A jackscrew system has a stationary base containing a standard right-hand threaded nut. Th… — Armed Services Vocational Aptitude Battery (ASVAB)

A jackscrew system has a stationary base containing a standard right-hand threaded nut. The screw, which has 4 threads per inch (TPI), supports a platform on its top. A technician uses a 1.25-foot long handle to turn the screw and lift a $720 \text{ lb}$ load exactly $2 \text{ inches}$ . Assuming $\pi \approx 3$ to simplify calculations and $100\%$ mechanical efficiency, which of the following correctly identifies the required rotation direction, effort force, and total work done?

Counterclockwise rotation; 2 lbs effort force; 1440 in-lbs total workAnswer
B
Clockwise rotation; 2 lbs effort force; 1440 in-lbs total work
C
Counterclockwise rotation; 24 lbs effort force; 1440 in-lbs total work
D
Counterclockwise rotation; 2 lbs effort force; 4 in-lbs total work

Answer

Counterclockwise rotation; 2 lbs effort force; 1440 in-lbs total work

Lifting the load requires backing the right-hand screw out of its nut, which means turning it counterclockwise. The IMA is calculated by matching units:

R = 15 \text{ inches}

and Pitch

= 0.25 \text{ inches}

. Using

\text{IMA} = \frac{2\pi R}{Pitch}

, the IMA is

\frac{90}{0.25} = 360

. The required effort force is the load divided by the IMA (

720 / 360 = 2 \text{ lbs}

). Finally, simple machines do not change total work; Work Out = Load

\times

Distance (

720 \text{ lbs} \times 2 \text{ inches} = 1440 \text{ in-lbs}

Step-by-Step Solution

Determine the required rotation direction for the screw.

Counterclockwise rotation.

A standard right-hand thread advances (moves inward/downward) when turned clockwise. To lift a load, the screw must back out of the base, which requires counterclockwise rotation.

Convert the handle length to inches and calculate the Ideal Mechanical Advantage (IMA).

R = 1.25 \text{ ft} = 15 \text{ inches}

. Pitch (

P

) =

\frac{1}{4} = 0.25 \text{ inches}

\text{IMA} = \frac{2\pi R}{P} = \frac{2(3)(15)}{0.25} = 360

Units must match (both in inches) to accurately calculate the mechanical advantage. IMA relates the circumference traveled by the effort to the linear pitch.

Calculate the required effort force.

\text{Effort Force} = \frac{\text{Load}}{\text{IMA}} = \frac{720 \text{ lbs}}{360} = 2 \text{ lbs}

The machine multiplies the effort force by a factor equal to the IMA (assuming

100\%

efficiency).

Calculate the total work done.

\text{Work} = \text{Load} \times \text{Vertical Distance} = 720 \text{ lbs} \times 2 \text{ inches} = 1440 \text{ in-lbs}

In any ideal machine, Work In equals Work Out. The total work performed against gravity depends only on the load weight and the distance it is lifted, regardless of the mechanical advantage.

Key Concept

Mechanics of Screws: Pitch, Mechanical Advantage, and Conservation of Work

Estimated Time:2m 0s

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