ASVAB Mechanical Comprehension Practice Test 481522 Results

Your Results Global Average
Questions 5 5
Correct 0 3.30
Score 0% 66%

Review

1 A = 5 ft., the green box weighs 35 lbs., and the blue box weighs 80 lbs. What does distance B need to be for this lever to balance?
65% Answer Correctly
0.55 ft.
16 ft.
2.19 ft.
175 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:

35 lbs. x 5 ft. = 80 lbs. x dB

dB = \( \frac{35 \times 5 ft⋅lb}{80 lbs.} \) = \( \frac{175 ft⋅lb}{80 lbs.} \) = 2.19 ft.

2

Friction resists movement in a direction __________ to the movement.

81% Answer Correctly

opposite

perpendicular

normal

parallel
Solution

Friction resists movement. Kinetic (also called sliding or dynamic) friction resists movement in a direction opposite to the movement. Because it opposes movement, kinetic friction will eventually bring an object to a stop. An example is a rock that's sliding across ice.

3

What type of load acts on a relatively small area of a structure?

75% Answer Correctly

concentrated load

dynamic load

non-uniformly distributed load

impact load
Solution

A concentrated load acts on a relatively small area of a structure, a static uniformly distributed load doesn't create specific stress points or vary with time, a dynamic load varies with time or affects a structure that experiences a high degree of movement, an impact load is sudden and for a relatively short duration and a non-uniformly distributed load creates different stresses at different locations on a structure.

4 If the green box weighs 5 lbs. and 65 lbs. of force is applied 5 ft. from the fulcrum at the blue arrow, how far from the fulcrum would the green box need to be placed to balance the lever?
55% Answer Correctly
32.5 ft.
1 ft.
65 ft.
25 ft.
Solution

To balance this lever the torques at the green box and the blue arrow must be equal. Torque is weight x distance from the fulcrum so the equation for equilibrium is:

Rada = Rbdb

where a represents the green box and b the blue arrow, R is resistance (weight/force) and d is the distance from the fulcrum.

Solving for da, our missing value, and plugging in our variables yields:

da = \( \frac{R_bd_b}{R_a} \) = \( \frac{65 lbs. \times 5 ft.}{5 lbs.} \) = \( \frac{325 ft⋅lb}{5 lbs.} \) = 65 ft.

5 How much work can a 7 hp engine do in 6 seconds?
52% Answer Correctly
23100 ft⋅lb
28 ft⋅lb
1 ft⋅lb
14 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:
\( W = 7 hp \times 550 \frac{ft⋅lb}{s} \times 6s = 23100 ft⋅lb \)