ASVAB Mechanical Comprehension Practice Test 405071 Results

	Your Results	Global Average
Questions	5	5
Correct	0	2.59
Score	0%	52%

Review

1 A 490 lb. barrel is rolled up a 13 ft. ramp to a platform that's 3 ft. tall. What effort is required to move the barrel?

53% Answer Correctly

	116.1 lbs.
	118.1 lbs.
	339.2 lbs.
	113.1 lbs.

Solution

This problem describes an inclined plane and, for an inclined plane, the effort force multiplied by the effort distance equals the resistance force multipied by the resistance distance:

F_ed_e = F_rd_r

Plugging in the variables from this problem yields:

F_e x 13 ft. = 490 lbs. x 3 ft.
F_e = \( \frac{1470 ft⋅lb}{13 ft.} \) = 113.1 lbs.

What defines the mechanical advantage of a first class lever?

65% Answer Correctly

	output force
	position of the fulcrum
	input force
	output distance

Solution

A first-class lever is used to increase force or distance while changing the direction of the force. The lever pivots on a fulcrum and, when a force is applied to the lever at one side of the fulcrum, the other end moves in the opposite direction. The position of the fulcrum also defines the mechanical advantage of the lever. If the fulcrum is closer to the force being applied, the load can be moved a greater distance at the expense of requiring a greater input force. If the fulcrum is closer to the load, less force is required but the force must be applied over a longer distance. An example of a first-class lever is a seesaw / teeter-totter.

3 If A = 9 ft., B = 1 ft., C = 4 ft., the green box weighs 50 lbs. and the blue box weighs 55 lbs., what does the orange box have to weigh for this lever to balance?

44% Answer Correctly

	49.38 lbs.
	98.75 lbs.
	50 lbs.
	296.25 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:

f_Ad_A = f_Bd_B + f_Cd_C

For this problem, this equation becomes:

50 lbs. x 9 ft. = 55 lbs. x 1 ft. + f_C x 4 ft.

450 ft. lbs. = 55 ft. lbs. + f_C x 4 ft.

f_C = \( \frac{450 ft. lbs. - 55 ft. lbs.}{4 ft.} \) = \( \frac{395 ft. lbs.}{4 ft.} \) = 98.75 lbs.

Boyle's law defines the relationship between pressure and volume as:

57% Answer Correctly

	\(\frac{P_1}{P_2} = \frac{V_1}{V_2}\)
	\(\frac{P_1}{P_2} = {V_1}{V_2}\)
	\(\frac{P_1}{P_2} = \frac{V_2}{V_1}\)
	\({P_1}{P_2} = {V_1}{V_2}\)

Solution

Boyle's law states that "for a fixed amount of an ideal gas kept at a fixed temperature, pressure and volume are inversely proportional". Expressed as a formula, that's \(\frac{P_1}{P_2} = \frac{V_2}{V_1}\)

The advantage of using a third-class lever is that it increases:

37% Answer Correctly

	the distance traveled by the load
	the speed of the load
	the force applied to the load
	the mechanical advantage of the lever

Solution

A third-class lever is used to increase distance traveled by an object in the same direction as the force applied. The fulcrum is at one end of the lever, the object at the other, and the force is applied between them. This lever does not impart a mechanical advantage as the effort force must be greater than the load but does impart extra speed to the load. Examples of third-class levers are shovels and tweezers.