Question

A monkey is sitting on a tree limb. The limb exerts a normal force of 48 N and a frictional force of 20 N. Find the magnitude of the total force exerted by the limb on the monkey.

Answer 1

Given: The limb of the tree exerts a normal force of 48 N and a frictional force of 20 N.



So, resultant magnitude of the force if given by
$$\begin{gathered} R=\sqrt{\left({48}^2+{20}^2\right)} \\ =\sqrt{2304+400} \\ =\sqrt{2704}=52\mathrm{N} \end{gathered}$$

∴ The magnitude of the total force exerted by the limb on the monkey is 52 N.