ASVAB Mechanical Comprehension Practice Test 774869 Results

Your Results Global Average
Questions 5 5
Correct 0 2.85
Score 0% 57%

Review

1

Collinear forces:

72% Answer Correctly

are unrelated to each other

act along the same line of action

pass through a common point

act in a common plane
Solution

Collinear forces act along the same line of action, concurrent forces pass through a common point and coplanar forces act in a common plane.

2 If the green box weighs 45 lbs. and is 6 ft. from the fulcrum, how much force would need to be applied at the blue arrow to balance the lever if the arrow's distance from the fulcrum is 8 ft.?
62% Answer Correctly
135 lbs.
5 lbs.
33.75 lbs.
101.25 lbs.
Solution

To balance this lever the torques at the green box and the blue arrow must be equal. Torque is weight x distance from the fulcrum so the equation for equilibrium is:

Rada = Rbdb

where a represents the green box and b the blue arrow, R is resistance (weight/force) and d is the distance from the fulcrum.

Solving for Rb, our missing value, and plugging in our variables yields:

Rb = \( \frac{R_ad_a}{d_b} \) = \( \frac{45 lbs. \times 6 ft.}{8 ft.} \) = \( \frac{270 ft⋅lb}{8 ft.} \) = 33.75 lbs.

3 If A = 10 ft., B = 1 ft., C = 8 ft., the green box weighs 40 lbs. and the blue box weighs 50 lbs., what does the orange box have to weigh for this lever to balance?
43% Answer Correctly
14.58 lbs.
4 lbs.
43.75 lbs.
131.25 lbs.
Solution
In order for this lever to balance, the torque acting on each side of the fulrum must be equal. So, the torque produced by A must equal the torque produced by B and C. Torque is weight x distance from the fulcrum which means that the following must be true for the lever to balance:

fAdA = fBdB + fCdC

For this problem, this equation becomes:

40 lbs. x 10 ft. = 50 lbs. x 1 ft. + fC x 8 ft.

400 ft. lbs. = 50 ft. lbs. + fC x 8 ft.

fC = \( \frac{400 ft. lbs. - 50 ft. lbs.}{8 ft.} \) = \( \frac{350 ft. lbs.}{8 ft.} \) = 43.75 lbs.

4

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

58% Answer Correctly

lever

gear

screw

pulley
Solution

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

5 If the handles of a wheelbarrow are 1.0 ft. from the wheel axle and the load is concentrated at a point 1.0 ft. from the axle, how many pounds of load will a 220 lbs. force lift?
47% Answer Correctly
321.5
-31.1
220
0
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 resistance force:
\( F_r = \frac{F_e d_e}{d_r} \)
\( F_r = \frac{220 \times 1.0}{1.0} \)
\( F_r = \frac{220.0}{1.0} \)
\( F_r = 220 \)