ASVAB Mechanical Comprehension Practice Test 328841 Results

	Your Results	Global Average
Questions	5	5
Correct	0	2.71
Score	0%	54%

Review

1 If a 30 lbs. weight is placed 5 ft. from the fulcrum at the blue arrow and the green box is 4 ft. from the fulcrum, how much would the green box have to weigh to balance the lever?

61% Answer Correctly

	9.38 lbs.
	37.5 lbs.
	112.5 lbs.
	6 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 = \( \frac{R_bd_b}{d_a} \) = \( \frac{30 lbs. \times 5 ft.}{4 ft.} \) = \( \frac{150 ft⋅lb}{4 ft.} \) = 37.5 lbs.

Which of the following will increase the mechanical advantage of this inclined plane?

59% Answer Correctly

	lengthen the ramp
	shorten the ramp
	lower the force acting at the blue arrow
	increase the force acting at the blue arrow

Solution

The mechanical advantage (MA) of an inclined plane is the effort distance divided by the resistance distance. In order to increase mechanical advantage, this ratio must increase which means making the effort distance longer and this can be accomplished by lengthening the length of the ramp.

Which of the following is not a characteristic of a ceramic?

61% Answer Correctly

	low density
	chemically stable
	low corrosive action
	high melting point

Solution

Ceramics are mixtures of metallic and nonmetallic elements that withstand exteme thermal, chemical, and pressure environments. They have a high melting point, low corrosive action, and are chemically stable. Examples include rock, sand, clay, glass, brick, and porcelain.

4 What is the power output of a 6 hp engine that's 50% efficient?

40% Answer Correctly

	8 \( \frac{ft⋅lb}{s} \)
	300 \( \frac{ft⋅lb}{s} \)
	6600 \( \frac{ft⋅lb}{s} \)
	1650 \( \frac{ft⋅lb}{s} \)

Solution

\( Efficiency = \frac{Power_{out}}{Power_{in}} \times 100 \)
Solving for power out: \( P_{o} = \frac{E \times P_{i}}{100} \)
Knowing that 1 hp = 550 \( \frac{ft⋅lb}{s} \), P_i becomes 6 hp x 550 \( \frac{ft⋅lb}{s} \) = 3300 \( \frac{ft⋅lb}{s} \)
\( P_{o} = \frac{E \times P_{i}}{100} = \frac{50 \times 3300 \frac{ft⋅lb}{s}}{100} \) \( = \frac{165000 \frac{ft⋅lb}{s}}{100} \) = 1650 \( \frac{ft⋅lb}{s} \)

Which of the following is not a type of structural load?

49% Answer Correctly

	dead load
	occupancy load
	wind load
	live load

Solution

Dead load is the weight of the building and materials, live load is additional weight due to occupancy or use, snow load is the weight of accumulated snow on a structure and wind load is the force of wind pressures against structure surfaces.