ASVAB Mechanical Comprehension Practice Test 753266 Results

Your Results Global Average
Questions 5 5
Correct 0 2.97
Score 0% 59%

Review

1

The standard unit of energy is the:

73% Answer Correctly

Watt

Joule

Horsepower

Volt
Solution

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

2 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 230 lbs. load concentrated at a point 1.5 ft. from the axle?
52% Answer Correctly
76.7
690
-90.2
1566.8
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:
\( F_e = \frac{F_r d_r}{d_e} \)
\( F_e = \frac{230 \times 1.5}{0.5} \)
\( F_e = \frac{345.0}{0.5} \)
\( F_e = 690 \)
3

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

58% Answer Correctly

lever

gear

pulley

screw
Solution

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

4 If a 25 lbs. weight is placed 7 ft. from the fulcrum at the blue arrow and the green box is 2 ft. from the fulcrum, how much would the green box have to weigh to balance the lever?
61% Answer Correctly
3 lbs.
21.88 lbs.
87.5 lbs.
262.5 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 Ra, our missing value, and plugging in our variables yields:

Ra = \( \frac{R_bd_b}{d_a} \) = \( \frac{25 lbs. \times 7 ft.}{2 ft.} \) = \( \frac{175 ft⋅lb}{2 ft.} \) = 87.5 lbs.

5 The radius of the axle is 5, the radius of the wheel is 10, and the blue box weighs 65 lbs. What is the effort force necessary to balance the load?
53% Answer Correctly
15 lbs.
32.5 lbs.
20 lbs.
10 lbs.
Solution

The mechanical advantage of a wheel and axle is the input radius divided by the output radius:

MA = \( \frac{r_i}{r_o} \)

In this case, the input radius (where the effort force is being applied) is 10 and the output radius (where the resistance is being applied) is 5 for a mechanical advantage of \( \frac{10}{5} \) = 2.0

MA = \( \frac{load}{effort} \) so effort = \( \frac{load}{MA} \) = \( \frac{65 lbs.}{2.0} \) = 32.5 lbs.