ASVAB Mechanical Comprehension Practice Test 46939 Results

Your Results Global Average
Questions 5 5
Correct 0 2.81
Score 0% 56%

Review

1 How much work can a 8 hp engine do in 10 seconds?
52% Answer Correctly
1 ft⋅lb
2 ft⋅lb
4 ft⋅lb
44000 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:
\( W = 8 hp \times 550 \frac{ft⋅lb}{s} \times 10s = 44000 ft⋅lb \)
2

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

all of these are correct

the portions of the system at an altitude above the input

everywhere in the system

the portions of the system at an altitude below 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.

3 If this lever is in equilibrium with an effort force of 10.0 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
0.6
0.4
0.2
0.07
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 = \( \frac{F_r}{F_e} \) = \( \frac{2 ft.}{10.0 ft.} \) = 0.2

In this case, the mechanical advantage is less than one meaning that each unit of effort force results in just 0.2 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.

4 If the force applied at the blue arrow over 9 ft. moves the green box 1.8 ft., what is the mechanical advantage of this lever?
56% Answer Correctly
7.5
2
15
5
Solution

Mechanical advantage (MA) can be calculated knowing only the distance the effort (blue arrow) moves and the distance the resistance (green box) moves. The equation is:

MA = \( \frac{E_d}{R_d} \)

where Ed is the effort distance and Rd is the resistance distance. For this problem, the equation becomes:

MA = \( \frac{9 ft.}{1.8 ft.} \) = 5

You might be wondering how having an effort distance of 5 times the resistance distance is an advantage. Remember the principle of moments. For a lever in equilibrium the effort torque equals the resistance torque. Because torque is force x distance, if the effort distance is 5 times the resistance distance, the effort force must be \( \frac{1}{5} \) the resistance force. You're trading moving 5 times the distance for only having to use \( \frac{1}{5} \) the force.

5 A = 9 ft., the green box weighs 45 lbs., and the blue box weighs 65 lbs. What does distance B need to be for this lever to balance?
65% Answer Correctly
24.92 ft.
6.23 ft.
0 ft.
1.56 ft.
Solution
In order for this lever to balance, the torque acting on side A must equal the torque acting on side B. Torque is weight x distance from the fulcrum which means that the following must be true for the lever to balance:

fAdA = fBdB

For this problem, the equation becomes:

45 lbs. x 9 ft. = 65 lbs. x dB

dB = \( \frac{45 \times 9 ft⋅lb}{65 lbs.} \) = \( \frac{405 ft⋅lb}{65 lbs.} \) = 6.23 ft.