ASVAB Mechanical Comprehension Practice Test 75708 Results

Your Results Global Average
Questions 5 5
Correct 0 2.68
Score 0% 54%

Review

1

What's the last gear in a gear train called?

38% Answer Correctly

driven gear

driver gear

idler gear

output gear
Solution

A gear train is two or more gears linked together. Gear trains are designed to increase or reduce the speed or torque outpout of a rotating system or change the direction of its output. The first gear in the chain is called the driver and the last gear in the chain the driven gear with the gears between them called idler gears.

2

When all forces acting on a system cancel each other out, this is called:

80% Answer Correctly

rest

potential energy

stasis

equilibrium
Solution

When a system is stable or balanced (equilibrium) all forces acting on the system cancel each other out. In the case of torque, equilibrium means that the sum of the anticlockwise moments about a center of rotation equal the sum of the clockwise moments.

3

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

49% Answer Correctly

occupancy load

dead load

live load

wind 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.

4 What is the power output of a 5 hp engine that's 35% efficient?
39% Answer Correctly
2887.5 \( \frac{ft⋅lb}{s} \)
1925 \( \frac{ft⋅lb}{s} \)
962.5 \( \frac{ft⋅lb}{s} \)
481.3 \( \frac{ft⋅lb}{s} \)
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 \( \frac{ft⋅lb}{s} \), Pi becomes 5 hp x 550 \( \frac{ft⋅lb}{s} \) = 2750 \( \frac{ft⋅lb}{s} \)
\( P_{o} = \frac{E \times P_{i}}{100} = \frac{35 \times 2750 \frac{ft⋅lb}{s}}{100} \) \( = \frac{96250 \frac{ft⋅lb}{s}}{100} \) = 962.5 \( \frac{ft⋅lb}{s} \)
5 If the green box weighs 25 lbs. and is 7 ft. from the fulcrum, how much weight would need to be placed at the blue arrow to balance the lever if the arrow's distance from the fulcrum is 4 ft.?
63% Answer Correctly
6 lbs.
14.58 lbs.
87.5 lbs.
43.75 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:

Rada = Rbdb

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

Rb = \( \frac{R_ad_a}{d_b} \) = \( \frac{25 lbs. \times 7 ft.}{4 ft.} \) = \( \frac{175 ft⋅lb}{4 ft.} \) = 43.75 lbs.