ASVAB Mechanical Comprehension Practice Test 734228 Results

	Your Results	Global Average
Questions	5	5
Correct	0	2.92
Score	0%	58%

Review

A fixed pulley is useful for which of the following?

53% Answer Correctly

	changing the direction of the input force
	multiplying the input force
	multiplying the input distance
	changing the direction of the output force

Solution

A fixed pulley is used to change the direction of a force and does not multiply the force applied. As such, it has a mechanical advantage of one. The benefit of a fixed pulley is that it can allow the force to be applied at a more convenient angle, for example, pulling downward or horizontally to lift an object instead of upward.

2 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 150 lbs. force lift?

47% Answer Correctly

	121.5
	-63
	150
	300

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:
\( F_r = \frac{F_e d_e}{d_r} \)
\( F_r = \frac{150 \times 1.0}{0.5} \)
\( F_r = \frac{150.0}{0.5} \)
\( F_r = 300 \)

3 If 45 lbs. of force is applied 5 ft. from the fulcrum at the blue arrow and the green box is 1 ft. from the fulcrum, how much would the green box have to weigh to balance the lever?

62% Answer Correctly

	450 lbs.
	56.25 lbs.
	675 lbs.
	225 lbs.

Solution

To balance this lever the torques at the green box and the blue arrow 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 green box and b the blue arrow, R is resistance (weight/force) and d is the distance from the fulcrum.

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

R_a = \( \frac{R_bd_b}{d_a} \) = \( \frac{45 lbs. \times 5 ft.}{1 ft.} \) = \( \frac{225 ft⋅lb}{1 ft.} \) = 225 lbs.

Concurrent forces:

55% Answer Correctly

	act in a common plane
	act along the same line of action
	pass through a common point
	act in a common dimension

Solution

Collinear forces act along the same line of action, concurrent forces pass through a common point and coplanar forces act in a common plane.

What type of load acts on a relatively small area of a structure?

75% Answer Correctly

	concentrated load
	impact load
	non-uniformly distributed load
	dynamic 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.