ASVAB Mechanical Comprehension Practice Test 499703 Results

Your Results Global Average
Questions 5 5
Correct 0 2.85
Score 0% 57%

Review

1

What's the first gear in a gear train called?

57% Answer Correctly

idler gear

input gear

driver gear

driven gear
Solution

A gear train is two or more gears linked together. Gear trains are designed to increase or reduce the speed or torque outpout of a rotating system or change the direction of its output. The first gear in the chain is called the driver and the last gear in the chain the driven gear with the gears between them called idler gears.

2 If the green box weighs 75 lbs. and is 7 ft. from the fulcrum, how much weight would need to be placed at the blue arrow to balance the lever if the arrow's distance from the fulcrum is 6 ft.?
63% Answer Correctly
12 lbs.
87.5 lbs.
262.5 lbs.
21.88 lbs.
Solution

To balance this lever the torques on each side of the fulcrum must be equal. Torque is weight x distance from the fulcrum so the equation for equilibrium is:

Rada = Rbdb

where a represents the left side of the fulcrum and b the right, R is resistance (weight) and d is the distance from the fulcrum.

Solving for Rb, our missing value, and plugging in our variables yields:

Rb = \( \frac{R_ad_a}{d_b} \) = \( \frac{75 lbs. \times 7 ft.}{6 ft.} \) = \( \frac{525 ft⋅lb}{6 ft.} \) = 87.5 lbs.

3

Which class of lever is used to increase force on an object in the same direction as the force is applied?

53% Answer Correctly

third

first

second

all of these
Solution

A second-class lever is used to increase force on an object in the same direction as the force is applied. This lever requires a smaller force to lift a larger load but the force must be applied over a greater distance. The fulcrum is placed at one end of the lever and mechanical advantage increases as the object being lifted is moved closer to the fulcrum or the length of the lever is increased. An example of a second-class lever is a wheelbarrow.

4

If the handles of a wheelbarrow are 3 ft. from the wheel axle, what force must you exert to lift the handles if it's carrying a 270 lb. load concentrated at a point 0.5 ft. from the axle?

56% Answer Correctly

90 lbs

0.83 lbs

810 lbs

45 lbs
Solution

This problem describes a second-class lever and, for a second class lever, the effort force multiplied by the effort distance equals the resistance force multipied by the resistance distance: Fede = Frdr. Plugging in the variables from this problem yields:

Fe x 3 ft. = 270 lbs x 0.5 ft
Fe = 135 ft-lb. / 3 ft 
Fe = 45 lbs

5 If the force applied at the blue arrow over 6 ft. moves the green box 0.67 ft., what is the mechanical advantage of this lever?
56% Answer Correctly
27
9.9
9
18
Solution

Mechanical advantage (MA) can be calculated knowing only the distance the effort (blue arrow) moves and the distance the resistance (green box) moves. The equation is:

MA = \( \frac{E_d}{R_d} \)

where Ed is the effort distance and Rd is the resistance distance. For this problem, the equation becomes:

MA = \( \frac{6 ft.}{0.67 ft.} \) = 9

You might be wondering how having an effort distance of 9 times the resistance distance is an advantage. Remember the principle of moments. For a lever in equilibrium the effort torque equals the resistance torque. Because torque is force x distance, if the effort distance is 9 times the resistance distance, the effort force must be \( \frac{1}{9} \) the resistance force. You're trading moving 9 times the distance for only having to use \( \frac{1}{9} \) the force.