Forza tal-frizzjoni statika u kinetika – problemi u soluzzjonijiet

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 u, aċċelerazzjoni tal-gravità huwa 9.8 m/s². Determine (a) The maximum force of the static friction (b) The minimum force of F 

Soluzzjoni

Magħruf:

Massa (m) = 1 kg

The coefficient static friction (μ_s) = 0.4

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

piż (w) = mg = (1 kg)(10 m/s²) = 10 kg m/s² = 10 Newton

Forza normali (N) = w = 10 Newton

Meħtieġ:

(A) The maximum force of the static friction (b) Il- minimum force of F

Soluzzjoni:

(A) The maximum force of the static friction

f_s = μ_s N

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

(b) Il- 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)

Magħruf:

The coefficient kinetic friction (μ_k) = 0.1

Box’s mass (m) = 1 kg

Aċċelerazzjoni tal-gravità (g) = 9.8 m/s²

Piż (w) = mg = (1 kg)(9.8 m/s²) = 9.8 kg m/s² = 9.8 Newton

Forza normali (N) = w = 9.8 Newton

Wanted : F

Soluzzjoni:

L-ewwel liġi ta’ Newton 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 veloċità kostanti, 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.

ΣF=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 pjan inklinat with constant velocity. Determine coefficient kinetic friction (μ_k). g = 9.8 m/s²

Soluzzjoni

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.

Magħruf:

Massa (m) = 1 kg

Aċċelerazzjoni tal-gravità (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

Forza normali (N) = w_y = 4.9√3 newton

Meħtieġ: coefficient kinetic friction (μ_k)

Soluzzjoni:

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

ΣF=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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