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

Solved problems in Newton’s laws of motionForce 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 

Force of static and kinetic friction – problems and solutions 1

Solution

Force of static and kinetic friction – problems and solutions 2

Known :

Mass (m) = 1 kg

The coefficient static frictions) = 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

fs = μs N

fs = (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.

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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)

Force of static and kinetic friction – problems and solutions 3

Known :

The coefficient kinetic friction (μk) = 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 – fk = 0

F = fk

The force of the kinetic friction :

fk = μk N = (0.1)(9.8 N) = 0.98 Newton

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

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

Force of static and kinetic friction – problems and solutions 4

Solution

Force of static and kinetic friction – problems and solutions 5

w = weight, wx = horizontal component of weight, points along the incline, wy = vertical component of weight, perpendicular to the inclined plane, N = normal force, fk = 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

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

wy = w cos 30o = (9.8 N)(0.5)3 = 4.93 Newton

Normal force (N) = wy = 4.93 Newton

Wanted : coefficient kinetic friction (μk)

Solution :

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

F = 0

wx – fk = 0

wx = fk

wx = μk N

5 = μk (53)

μk = 5 / 53

μk = 1 /3

μk = 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