ASVAB Mechanical Comprehension Practice Test 154962 Results

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

Review

Which of these is the formula for force?

76% Answer Correctly

	F = m/a
	F = am²
	F = a/m
	F = ma

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.

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

45% Answer Correctly

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

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.

3 If the green box weighs 60 lbs. and is 3 ft. from the fulcrum, how far from the fulcrum would a 50 lbs. weight need to be placed to balance the lever?

61% Answer Correctly

	3.6 ft.
	0.9 ft.
	180 ft.
	20 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 = \( \frac{R_ad_a}{R_b} \) = \( \frac{60 lbs. \times 3 ft.}{50 lbs.} \) = \( \frac{180 ft⋅lb}{50 lbs.} \) = 3.6 ft.

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
	45 lbs
	810 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: F_ed_e = F_rd_r. Plugging in the variables from this problem yields:

F_e x 3 ft. = 270 lbs x 0.5 ft
F_e = 135 ft-lb. / 3 ft
F_e= 45 lbs

Torque involves a perpendicular force applied to a lever arm that moves around a center of rotation. Increasing the length of the lever arm will do which of the following?

54% Answer Correctly

	increase torque
	decrease applied force
	increase applied force
	decrease torque

Solution

Torque measures force applied during rotation: τ = rF. Torque (τ, the Greek letter tau) = the radius of the lever arm (r) multiplied by the force (F) applied. Radius is measured from the center of rotation or fulcrum to the point at which the perpendicular force is being applied. The resulting unit for torque is newton-meter (N-m) or foot-pound (ft-lb).