ASVAB Mechanical Comprehension Practice Test 807243 Results

Your Results Global Average
Questions 5 5
Correct 0 3.01
Score 0% 60%

Review

1

Tension is a force that does which of the following?

75% Answer Correctly

slows an object

stretches an object

compacts an object

heats up an object


Solution

Tension is a force that stretches or elongates something. When a cable or rope is used to pull an object, for example, it stretches internally as it accepts the weight that it's moving. Although tension is often treated as applying equally to all parts of a material, it's greater at the places where the material is under the most stress.


2

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

57% Answer Correctly

driven gear

input gear

idler gear

driver 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.


3 If the green box weighs 30 lbs. and is 4 ft. from the fulcrum, how much force would need to be applied at the blue arrow to balance the lever if the arrow's distance from the fulcrum is 5 ft.?
62% Answer Correctly
12 lbs.
8 lbs.
24 lbs.
0 lbs.

Solution

To balance this lever the torques at the green box and the blue arrow must be equal. Torque is weight x distance from the fulcrum so the equation for equilibrium is:

Rada = Rbdb

where a represents the green box and b the blue arrow, R is resistance (weight/force) and d is the distance from the fulcrum.

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

Rb = \( \frac{R_ad_a}{d_b} \) = \( \frac{30 lbs. \times 4 ft.}{5 ft.} \) = \( \frac{120 ft⋅lb}{5 ft.} \) = 24 lbs.


4 A 310 lb. barrel is rolled up a 7 ft. ramp to a platform that's 4 ft. tall. What effort is required to move the barrel?
53% Answer Correctly
177.1 lbs.
265.7 lbs.
178.6 lbs.
180.1 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 7 ft. = 310 lbs. x 4 ft.
Fe = \( \frac{1240 ft⋅lb}{7 ft.} \) = 177.1 lbs.


5 If the handles of a wheelbarrow are 0.5 ft. from the wheel axle, how many pounds of force must you exert to lift the handles if it's carrying a 150 lbs. load concentrated at a point 1.5 ft. from the axle?
52% Answer Correctly
None of these is correct
450
919.1
225

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. In this problem we're looking for effort force:
\( F_e = \frac{F_r d_r}{d_e} \)
\( F_e = \frac{150 \times 1.5}{0.5} \)
\( F_e = \frac{225.0}{0.5} \)
\( F_e = 450 \)