ASVAB Mechanical Comprehension Practice Test 775830 Results

Your Results Global Average
Questions 5 5
Correct 0 3.12
Score 0% 62%

Review

1

The standard unit of energy is the:

73% Answer Correctly

Watt

Joule

Volt

Horsepower
Solution

The Joule (J) is the standard unit of energy and has the unit \({kg \times m^2} \over s^2\).

2 The green box weighs 35 lbs. and a 70 lbs. weight is placed 1 ft. from the fulcrum at the blue arrow. How far from the fulcrum would the green box need to be placed to balance the lever?
57% Answer Correctly
0 ft.
2 ft.
0.5 ft.
1 ft.
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 da, our missing value, and plugging in our variables yields:

da = \( \frac{R_bd_b}{R_a} \) = \( \frac{70 lbs. \times 1 ft.}{35 lbs.} \) = \( \frac{70 ft⋅lb}{35 lbs.} \) = 2 ft.

3

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

45 lbs

810 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

4 A = 9 ft., the green box weighs 35 lbs., and the blue box weighs 65 lbs. What does distance B need to be for this lever to balance?
65% Answer Correctly
3 ft.
9.69 ft.
1.62 ft.
4.85 ft.
Solution
In order for this lever to balance, the torque acting on side A must equal the torque acting on side B. Torque is weight x distance from the fulcrum which means that the following must be true for the lever to balance:

fAdA = fBdB

For this problem, the equation becomes:

35 lbs. x 9 ft. = 65 lbs. x dB

dB = \( \frac{35 \times 9 ft⋅lb}{65 lbs.} \) = \( \frac{315 ft⋅lb}{65 lbs.} \) = 4.85 ft.

5

What type of load creates different stresses at different locations on a structure?

60% Answer Correctly

non-uniformly distributed load

static uniformly distributed load

dynamic load

impact load
Solution

A concentrated load acts on a relatively small area of a structure, a static uniformly distributed load doesn't create specific stress points or vary with time, a dynamic load varies with time or affects a structure that experiences a high degree of movement, an impact load is sudden and for a relatively short duration and a non-uniformly distributed load creates different stresses at different locations on a structure.