ASVAB Mechanical Comprehension Practice Test 576044 Results

	Your Results	Global Average
Questions	5	5
Correct	0	3.21
Score	0%	64%

Review

1 If the handles of a wheelbarrow are 1.0 ft. from the wheel axle and the load is concentrated at a point 0.5 ft. from the axle, how many pounds of load will a 130 lbs. force lift?

47% Answer Correctly

	260
	130
	37
	0

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 resistance force:
\( F_r = \frac{F_e d_e}{d_r} \)
\( F_r = \frac{130 \times 1.0}{0.5} \)
\( F_r = \frac{130.0}{0.5} \)
\( F_r = 260 \)

Which of these is the formula for force?

77% Answer Correctly

	F = ma
	F = m/a
	F = a/m
	F = am²

Solution

Newton's Second Law of Motion states that "The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object." This Law describes the linear relationship between mass and acceleration when it comes to force and leads to the formula F = ma or force equals mass multiplied by rate of acceleration.

3 The green box weighs 25 lbs. and a 20 lbs. weight is placed 9 ft. from the fulcrum at the blue arrow. How far from the fulcrum would the green box need to be placed to balance the lever?

57% Answer Correctly

	2.4 ft.
	14.4 ft.
	7.2 ft.
	28.8 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_a, our missing value, and plugging in our variables yields:

d_a = \( \frac{R_bd_b}{R_a} \) = \( \frac{20 lbs. \times 9 ft.}{25 lbs.} \) = \( \frac{180 ft⋅lb}{25 lbs.} \) = 7.2 ft.

Normal force is generally equal to the __________ of an object.

61% Answer Correctly

	coefficient of friction
	mass
	density
	weight

Solution

Normal force arises on a flat horizontal surface in response to an object's weight pressing it down. Consequently, normal force is generally equal to the object's weight.

5 If the green arrow in this diagram represents 600 ft⋅lb of work, how far will the box move if it weighs 150 pounds?

72% Answer Correctly

	4 ft.
	8 ft.
	600 ft.
	16 ft.

Solution

The Law of Work states that the work put into a machine is equal to the work received from the machine under ideal conditions. In equation form, that's:

W_in = W_out
F_effort x d_effort = F_resistance x d_resistance

In this problem, the effort work is 600 ft⋅lb and the resistance force is 150 lbs. and we need to calculate the resistance distance:

W_in = F_resistance x d_resistance
600 ft⋅lb = 150 lbs. x d_resistance
d_resistance = \( \frac{600ft⋅lb}{150 lbs.} \) = 4 ft.