Answer to Question #170350 in Mechanics | Relativity for kush

Question

A man is walking over a dome of 10 m in radius. how far can he descend from the dome without slipping? take the coefficient of friction between the surface of the dome and the shoe of the man 0.6.

Answer 1

Explanations & Calculations

• Refer to the sketch attached
• Consider a situation when the person is at an arbitrary point on the dome as shown & write equations for his equilibrium.

• The contact force on him at that point from the dome

$\qquad\qquad \begin{aligned} \small R&= \small mg\cos\theta \end{aligned}$

• The person can safely descend some distance down until the component of his weight becomes equal to the maximum static friction available at that point.

• Then,

$\qquad\qquad \begin{aligned} \small f_{s.max}&\geq\small mg\sin\theta\\ \small \mu R&\geq \small mg\sin\theta\\ \small \mu\cdot mg\cos\theta&\geq\small mg\sin\theta \\ \small \tan\theta&\leq\small \mu\\ \theta&\leq\small \tan^{-1}(0.6)\\ &\leq\small 30.96^0 \end{aligned}$

• As mentioned above he can safely come down up to a maximum angle of 30.96 degrees from the verticle
• Then

$\qquad\qquad \begin{aligned} \small \theta _{max}&= \small 30.96^0=0.540rad \end{aligned}$

• The distance he can come down is (distance measured from the top of the dome)

$\qquad\qquad \begin{aligned} \small S&= \small r\theta\\ &= \small 10m\times 0.540\\ &= \small \bold{5.4\,m} \end{aligned}$