Question

Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 9.0 kg gibbon has an arm length (hand to shoulder) of 0.60 m. We can model its motion as 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 3.5 m/s. What upward force must a branch provide to support the swinging gibbon

Answer 1

The correct solution will be "271.95 N".

Step-by-step explanation:

The given values are:

velocity

v = 3.5 m/s

mass

m = 9.0 kg

r = 0.6 m

According to the question:

⇒
`F_(branch)=F_(gravity)+F_(centrifugal)`

⇒
`=mg+(mv^2)/(r)`

On substituting the values, we get

⇒
`=9* 9.8+(9* (3.5)^2)/(0.6)`

⇒
`=88.2+(110.35)/(0.6)`

⇒
`=271.95 \ N`