Q11-45P - University Physics with Modern Physics

Question

Mountaineers often use a rope to lower themselves down r = the face of a cliff (this is called rappelling). They do this with their body nearly horizontal and their feet pushing against the cliff. Suppose that an 82-0-kg climber, who is 1.90 m tall and has a center of gravity 1.1 mm from his feet, rappels down a vertical cliff with his body raised 35.0^o above the horizontal. He holds the rope 1.40 m from his feet, and it makes a 25.0^o angle with the cliff face. (a) What tension does his rope need to support? (b) Find the horizontal and vertical components of the force that the cliff face exerts on the climber’s feet. (c) What minimum coefficient of static friction is needed to prevent the climber’s feet from slipping on the cliff face if he has one foot at a time against the cliff?

Step-by-Step Solution

Answer

1. The tension in the cable is 525 N.
2. The horizontal component is 222 N and the vertical component is 328 N.
3. The coefficient of static friction is 1.48

Step 1: The given data

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

Step 2: Formula used

Torque $\mathit{\tau }\mathbf{=}\mathit{F}\mathit{L}$

Where F is force exerted and L is distance.

(a) Step 3: Find the tension in the rope

Let T be the tension needed by the rope.

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

Applying equilibrium condition on torque

This gives torque $\tau =0 N·m$

Therefore, 

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

Tension in the rope is 525 N

(b) Step 4: Find horizontal and vertical components of force

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 $\sum F_x=0 and \sum F_y=0$ 

$\sum F_x=0$ implies

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

$\sum F_y=0$implies

$f_y+T\cos 25°-w=0$

Thus

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

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

(c) Step 5: Find the coefficient of friction

Let the coefficient of friction be $\mu$

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

Hence, the coefficient of friction is 1.48