91Ó°ÊÓ

Given that an 82.0-kg climber, who is 1.90 m tall and has a center of gravity 1.1 m from his feet, rappels down a vertical cliff with his body raised 35.0^o above the horizontal. He holds the rope1.40 m from his feet, and it makes a 25.0^o angle with the cliff face.

Mass of climber m = 82kg

Torque $\mathit{Ï„}\mathbf{=}\mathit{F}\mathit{L}$

Where Fis force exerted and L is distance.

Let T be the tension needed by the rope.

The moment arm for the force T is $\left(1.4m\right)\cos 10°$.

Applying equilibrium condition on torque

This gives torque$\mathrm{τ}=0N·m$

Therefore,

role="math" localid="1668246471410" $$\begin{gathered} T\left(1.4m\right)cos10°-w\left(1.1m\right)cos35°=0 \\ T=525N \end{gathered}$$

Tension in the rope is 525 N

Let the horizontal component of the force that the cliff face exerts on the climber’s feet is $f_x$and the vertical component is $f_y$.

Applying equilibrium condition

Net horizontal and vertical force is zero.

So $∑F_x=0and∑F_y=0$

$∑F_x=0$implies

$\begin{array}{rcl}{f}_x& =& T\sin 25°\\ & =& 222\text{N}\end{array}$

$∑F_y=0$implies

$f_y+T\cos 25°-w=0$

Thus

$\begin{array}{rcl}{f}_y& =& \left(82kg\right)\left(9.8m/s^2\right)-\left(525N\right)\cos {25}^o\\ & =& 328N\end{array}$

Hence horizontal component is 222 N and the vertical component is 328 N.

Let the coefficient of friction be$\mathrm{μ}$

$$\begin{gathered} \mathrm{μ}=\frac{f_y}{f_x} \\ \mathrm{μ}=\frac{328N}{222N}=1.48 \end{gathered}$$

Hence, the coefficient of friction is 1.48