ASVAB Mechanical Comprehension Other Forces Practice Test 918330 Results

Your Results Global Average
Questions 5 5
Correct 0 3.38
Score 0% 68%

Review

1 A = 6 ft., the green box weighs 35 lbs., and the blue box weighs 65 lbs. What does distance B need to be for this lever to balance?
65% Answer Correctly
9.69 ft.
3.23 ft.
210 ft.
12.92 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:

35 lbs. x 6 ft. = 65 lbs. x dB

dB = \( \frac{35 \times 6 ft⋅lb}{65 lbs.} \) = \( \frac{210 ft⋅lb}{65 lbs.} \) = 3.23 ft.

2 If A = 1 ft. and the green box weighs 40 lbs. what is the torque acting on the A side of this lever?
76% Answer Correctly
160 ft⋅lb
0 ft⋅lb
40 ft⋅lb
13 ft⋅lb
Solution
For a lever, torque is weight x distance from the fulcrum which, in this case, is: 40 ft. x 1 lbs. = 40 ft⋅lb
3 A = 6 ft., the green box weighs 35 lbs., and the blue box weighs 65 lbs. What does distance B need to be for this lever to balance?
65% Answer Correctly
9.69 ft.
3.23 ft.
210 ft.
12.92 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:

35 lbs. x 6 ft. = 65 lbs. x dB

dB = \( \frac{35 \times 6 ft⋅lb}{65 lbs.} \) = \( \frac{210 ft⋅lb}{65 lbs.} \) = 3.23 ft.

4 A = 6 ft., the green box weighs 35 lbs., and the blue box weighs 65 lbs. What does distance B need to be for this lever to balance?
65% Answer Correctly
9.69 ft.
3.23 ft.
210 ft.
12.92 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:

35 lbs. x 6 ft. = 65 lbs. x dB

dB = \( \frac{35 \times 6 ft⋅lb}{65 lbs.} \) = \( \frac{210 ft⋅lb}{65 lbs.} \) = 3.23 ft.

5 A = 6 ft., the green box weighs 35 lbs., and the blue box weighs 65 lbs. What does distance B need to be for this lever to balance?
65% Answer Correctly
9.69 ft.
3.23 ft.
210 ft.
12.92 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:

35 lbs. x 6 ft. = 65 lbs. x dB

dB = \( \frac{35 \times 6 ft⋅lb}{65 lbs.} \) = \( \frac{210 ft⋅lb}{65 lbs.} \) = 3.23 ft.