Answer to Question #91266 in Mechanics | Relativity for Umar

Question

A block weight 200N rest on a rough horizontal table. When the block just begins to slide the pulling force is found to be 40N. What is the coefficient of friction between the two surfaces

Answer 1

Applying the Newton’s Second Law of Motion, we get (at the moment when the block begins to slide the acceleration is zero):

$$\sum F_x = ma_x = 0,$$ $$F_{pull} - F_{s.fr} = 0,$$ $$F_{pull} = F_{s.fr} = \mu_s N = \mu_s W,$$

here, $F_{pull} = 40N$ is the pulling force required to start the block of weight $W = 200 N$ to slide, $F_{s.fr}$ is the force of static friction, $\mu_s$ is the coefficient of static friction, $N$ is the normal force.

Then, from the last formula we can calculate the coefficient of static friction:

$$\mu_s = \dfrac{F_{pull}}{W} = \dfrac{40 N}{200 N} = 0.2$$

Answer:

$\mu_s = 0.2$