Question

An object with a mass of 45.01 kg rests on a plane inclined 31.52° from horizontal. What is the force of static friction?Hint *** to get the approximate static coefficient of friction use tan (O)

Answer 1

The coefficient of friction can be given as,

$\mu=\tan \theta$

Plug in the known value,

$\begin{gathered} \mu=\tan 31.52^{\circ} \\ =0.613 \end{gathered}$

The force of static friction is given as,

$f=\mu mg$

Substitute the known values,

$\begin{gathered} f=(0.613)(45.01kg)(9.8m/s^2)(\frac{1\text{ N}}{1kgm/s^2}) \\ =270.4\text{ N} \end{gathered}$

Thus, the force of static friction is 270.4 N.