ASVAB Mechanical Comprehension Practice Test 333301 Results

	Your Results	Global Average
Questions	5	5
Correct	0	2.75
Score	0%	55%

Review

1 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 100 lbs. load concentrated at a point 1.0 ft. from the axle?

52% Answer Correctly

	40
	35.3
	77.5
	59.2

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:
\( F_e = \frac{F_r d_r}{d_e} \)
\( F_e = \frac{100 \times 1.0}{2.5} \)
\( F_e = \frac{100.0}{2.5} \)
\( F_e = 40 \)

Which of these will have the most impact on the kinetic energy of an object?

54% Answer Correctly

	its speed
	its weight
	its direction
	its mass

Solution

Kinetic energy is the energy of movement and is a function of the mass of an object and its speed: \(KE = {1 \over 2}mv^2\) where m is mass in kilograms, v is speed in meters per second, and KE is in joules. The most impactful quantity to kinetic energy is velocity as an increase in mass increases KE linearly while an increase in speed increases KE exponentially.

For a hydraulic system, pressure applied to the input of the system will increase the pressure in which parts of the system?

58% Answer Correctly

	the portions of the system at an altitude below the input
	everywhere in the system
	all of these are correct
	the portions of the system at an altitude above the input

Solution

Pascal's law states that a pressure change occurring anywhere in a confined incompressible fluid is transmitted throughout the fluid such that the same change occurs everywhere. For a hydraulic system, this means that a pressure applied to the input of the system will increase the pressure everywhere in the system.

4 If the green box weighs 35 lbs. and is 7 ft. from the fulcrum, how far from the fulcrum would a 75 lbs. force need to be applied to balance the lever?

58% Answer Correctly

	5 ft.
	3.27 ft.
	1.63 ft.
	13.07 ft.

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

d_b = \( \frac{R_ad_a}{R_b} \) = \( \frac{35 lbs. \times 7 ft.}{75 lbs.} \) = \( \frac{245 ft⋅lb}{75 lbs.} \) = 3.27 ft.

Which class of lever is used to increase force on an object in the same direction as the force is applied?

53% Answer Correctly

	first
	all of these
	second
	third

Solution

A second-class lever is used to increase force on an object in the same direction as the force is applied. This lever requires a smaller force to lift a larger load but the force must be applied over a greater distance. The fulcrum is placed at one end of the lever and mechanical advantage increases as the object being lifted is moved closer to the fulcrum or the length of the lever is increased. An example of a second-class lever is a wheelbarrow.