ASVAB Mechanical Comprehension Practice Test 342018 Results

	Your Results	Global Average
Questions	5	5
Correct	0	2.75
Score	0%	55%

Review

Which of the following is not a type of structural load?

50% Answer Correctly

	dead load
	live load
	wind load
	occupancy load

Solution

Dead load is the weight of the building and materials, live load is additional weight due to occupancy or use, snow load is the weight of accumulated snow on a structure and wind load is the force of wind pressures against structure surfaces.

2 If the green box weighs 60 lbs. and is 7 ft. from the fulcrum, how far from the fulcrum would a 60 lbs. weight need to be placed to balance the lever?

61% Answer Correctly

	21 ft.
	1.75 ft.
	7 ft.
	3.5 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:

R_ad_a = R_bd_b

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 d_b, our missing value, and plugging in our variables yields:

d_b = \( \frac{R_ad_a}{R_b} \) = \( \frac{60 lbs. \times 7 ft.}{60 lbs.} \) = \( \frac{420 ft⋅lb}{60 lbs.} \) = 7 ft.

3 A = 7 ft., the green box weighs 40 lbs., and the blue box weighs 70 lbs. What does distance B need to be for this lever to balance?

65% Answer Correctly

	12 ft.
	4 ft.
	1.33 ft.
	1 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:

f_Ad_A = f_Bd_B

For this problem, the equation becomes:

40 lbs. x 7 ft. = 70 lbs. x d_B

d_B = \( \frac{40 \times 7 ft⋅lb}{70 lbs.} \) = \( \frac{280 ft⋅lb}{70 lbs.} \) = 4 ft.

Which of the following will increase the mechanical advantage of a second-class lever?

55% Answer Correctly

	move the object being lifted closer to the fulcrum
	move the object being lifted farther away from the fulcrum
	decrease the length of the lever
	move the fulcrum between the force and the object being lifted

Solution

A second-class lever is used to increase force on an object in the same direction as the force is applied. This lever requires a smaller force to lift a larger load but the force must be applied over a greater distance. The fulcrum is placed at one end of the lever and mechanical advantage increases as the object being lifted is moved closer to the fulcrum or the length of the lever is increased. An example of a second-class lever is a wheelbarrow.

Depending on where you apply effort and resistance, the wheel and axle can multiply:

45% Answer Correctly

	speed or power
	force or distance
	force or speed
	power or distance

Solution

If you apply the resistance to the axle and the effort to the wheel, the wheel and axle will multiply force and if you apply the resistance to the wheel and the effort to the axle, it will multiply speed.