91影视

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 above the horizontal. He holds the rope1.40 m from his feet, and it makes a angle with the cliff face.

Mass of climber m = 82 kg

Torque $\mathbf{蟿}\mathbf{=}\mathbf{FL}$

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$\left(1.4\mathrm{m}\right)\cos 10掳$ .

Applying equilibrium condition on torque

This gives torque $\mathrm{蟿}=0\mathrm{N}.\mathrm{m}$

Therefore,

$\begin{array}{r}\mathrm{T}\left(1.4\mathrm{m}\right)\cos 鈦�{10}^{鈭�}鈭�\mathrm{w}\left(1.1\mathrm{m}\right)\cos 鈦�{35}^{鈭�}=0\\ \mathrm{T}=525\mathrm{N}\end{array}$

Tension in the rope is 525 N

Let the horizontal component of the force that the cliff face exerts on the climber鈥檚 feet is ${\mathrm{f}}_{\mathrm{x}}$and the vertical component is ${\mathrm{f}}_{\mathrm{y}}$.

Applying equilibrium condition

Net horizontal and vertical force is zero.

So, $鈭�F_x=0\text{and}鈭�F_y=0$

$$\begin{gathered} \underset{}{鈭�}{\mathrm{F}}_{\mathrm{x}}=0\mathrm{implies} \\ {\mathrm{f}}_{\mathrm{x}}=\mathrm{T}\sin 25掳 \\ =222\mathrm{N} \\ \underset{}{鈭�}{\mathrm{F}}_{\mathrm{y}}=0\mathrm{implies}{\mathrm{f}}_{\mathrm{y}}+\mathrm{T}\cos 25掳-\mathrm{w}=0 \end{gathered}$$
Thus

$\begin{array}{r}{\mathrm{f}}_{\mathrm{y}}=\left(82\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)鈭�\left(525\mathrm{N}\right)\cos 鈦�{25}^{鈭�}\\ =328\mathrm{N}\end{array}$

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

Let the coefficient of friction be $渭$

$\begin{array}{r}渭=\frac{f_y}{f_x}\\ 渭=\frac{328N}{222N}=1.48\end{array}$

Hence, the coefficient of friction is 1.48