Question

a screw jack is used to lift a small car. the handle used to turn the screw has a length (radius) of 20 cm. the screw advances 0.5 cm for each full rotation (pitch of the screw). the worker applies an effort force of 150 n at the handle. part a. determine the ideal mechanical advantage (ima) of the screw. show your work to receive full credit. hint: ima = circumference of handle’s circle / pitch of screw type response here g: 20,.5,150 u: ima e: circumference of handles circle/pitch of screw s:

Explanation:

Step1: Calculate circumference of handle's circle

The formula for the circumference of a circle is $C = 2\pi r$. Given $r = 20$ cm, so $C=2\pi\times20 = 40\pi$ cm.

Step2: Calculate Ideal Mechanical Advantage (IMA)

Using the formula $IMA=\frac{\text{Circumference of Handle's Circle}}{\text{ Pitch of Screw}}$. Substitute $C = 40\pi$ cm and pitch = 0.5 cm. So $IMA=\frac{40\pi}{0.5}=\frac{40\times3.14}{0.5}=251.2$.

Answer:

251.2