleration on rough horizontal surface with friction force
7. Motion on inclined plane without friction force
8. Motion on rough inclined plane with friction force
9. Motion in an elevator
10. Motion of bodies connected by cord and pulley
11. Two bodies with the same magnitude of accelerations
12. Rounding a flat curve – dynamics of circular motion
13. Rounding a banked curve – dynamics of circular motion
14. Uniform motion in a horizontal circle
15. Centripetal force in uniform circular motion