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 71° 50? (b) through what angle, to the nearest minute, must the pulley be rotated to raise the weight 3 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 $71^{\circ}50'$ to degrees. Since $1^{\circ} = 60'$, then $50'=\frac{50}{60}\approx0.833^{\circ}$. So $71^{\circ}50'\approx71.833^{\circ}$. To convert to radians, use the formula $\theta_{rad}=\theta_{deg}\times\frac{\pi}{180}$. So $\theta = 71.833\times\frac{\pi}{180}\approx1.254$ radians.

Step2: Use arc - length formula

The arc - length formula is $s = r\theta$, where $r$ is the radius of the pulley and $\theta$ is the angle in radians. Given $r = 9.28$ inches and $\theta\approx1.254$ radians. Then $s=9.28\times1.254 = 11.64712$ inches. Rounding to the nearest tenth, $s\approx11.6$ inches.

Step3: For part (b), use arc - length formula to find angle

We know $s = 3$ inches and $r = 9.28$ inches. From $s = r\theta$, we can solve for $\theta$ (in radians), so $\theta=\frac{s}{r}=\frac{3}{9.28}\approx0.323$ radians.

Step4: Convert radians to degrees and minutes

To convert radians to degrees, use $\theta_{deg}=\theta_{rad}\times\frac{180}{\pi}$. So $\theta_{deg}=0.323\times\frac{180}{\pi}\approx18.5^{\circ}$. Since $1^{\circ} = 60'$, the decimal part $0.5^{\circ}\times60 = 30'$. So the angle is $18^{\circ}30'$.

Answer:

(a) $11.6$
(b) $18^{\circ}30'$