ASVAB Mechanical Comprehension Practice Test 687774 Results

Your Results Global Average
Questions 5 5
Correct 0 3.09
Score 0% 62%

Review

1 A = 7 ft., the green box weighs 25 lbs., and the blue box weighs 40 lbs. What does distance B need to be for this lever to balance?
65% Answer Correctly
1.09 ft.
280 ft.
17.5 ft.
4.38 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:

25 lbs. x 7 ft. = 40 lbs. x dB

dB = \( \frac{25 \times 7 ft⋅lb}{40 lbs.} \) = \( \frac{175 ft⋅lb}{40 lbs.} \) = 4.38 ft.

2

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

53% Answer Correctly

static friction

normal friction

coefficient of friction

kinetic 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 The green box weighs 30 lbs. and a 25 lbs. weight is placed 7 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
11.67 ft.
2.92 ft.
5.83 ft.
23.33 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 = \( \frac{R_bd_b}{R_a} \) = \( \frac{25 lbs. \times 7 ft.}{30 lbs.} \) = \( \frac{175 ft⋅lb}{30 lbs.} \) = 5.83 ft.

4 60 lbs. of effort is used by a machine to lift a 480 lbs. box. What is the mechanical advantage of the machine?
84% Answer Correctly
4
10
2
8
Solution

Mechanical advantage is resistance force divided by effort force:

MA = \( \frac{F_r}{F_e} \) = \( \frac{480 lbs.}{60 lbs.} \) = 8

5 How much work can a 1 hp engine do in 9 seconds?
52% Answer Correctly
9 ft⋅lb
4950 ft⋅lb
0 ft⋅lb
2 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:
\( W = 1 hp \times 550 \frac{ft⋅lb}{s} \times 9s = 4950 ft⋅lb \)