ASVAB Mechanical Comprehension Practice Test 477685 Results

Your Results Global Average
Questions 5 5
Correct 0 2.99
Score 0% 60%

Review

1

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

90 lbs

810 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

2 If A = 12 ft., B = 1 ft., C = 10 ft., the green box weighs 50 lbs. and the blue box weighs 60 lbs., what does the orange box have to weigh for this lever to balance?
43% Answer Correctly
600 lbs.
162 lbs.
54 lbs.
4 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:

50 lbs. x 12 ft. = 60 lbs. x 1 ft. + fC x 10 ft.

600 ft. lbs. = 60 ft. lbs. + fC x 10 ft.

fC = \( \frac{600 ft. lbs. - 60 ft. lbs.}{10 ft.} \) = \( \frac{540 ft. lbs.}{10 ft.} \) = 54 lbs.

3 If the green box weighs 55 lbs. and is 5 ft. from the fulcrum, how far from the fulcrum would a 35 lbs. force need to be applied to balance the lever?
58% Answer Correctly
1.96 ft.
7.86 ft.
0 ft.
3.93 ft.
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 db, our missing value, and plugging in our variables yields:

db = \( \frac{R_ad_a}{R_b} \) = \( \frac{55 lbs. \times 5 ft.}{35 lbs.} \) = \( \frac{275 ft⋅lb}{35 lbs.} \) = 7.86 ft.

4

What's the first gear in a gear train called?

57% Answer Correctly

input gear

driven gear

idler gear

driver gear
Solution

A gear train is two or more gears linked together. Gear trains are designed to increase or reduce the speed or torque outpout of a rotating system or change the direction of its output. The first gear in the chain is called the driver and the last gear in the chain the driven gear with the gears between them called idler gears.

5 40 lbs. of effort is used by a machine to lift a 80 lbs. box. What is the mechanical advantage of the machine?
84% Answer Correctly
1.8
3
6
2
Solution

Mechanical advantage is resistance force divided by effort force:

MA = \( \frac{F_r}{F_e} \) = \( \frac{80 lbs.}{40 lbs.} \) = 2