ASVAB Mechanical Comprehension Practice Test 43414 Results

	Your Results	Global Average
Questions	5	5
Correct	0	2.93
Score	0%	59%

Review

Power is the rate at which:

62% Answer Correctly

	potential energy is converted into kinetic energy
	work is done
	friction is overcome
	input force is transferred to output force

Solution

Power is the rate at which work is done, P = w/t, or work per unit time. The watt (W) is the unit for power and is equal to 1 joule (or newton-meter) per second. Horsepower (hp) is another familiar unit of power used primarily for rating internal combustion engines. 1 hp equals 746 watts.

What type of load doesn't create specific stress points or vary with time?

60% Answer Correctly

	concentrated load
	non-uniformly distributed load
	impact load
	static uniformly distributed load

Solution

A concentrated load acts on a relatively small area of a structure, a static uniformly distributed load doesn't create specific stress points or vary with time, a dynamic load varies with time or affects a structure that experiences a high degree of movement, an impact load is sudden and for a relatively short duration and a non-uniformly distributed load creates different stresses at different locations on a structure.

Which of the following will increase the mechanical advantage of this inclined plane?

58% Answer Correctly

	lower the force acting at the blue arrow
	shorten the ramp
	increase the force acting at the blue arrow
	lengthen the ramp

Solution

The mechanical advantage (MA) of an inclined plane is the effort distance divided by the resistance distance. In order to increase mechanical advantage, this ratio must increase which means making the effort distance longer and this can be accomplished by lengthening the length of the ramp.

4 If a 5 lbs. weight is placed 3 ft. from the fulcrum at the blue arrow and the green box is 6 ft. from the fulcrum, how much would the green box have to weigh to balance the lever?

61% Answer Correctly

	2.5 lbs.
	1 lbs.
	1.25 lbs.
	7.5 lbs.

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

R_a = \( \frac{R_bd_b}{d_a} \) = \( \frac{5 lbs. \times 3 ft.}{6 ft.} \) = \( \frac{15 ft⋅lb}{6 ft.} \) = 2.5 lbs.

5 If the handles of a wheelbarrow are 0.5 ft. from the wheel axle, how many pounds of force must you exert to lift the handles if it's carrying a 140 lbs. load concentrated at a point 1.0 ft. from the axle?

52% Answer Correctly

	70
	0
	-4
	280

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: F_ed_e = F_rd_r. In this problem we're looking for effort force:
\( F_e = \frac{F_r d_r}{d_e} \)
\( F_e = \frac{140 \times 1.0}{0.5} \)
\( F_e = \frac{140.0}{0.5} \)
\( F_e = 280 \)