ASVAB Mechanical Comprehension Practice Test 485457 Results

	Your Results	Global Average
Questions	5	5
Correct	0	3.39
Score	0%	68%

Review

1 90 lbs. of effort is used by a machine to lift a 270 lbs. box. What is the mechanical advantage of the machine?

84% Answer Correctly

	-1
	9
	3
	3.3

Solution

Mechanical advantage is resistance force divided by effort force:

MA = \( \frac{F_r}{F_e} \) = \( \frac{270 lbs.}{90 lbs.} \) = 3

Friction between two or more solid objects that are not moving relative to each other is called:

74% Answer Correctly

	static friction
	gravitational friction
	dynamic friction
	kinetic friction

Solution

Static friction is friction between two or more solid objects that are not moving relative to each other. An example is the friction that prevents a box on a sloped surface from sliding farther down the surface.

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

74% Answer Correctly

	impact load
	dynamic load
	concentrated load
	non-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.

4 The radius of the axle is 5, the radius of the wheel is 8, and the blue box weighs 50 lbs. What is the effort force necessary to balance the load?

53% Answer Correctly

	31.25 lbs.
	9.6 lbs.
	13 lbs.
	12.8 lbs.

Solution

The mechanical advantage of a wheel and axle is the input radius divided by the output radius:

MA = \( \frac{r_i}{r_o} \)

In this case, the input radius (where the effort force is being applied) is 8 and the output radius (where the resistance is being applied) is 5 for a mechanical advantage of \( \frac{8}{5} \) = 1.6

MA = \( \frac{load}{effort} \) so effort = \( \frac{load}{MA} \) = \( \frac{50 lbs.}{1.6} \) = 31.25 lbs.

5 If the handles of a wheelbarrow are 2.5 ft. from the wheel axle, how many pounds of force must you exert to lift the handles if it's carrying a 240 lbs. load concentrated at a point 1.5 ft. from the axle?

52% Answer Correctly

	-56
	360
	144
	600

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:
\( F_e = \frac{F_r d_r}{d_e} \)
\( F_e = \frac{240 \times 1.5}{2.5} \)
\( F_e = \frac{360.0}{2.5} \)
\( F_e = 144 \)