Answer to Question #180558 in Physics for Mai Bui

Question

A 14kg is being pulled with a force of 40N at a 22^o angle. What is the acceleration if coefficient of friction is 0.35?

Answer 1

By defintion, the friction force is:

$$F_{fr} = \mu mg\\ F_{fr} = 0.35\cdot 15\cdot 9.8 = 51.45N$$

where $\mu = 0.35$ is the coefficient of friction, $m = 14kg$ is the mass of the body, and $g = 9.8 m/s^2$ is the gravitational acceleration.

The horizontal projection of the pulling force is:

$$F_x = F\cos \theta\\ F_x = 40N\cdot \cos 22\degree \approx 37N$$

where $F = 40N$, and $\theta = 22\degree$.

Since the friction force is greater than the pulling force:

$$F_{fr} > F_x$$

the body will not move (the acceleration will be 0).

Answer. 0.