ASVAB Mechanical Comprehension Practice Test 451548 Results

Your Results Global Average
Questions 5 5
Correct 0 3.37
Score 0% 67%

Review

1

Which of the following surfaces would have the lowest coefficient of friction?

86% Answer Correctly

leather

concrete

tile

ice
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.

2 If you have a gear train with two gears, the first with 36 teeth and the second with 8 teeth, how many revolutions does the second gear make for each revolution of the first gear?
78% Answer Correctly
2.3
4.5
6.5
8.5
Solution

The gear ratio (Vr) of a gear train is the product of the gear ratios between the pairs of meshed gears. Let N represent the number of teeth for each gear:

Vr = \( \frac{N_1}{N_2} \) \( \frac{N_2}{N_3} \) \( \frac{N_3}{N_4} \) ... \( \frac{N_n}{N_{n+1}} \)

In this problem, we have only two gears so the equation becomes:

Vr = \( \frac{N_1}{N_2} \) = \( \frac{36}{8} \) = 4.5

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

810 lbs

90 lbs

45 lbs

0.83 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 How much work can a 2 hp engine do in 8 seconds?
52% Answer Correctly
1 ft⋅lb
0 ft⋅lb
8800 ft⋅lb
8 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:
\( W = 2 hp \times 550 \frac{ft⋅lb}{s} \times 8s = 8800 ft⋅lb \)
5 If the green box weighs 35 lbs. and is 1 ft. from the fulcrum, how much weight would need to be placed at the blue arrow to balance the lever if the arrow's distance from the fulcrum is 2 ft.?
63% Answer Correctly
17.5 lbs.
35 lbs.
4.38 lbs.
70 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 Rb, our missing value, and plugging in our variables yields:

Rb = \( \frac{R_ad_a}{d_b} \) = \( \frac{35 lbs. \times 1 ft.}{2 ft.} \) = \( \frac{35 ft⋅lb}{2 ft.} \) = 17.5 lbs.