ASVAB Mechanical Comprehension Simple Machines Practice Test 687282 Results

Your Results Global Average
Questions 5 5
Correct 0 2.81
Score 0% 56%

Review

1 If the handles of a wheelbarrow are 2.0 ft. from the wheel axle, how many pounds of force must you exert to lift the handles if it's carrying a 190 lbs. load concentrated at a point 1.0 ft. from the axle?
52% Answer Correctly
None of these is correct
58
0
95
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:
\( F_e = \frac{F_r d_r}{d_e} \)
\( F_e = \frac{190 \times 1.0}{2.0} \)
\( F_e = \frac{190.0}{2.0} \)
\( F_e = 95 \)
2 If the radius of the axle is 3 and the radius of the wheel is 6, what is the mechanical advantage of this wheel and axle configuration?
52% Answer Correctly
-3
2.0
3
0.5
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 6 and the output radius (where the resistance is being applied) is 3 for a mechanical advantage of \( \frac{6}{3} \) = 2.0

3 The radius of the axle is 5, the radius of the wheel is 6, and the blue box weighs 45 lbs. What is the effort force necessary to balance the load?
53% Answer Correctly
5 lbs.
6 lbs.
30 lbs.
37.5 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 6 and the output radius (where the resistance is being applied) is 5 for a mechanical advantage of \( \frac{6}{5} \) = 1.2

MA = \( \frac{load}{effort} \) so effort = \( \frac{load}{MA} \) = \( \frac{45 lbs.}{1.2} \) = 37.5 lbs.

4 If the green box weighs 20 lbs. and is 9 ft. from the fulcrum, how far from the fulcrum would a 75 lbs. weight need to be placed to balance the lever?
61% Answer Correctly
2.4 ft.
1.2 ft.
0 ft.
0.6 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 db, our missing value, and plugging in our variables yields:

db = \( \frac{R_ad_a}{R_b} \) = \( \frac{20 lbs. \times 9 ft.}{75 lbs.} \) = \( \frac{180 ft⋅lb}{75 lbs.} \) = 2.4 ft.

5 If the green box weighs 75 lbs. and is 7 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 8 ft.?
63% Answer Correctly
32.81 lbs.
65.63 lbs.
16.41 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 Rb, our missing value, and plugging in our variables yields:

Rb = \( \frac{R_ad_a}{d_b} \) = \( \frac{75 lbs. \times 7 ft.}{8 ft.} \) = \( \frac{525 ft⋅lb}{8 ft.} \) = 65.63 lbs.