Question

refer to the figure to the right. (a) how many inches will the weight in the figure rise if the pulley is rotated through an angle of 72° 40′? (b) through what angle, to the nearest minute, must the pulley be rotated to raise the weight 6 in.? (a) the weight in the figure will rise inches. (do not round until the final answer. then round to the nearest tenth as needed.)

Explanation:

Step1: Convert angle to radians

First, convert $72^{\circ}40'$ to decimal - degrees. $40'=\frac{40}{60}\approx0.6667^{\circ}$, so $72^{\circ}40'\approx72.6667^{\circ}$. Then convert to radians: $\theta = 72.6667\times\frac{\pi}{180}\approx1.268$ radians. The formula for arc - length $s = r\theta$, where $r = 9.65$ inches.

Step2: Calculate arc - length for part (a)

Substitute $r = 9.65$ inches and $\theta\approx1.268$ radians into the arc - length formula $s=r\theta$. So $s=9.65\times1.268 = 12.2462\approx12.2$ inches.

Step3: Solve for the angle in part (b)

We know $s = 6$ inches and $r = 9.65$ inches. Using the formula $s = r\theta$, we can solve for $\theta$: $\theta=\frac{s}{r}=\frac{6}{9.65}\approx0.622$ radians. Convert this angle to degrees: $\theta_d=0.622\times\frac{180}{\pi}\approx35.64^{\circ}$. Convert the decimal part of the degree to minutes. $0.64\times60 = 38.4\approx38'$. So the angle is $35^{\circ}38'$.

Answer:

(a) $12.2$
(b) $35^{\circ}38'$