Motion on the rough inclined plane with the friction force – application of Newton’s law of motion problems and solutions

1. Object’s mass = 2 kg, acceleration due to gravity = 9.8 m/s2, coefficient of the static friction = 0.2, coefficient of the kinetic friction = 0.1. Is the object at rest or accelerating? If the object is accelerated, find (a) the net force (b) magnitude and direction of the box’s acceleration!

Motion on rough incline plane with friction force - application of Newton's law of motion problems and solutions 1

Solution

Motion on rough incline plane with friction force - application of Newton's law of motion problems and solutions 2

Known :

Mass (m) = 2 kg

Acceleration due to gravity (g) = 9.8 m/s2

Coefficient of the static friction (μs) = 0.2

Coefficient of the kinetic friction (μk) = 0.1

Weight (w) = m g = (2)(9.8) = 19.6 Newton

The horizontal component of the weight (wx) = w sin 30o = (19.6)(0.5) = 9.8 Newton

The vertical component of th weight (wy) = w cos 30o = (19.6)(0.5√3) = 9.8√3 Newton

The normal force (N) = wy = 9.8√3 Newton

Force of the static friction (fs) = (0.2)(9.8√3) = 1.96√3 Newton = 3.39 Newton

Force of the kinetic friction (fk) = (0.1)(9.8√3) = 0.98√3 Newton = 1.69 Newton

Solution :

Object is at rest if wx < fs, object is moving down if wx > fs.

wx = 9.8 Newton and fs = 3.39 Newton.

(a) the net force

F = wx – fk = 9.8 – 1.69 = 8.11 Newton

(b) magnitude and direction of the acceleration

F = m a

8.11 = (2) a

a = 4.05

Magnitude of the acceleration = 4.05 m/s2 and direction of the acceleration = downward.

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2. Object’s mass = 4 kg, acceleration due to gravity = 9,8 m/s2. Coefficient of the kinetic friction = 0.2 and coefficient of the static friction = 0.4. Magnitude of the force F = 40 Newton. The object is at rest or slides down ? If the object slides down, find (a) the net force (b) magnitude and direction of the acceleration!

Motion on rough incline plane with friction force - application of Newton's law of motion problems and solutions 3

Solution

Motion on rough incline plane with friction force - application of Newton's law of motion problems and solutions 4

Known :

Mass (m) = 4 kg

Acceleration due to gravity (g) = 9.8 m/s2

The coefficient of the static friction (μs) = 0.4

The coefficient of the kinetic friction (μk) = 0.2

Weight (w) = m g = (4)(9.8) = 39.2 Newton

The horizontal component of the weight (wx) = w sin 30o = (39.2)(0.5) = 19.6 Newton

The vertical component of the weight (wy) = w cos 30o = (392)(0..5√3) = 19.6√3 Newton

The normal force (N) = wy = 19.6√3 Newton = 33.95 Newton

the static friction force (fs) = μs N = (0,4)(33.95) = 13.58 Newton

The kinetic friction force (fk) = μk N = (0.2)(33.95) = 6.79 Newton

F = 40 Newton

Solution :

The object slides down if F < wx + fs. The object slides up if F > wx + fs.

F = 40 Newton, wx = 19.6 Newton and fs = 13.58 Newton.

F is greater than wx + fs so the object slides up.

(a) The net force

F = F – wxfk = 40 – 19.6 – 6.79 = 13.61 Newton

(b) The magnitude and direction of the acceleration

F = m a

6.4 = (4) a

a = 1.6

The magnitude of the acceleration is 1.6 m/s2 and direction of the acceleration is upward.

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  2. Normal force
  3. Newton’s second law of motion
  4. Friction force
  5. Motion on the horizontal surface without friction force
  6. The motion of two bodies with the same acceleration on the rough horizontal surface with the friction force
  7. Motion on the inclined plane without friction force
  8. Motion on the rough inclined plane with the friction force
  9. Motion in an elevator
  10. The 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