ASVAB Mechanical Comprehension Practice Test 236129 Results

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

Review

1 If this lever is in equilibrium with an effort force of 3.33 ft. lb. at the blue arrow and a resistance force of 2 ft. lb. at the green box, what is its mechanical advantage?
48% Answer Correctly
1.8
-7.4
0.6
0.2
Solution

Mechanical advantage (MA) is the ratio by which effort force relates to resistance force. If both forces are known, calculating MA is simply a matter of dividing resistance force by effort force:

MA = \( \frac{F_r}{F_e} \) = \( \frac{2 ft.}{3.33 ft.} \) = 0.6

In this case, the mechanical advantage is less than one meaning that each unit of effort force results in just 0.6 units of resistance force. However, a third class lever like this isn't designed to multiply force like a first class lever. A third class lever is designed to multiply distance and speed at the resistance by sacrificing force at the resistance. Different lever styles have different purposes and multiply forces in different ways.

2

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

38% Answer Correctly

driver gear

driven gear

output gear

idler 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.

3

The force exerted on an object due to gravity is called:

70% Answer Correctly

potential energy

mass

density

weight
Solution

Mass is an intrinsic property of matter and does not vary. Weight is the force exerted on the mass of an object due to gravity and a specific case of Newton's Second Law of Motion. Replace force with weight and acceleration with acceleration due to gravity on Earth (g) and the result is the formula for weight: W = mg or, substituting for g, weight equals mass multiplied by 9.8 m/s2.

4 How much work can a 3 hp engine do in 6 seconds?
52% Answer Correctly
12 ft⋅lb
6 ft⋅lb
9900 ft⋅lb
1 ft⋅lb
Solution
Horsepower (hp) is a common measure of power output for complex machines. By definition, a 1 hp machine does 550 ft⋅lb of work in 1 second: 1 hp = 550 ft⋅lb/s. Substituting the variables for this problem gives us:
\( W = 3 hp \times 550 \frac{ft⋅lb}{s} \times 6s = 9900 ft⋅lb \)
5 If a 25 lbs. weight is placed 1 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
2.08 lbs.
12.5 lbs.
4.17 lbs.
16.67 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 Ra, our missing value, and plugging in our variables yields:

Ra = \( \frac{R_bd_b}{d_a} \) = \( \frac{25 lbs. \times 1 ft.}{6 ft.} \) = \( \frac{25 ft⋅lb}{6 ft.} \) = 4.17 lbs.