ASVAB Mechanical Comprehension Practice Test 333143 Results

	Your Results	Global Average
Questions	5	5
Correct	0	3.12
Score	0%	62%

Review

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

	everywhere in the system
	the portions of the system at an altitude below the input
	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.

A box is resting on a smooth floor. Static friction is present:

59% Answer Correctly

	at all times
	if the coefficient of friction is greater than one
	only if normal force is present
	when an attempt is made to move the box

Solution

For any given surface, the coefficient of static friction is higher than the coefficient of kinetic friction. More force is required to initally get an object moving than is required to keep it moving. Additionally, static friction only arises in response to an attempt to move an object (overcome the normal force between it and the surface).

3 If a 5 lbs. weight is placed 3 ft. from the fulcrum at the blue arrow and the green box is 6 ft. from the fulcrum, how much would the green box have to weigh to balance the lever?

61% Answer Correctly

	0.63 lbs.
	15 lbs.
	1.25 lbs.
	2.5 lbs.

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

R_a = \( \frac{R_bd_b}{d_a} \) = \( \frac{5 lbs. \times 3 ft.}{6 ft.} \) = \( \frac{15 ft⋅lb}{6 ft.} \) = 2.5 lbs.

4 If the green arrow in this diagram represents 690 ft⋅lb of work, how far will the box move if it weighs 230 pounds?

72% Answer Correctly

	3 ft.
	460 ft.
	57 ft.
	12 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 690 ft⋅lb and the resistance force is 230 lbs. and we need to calculate the resistance distance:

W_in = F_resistance x d_resistance
690 ft⋅lb = 230 lbs. x d_resistance
d_resistance = \( \frac{690ft⋅lb}{230 lbs.} \) = 3 ft.

Coplanar forces:

62% Answer Correctly

	act in a common plane
	act along the same line of action
	have opposite dimensions
	pass through a common point

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.