ASVAB Mechanical Comprehension Practice Test 997334 Results

Your Results Global Average
Questions 5 5
Correct 0 3.06
Score 0% 61%

Review

1

Which of the following is not a type of simple machine?

58% Answer Correctly

lever

gear

screw

pulley
Solution

The six types of simple machines are the lever, wheel and axle, pulley, inclined plane, wedge, and screw.

2 30 lbs. of effort is used by a machine to lift a 150 lbs. box. What is the mechanical advantage of the machine?
84% Answer Correctly
1
5
6.5
8
Solution

Mechanical advantage is resistance force divided by effort force:

MA = \( \frac{F_r}{F_e} \) = \( \frac{150 lbs.}{30 lbs.} \) = 5

3

The mechanical advantage of a wheel and axle is equal to the:

61% Answer Correctly

difference in the diameters of the wheels

length of the axle

ratio of the diameters of the wheels

difference in the lengths of the axles
Solution

A wheel and axle uses two different diameter wheels mounted to a connecting axle. Force is applied to the larger wheel and large movements of this wheel result in small movements in the smaller wheel. Because a larger movement distance is being translated to a smaller distance, force is increased with a mechanical advantage equal to the ratio of the diameters of the wheels. An example of a wheel and axle is the steering wheel of a car.

4 How much work can a 6 hp engine do in 3 seconds?
52% Answer Correctly
2 ft⋅lb
12 ft⋅lb
24 ft⋅lb
9900 ft⋅lb
Solution
Horsepower (hp) is a common measure of power output for complex machines. By definition, a 1 hp machine does 550 ft⋅lb of work in 1 second: 1 hp = 550 ft⋅lb/s. Substituting the variables for this problem gives us:
\( W = 6 hp \times 550 \frac{ft⋅lb}{s} \times 3s = 9900 ft⋅lb \)
5 A 460 lb. barrel is rolled up a 19 ft. ramp to a platform that's 1 ft. tall. What effort is required to move the barrel?
53% Answer Correctly
24.2 lbs.
26.6 lbs.
22.2 lbs.
25.7 lbs.
Solution

This problem describes an inclined plane and, for an inclined plane, 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 19 ft. = 460 lbs. x 1 ft.
Fe = \( \frac{460 ft⋅lb}{19 ft.} \) = 24.2 lbs.