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 76°50? (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: Recall arc - length formula

The formula for the arc - length $s$ of a circle is $s = r\theta$, where $r$ is the radius of the circle and $\theta$ is the central angle in radians.

Step2: Convert angle to radians for part (a)

The given angle $\theta=76^{\circ}50'$. First, convert $50'$ to degrees: $50'=\frac{50}{60}\approx0.833^{\circ}$, so $\theta = 76.833^{\circ}$. Then convert to radians: $\theta = 76.833\times\frac{\pi}{180}\text{ radians}$. The radius $r = 9.01$ inches. Using the arc - length formula $s=r\theta$, we have $s=9.01\times76.833\times\frac{\pi}{180}$.
\[s = 9.01\times\frac{76.833\pi}{180}\approx9.01\times\frac{76.833\times3.14159}{180}\approx12.1\]

Step3: Solve for the angle in part (b)

We know $s = 6$ inches and $r = 9.01$ inches. From $s = r\theta$, we can solve for $\theta$ (in radians): $\theta=\frac{s}{r}=\frac{6}{9.01}\text{ radians}$. Then convert to degrees and minutes. $\theta=\frac{6}{9.01}\times\frac{180}{\pi}\text{ degrees}$.
\[\theta=\frac{6\times180}{9.01\pi}\text{ degrees}\approx\frac{1080}{9.01\times3.14159}\text{ degrees}\approx38.2^{\circ}\]
$0.2^{\circ}\times60 = 12'$, so $\theta\approx38^{\circ}12'$

Answer:

(a) $12.1$ inches
(b) $38^{\circ}12'$