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

Answer:

I think the answer is 28

Answer 2

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

Explanation:

Given Data

The normal force exerted by the limb of the tree $F_n$  = 48 N

Frictional force between the limb and monkey  $F_f$ = 20 N

To find - the magnitude of total force exerted by the limb on the monkey

Magnitude of the total force exerted by the limb on the monkey is equals to the square root of sum of the squares of the normal force exerted by the limb and frictional force between the limb and monkey.

Magnitude of the total force is also called as Resultant Force.

$\text {Magnitude of the total force exerted by the limb on the monkey} = \sqrt{F_n ^{2} + F_f ^{2}}$

Substitute the respective values in above equation

$\text {Magnitude of the total force} = \sqrt{48 ^{2} + 20^{2}}$

$\text {Magnitude of the total force} = \sqrt{2304 +400}$

$\text {Magnitude of the total force} = \sqrt{2704}$

Magnitude of the total force = 52 N

Therefore total force exerted by the limb on the monkey is 52 N when the normal force exerted by the limb is 48 N and the frictional force between limb and monkey is 20 N.