Question

1.5 pulley drives and
sprockets practice problems
pltw
pulleys and belts
in a pulley system, pulley a is moving at 1500 rpm and has a diameter of 15 in.
three pulleys, b, c, and d, all of different sizes, are attached to a single output axle.
speed and torque output are changed within the system by moving the drive belt
between pulleys b, c, and d.

1. sketch and annotate the drive train described above.
2. a speed of 1750 rpm is required when the drive belt is connected to pulley b. what

is the diameter of pulley b?

Explanation:

Step1: Write the formula for pulley speed and diameter

The relationship between the speed (\(N\)) and diameter (\(D\)) of two pulleys is given by \(N_1D_1 = N_2D_2\). Here, \(N_1\) and \(D_1\) are the speed and diameter of pulley \(A\), and \(N_2\) and \(D_2\) are the speed and diameter of pulley \(B\).

Step2: Substitute the known values into the formula

We know that \(N_1 = 1500\) rpm, \(D_1=15\) in, and \(N_2 = 1750\) rpm. Substituting these values into the formula \(N_1D_1 = N_2D_2\), we get \(1500\times15=1750\times D_2\).

Step3: Solve for \(D_2\)

First, calculate \(1500\times15 = 22500\). Then, the equation becomes \(22500=1750\times D_2\). To find \(D_2\), we use \(D_2=\frac{22500}{1750}\). Simplifying \(\frac{22500}{1750}=\frac{2250}{175}=\frac{450}{35}=\frac{90}{7}\approx12.86\) in.

Answer:

The diameter of pulley \(B\) is approximately \(12.86\) inches.