Question

If Mr. Mangan is pushing a 90 kg sled across the ground with a force of300 N and it is just about to move, what is the coefficient of staticfriction between the sled and the ground?

Answer 1

Given data:

* The mass of the sled is 90 kg.

* The force applied to move the sled is 300 N.

Solution:

The normal force acting on the sled is,

$F_N=mg$

where m is the mass of sled, g is the acceleration due to gravity,

Substituting the known values,

$\begin{gathered} F_N=90\times9.8 \\ F_N=882\text{ N} \end{gathered}$

The force acting on the sled is the same as static friction force on the sled.

Thus,

$F=F_s$

where F is the applied force, and F_s is the static friction,

The static friction of the sled is,

$F_s=\mu_sF_N$

Thus, the coefficient of static friction is,

$\begin{gathered} F=\mu_sF_N \\ \mu_s=\frac{F}{F_N} \end{gathered}$

Substituting the known values,

$\begin{gathered} \mu_s=\frac{300}{882} \\ \mu_s=\text{0}.34 \end{gathered}$

Thus, the coefficient of static friction is 0.34.