Question

1. you have a 0.234-m-long wrench. a job requires a torque of 32.4 n·m, and you can exert a force of 232 n

a. what is the smallest angle, with respect to the handle of the wrench, at which you can pull on the wrench and get the job
done?
b. a friend can exert 275 n. what is the smallest angle she can use to accomplish the job?

Explanation:

Part a

Step1: Recall Torque Formula

Torque \(\tau = rF\sin\theta\), where \(r\) is the length of the wrench, \(F\) is the force applied, and \(\theta\) is the angle between the force and the wrench handle. We need to find the smallest \(\theta\) (so \(\sin\theta\) is maximized when \(\theta = 90^\circ\), but here we solve for \(\theta\) given \(\tau\), \(r\), \(F\)). Rearranging for \(\sin\theta\): \(\sin\theta=\frac{\tau}{rF}\)
Given \(\tau = 32.4\space N\cdot m\), \(r = 0.234\space m\), \(F = 232\space N\)

Step2: Calculate \(\sin\theta\)

Substitute values: \(\sin\theta=\frac{32.4}{0.234\times232}\)
First calculate denominator: \(0.234\times232 = 0.234\times200+0.234\times32 = 46.8 + 7.488 = 54.288\)
Then \(\sin\theta=\frac{32.4}{54.288}\approx0.597\)

Step3: Find \(\theta\)

\(\theta=\arcsin(0.597)\approx36.7^\circ\) (using calculator for arcsin)

Step1: Recall Torque Formula

Again, \(\tau = rF\sin\theta\), rearrange for \(\sin\theta=\frac{\tau}{rF}\)
Given \(\tau = 32.4\space N\cdot m\), \(r = 0.234\space m\), \(F = 275\space N\)

Step2: Calculate \(\sin\theta\)

Substitute values: \(\sin\theta=\frac{32.4}{0.234\times275}\)
Denominator: \(0.234\times275 = 0.234\times200+0.234\times75 = 46.8+17.55 = 64.35\)
\(\sin\theta=\frac{32.4}{64.35}\approx0.5035\)

Step3: Find \(\theta\)

\(\theta=\arcsin(0.5035)\approx30.2^\circ\) (using calculator for arcsin)

Answer:

The smallest angle is approximately \(36.7^\circ\)

Part b