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Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 8.7 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 3.7 m/s. What upward force must a branch provide to support the swinging gibbon?

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MechEngineer

Answer:

198.505  N

Explanation:

Since this can be modeled as a point mass swinging system with the radius being the arm length. So r = 0.6m.

Also at it's lowest point the ape has a horizontal velocity v = 3.7 m/s. This velocity could result in a centripetal acceleration a, creating tension on the branch.

[tex] a = \frac{v^2}{r}[/tex]

According to Newton's 2nd law, F = ma. So the tension is:

[tex] F = ma = \frac{mv^2}{r} = \frac{8.7*3.7^2}{0.6} = 198.5N[/tex]

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