Question

A block of mass 20kg place on horizontal plane A force of 40N was applied to cause the block of start moving calculate the co efficient of static friction

Answer 1

Approximately
`0.20`.

Assumptions:
`g = 9.81\; {\rm N\cdot kg^(-1)}`, and
`40\; {\rm N}` refers to the minimum amount of force in the horizontal direction required to move the block.

Step-by-step explanation:

Assume that two surfaces are in contact with one another. As long as the two surfaces are stationary relative to each other, the static friction between them would oppose external forces that would have started a motion between the two surfaces.

For an external force parallel to the two surfaces to start a motion, the minimum magnitude of this force required is equal to the maximum possible magnitude of static friction between the two surfaces. This value is equal to the product of:

• magnitude of the normal force between the two surfaces, and
• the coefficient of static friction
  `\mu_(\rm s)`.

In this question, the coefficient of static friction can be found in the following steps:

• Deduce the maximum possible magnitude of static friction between the two surfaces.
• Find the normal force between the two surfaces.
• Divide the maximum possible magnitude of static friction by the magnitude of the normal force to find the constant of static friction.

Assume that the force on the block in this question is horizontal (parallel to the contact surface,) and that the magnitude of this force is the minimum value required to start a motion. The maximum possible magnitude of static friction between the two surfaces would be equal to the magnitude of this external force:
`40\; {\rm N}`.

Let
`m = 20\; {\rm kg}` denote the mass of this block. Since the block in this question is on a horizontal surface, forces in the vertical direction would be balanced. Hence, the normal force between the two surfaces should be equal in magnitude to the weight of this block,
`m\, g`.

To find the coefficient of static friction, divide the maximum magnitude of static friction by the magnitude of the normal force:

`\begin{aligned}\mu_(\rm s) &= \frac{(\text{maximum magnitude of friction})}{(\text{magnitude of normal force})} \\ &= \frac{(40\; {\rm N})}{(20\; {\rm kg})\, (9.81\; {\rm N\cdot kg^(-1)})} \\ &\approx 0.20\end{aligned}`.

In other words, the coefficient of static friction between the two surfaces is approximately
`0.20` under the assumptions.