ASVAB Mechanical Comprehension Practice Test 648579 Results

Your Results Global Average
Questions 5 5
Correct 0 2.77
Score 0% 55%

Review

1 If a 55 lbs. weight is placed 5 ft. from the fulcrum at the blue arrow and the green box is 6 ft. from the fulcrum, how much would the green box have to weigh to balance the lever?
61% Answer Correctly
11.46 lbs.
0 lbs.
45.83 lbs.
91.67 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 = \( \frac{R_bd_b}{d_a} \) = \( \frac{55 lbs. \times 5 ft.}{6 ft.} \) = \( \frac{275 ft⋅lb}{6 ft.} \) = 45.83 lbs.

2

Which of the following represents how much two materials resist sliding across each other?

54% Answer Correctly

kinetic friction

static friction

normal friction

coefficient of friction
Solution

Coefficient of friction (μ) represents how much two materials resist sliding across each other.  Smooth surfaces like ice have low coefficients of friction while rough surfaces like concrete have high μ.

3

Specific gravity is a comparison of the density of an object with the density of:

57% Answer Correctly

oil

water

air

carbon
Solution

Specific gravity is the ratio of the density of equal volumes of a substance and water and is measured by a hyrdometer.

4 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 270 lbs. load concentrated at a point 0.5 ft. from the axle?
52% Answer Correctly
None of these is correct
1080
135
270
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:
\( F_e = \frac{F_r d_r}{d_e} \)
\( F_e = \frac{270 \times 0.5}{0.5} \)
\( F_e = \frac{135.0}{0.5} \)
\( F_e = 270 \)
5 The radius of the axle is 3, the radius of the wheel is 8, and the blue box weighs 85 lbs. What is the effort force necessary to balance the load?
53% Answer Correctly
31.84 lbs.
21.36 lbs.
24 lbs.
8.01 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 8 and the output radius (where the resistance is being applied) is 3 for a mechanical advantage of \( \frac{8}{3} \) = 2.67

MA = \( \frac{load}{effort} \) so effort = \( \frac{load}{MA} \) = \( \frac{85 lbs.}{2.67} \) = 31.84 lbs.