Question

Ian has one cam that is 1 inch wide, with a point where

the rope attaches on its right corner, so that the device

pivots on its left corner. In other words, the left corner

of the cam becomes the center of a circle, and the point

where the rope attaches follow the curve

(a) If a crack is wide, at what angle must Ian

turn the device in order to put it into the crack?

(Hint: compute the x-coordinate of the right edge of the

device.)

(b) If Ian falls, the rope will pull on the right edge of the

cam while the left edge remains wedged in a fixed

position. What happens to the width of the cam in

the crack?

Short answer

part a) The angle that the arms make with the rope is $\theta ={\cos }^{-1}0.75$

Part b) The width of the cam decreases.

Step-by-step solution

Part (a) Step 1: The objective is to find the angle that the arms make with the rope.

Let $\theta$is an angle.

Assume that the cam's arms are symmetrically positioned.

Each$1inch$arm must then fit inside a $0.75inch$crack.

As a result, the $x$coordinates of the two arm ends are as follows.

Thus,

$$\begin{gathered} 1\cos \theta =0.75 \\ \Rightarrow \cos \theta =0.75 \\ \Rightarrow \theta ={\cos }^{-1}0.75 \end{gathered}$$

Therefore, the angle that the arms make with the rope is $\theta ={\cos }^{-1}0.75$

 Part (b) Step 2: What happens to the width of the cam in the crack?

If lan descends, the rope pulls on the right edge of the cam, which causes the cam to push even harder against the crack's walls. The cam's breadth narrows as it progresses.