Question

Refer to the figure to the right.

9.67 in

(a) How many inches will the weight in the figure rise if the
pulley is rotated through an angle of 77° 50'?

(b) Through what angle, to the nearest minute, must the
pulley be rotated to raise the weight 4 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.)

Answer 1

Step-by-step explanation:

Use arc length:

s = rθ

where r is the radius and θ is the angle in radians.

a) First, convert the angle to radians (1 degree = 60 minutes).

77 ⁵⁰/₆₀° × (π radian / 180°) = 1.358 radians

s = rθ

s = (9.67 in) (1.358 rad)

s = 13.1 in

b) s = rθ

4 in = (9.67 in) θ

θ = 0.414 radians

θ = 23.7°

θ = 23° 42'

Answer 2

The weight attached to the pulley is an illustration of arc length.

• The weight in the figure will rise 13.2 inches
• The angle of rotation is $23^o, 42'$

The given parameters are:

$r = 9.67in$ --- the radius

(a) The distance moved by the weight

Here, we have:

$\theta = 77^o50'$

Convert to degrees

$\theta = 77 + \frac{50}{60}$

$\theta = 77 + \frac{5}{6}$

$\theta = \frac{77 \times 6 + 5}{6}$

$\theta = \frac{467}{6}$

Convert to radians

$\theta = \frac{467}{6} \times \frac{\pi}{180} rad$

$\theta = 1.36\ rad$

The distance moved is calculated using the length of an arc formula

$L =r\theta$

So, we have:

$L =9.67in \times 1.36$

$L =13.1512\ in$

$L =13.2\ in$ --- approximated

The weight in the figure will rise 13.2 inches

(b) The angle, when the weight raises 4  inches

$L =r\theta$

This means that:

$L = 4$

So, we have:

$4 = 9.67 \times \theta$

Solve for $\theta$

$\theta = \frac{4}{9.67}$

$\theta = 0.4136$

Convert to degrees

$\theta =0.4136 \times \frac{180}{\pi}$

$\theta =23.70^o$

Convert to minutes

$\theta =23^o , 0.70 \times 60'$

$\theta =23^o , 42'$

The angle of rotation is $23^o, 42'$

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