Question

A bear sling (Fig. 4-40) is used in some national parks for placing backpackers’ food out of the reach of bears. As a backpacker raises the pack by pulling down on the rope, the force F needed

(a) decreases as the pack rises until the rope is straight across.

(b) doesn’t change.

(c) increases until the rope is straight.

(d) increases, but the rope always sags where the pack hangs.

Short answer

The correct option is (d).

Step-by-step solution

Step 1. Forces on the rope and the backpackers

The tension is the same everywhere on the rope.

Given data:

The force to pull down the rope is F.

Assumption:

Let W be the weight of the backpackers.

Let $\theta$be the angle of the rope with the horizontal plane.

Step 2. Maintain the equilibrium condition of the weight

The force on the rope will act on the rope along its length on both sides of the backpacker.

Then, using the component method,

$\begin{array}{c}2F\sin \theta =W\\ F=\frac{W}{2\sin \theta }\end{array}$

Now, when the backpacker rises, the angle $\theta$decreases. You know that when the value of $\theta$decreases, the value of $\sin \theta$also decreases.

F should increase when $\sin \theta$decreases to maintain the equilibrium.

Also, for equilibrium, there should be a vertical component of the force equal to the weight of the backpackers.

Hence, option (c) is correct.