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

Answer:

The correct solution will be "271.95 N".

Explanation:

The given values are:

velocity  

v = 3.5 m/s

mass

m = 9.0 kg

r = 0.6 m

According to the question:

⇒ $F_{branch}=F_{gravity}+F_{centrifugal}$

⇒             $=mg+\frac{mv^2}{r}$

On substituting the values, we get

⇒             $=9\times 9.8+\frac{9\times (3.5)^2}{0.6}$

⇒             $=88.2+\frac{110.35}{0.6}$

⇒             $=271.95 \ N$