Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to thenext handhold. A 9.3kggibbon has an arm length (hand to shoulder) of 0.60m. We can model its motionas that of a point mass swinging at the end of a 0.60-m-long, massless rod. At the lowest point of its swing, the gibbon is moving at 3.3m/s. What upward force must a branch provide to support the swinging gibbon?

Upward force provided by the branch is 260 N

Explanation:

Given -

Mass of Gibbon, m = 9.3 kg

Length of the branch, r = 0.6 m

Speed of the movement, v = 3.3 m/s

Upward force, T = ?

The tension force in the rod must be greater than the weight at the bottom of the swing in order to provide an upward centripetal acceleration.

Therefore,

F net = T - mg

F net = ma = mv²/r

Thus,

T = mv²/r + mg

T = m ( v²/r + g)

T = 9.3 [ ( 3.3)² / 0.6 + 9.8]

T = 259.9 N ≈ 260 N

Therefore, upward force provided by the branch is 260 N

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.