ASVAB Mechanical Comprehension Practice Test 132844 Results

Your Results Global Average
Questions 5 5
Correct 0 2.54
Score 0% 51%

Review

1

The mechanical advantage of a third class lever is always:

37% Answer Correctly

equal to one

less than one

not equal to one

greater than one
Solution

A third class lever is designed to multiply distance and speed at the expense of effort force. Because the effort force is greater than the resistance, the mechanical advantage of a third class lever is always less than one.

An example of a third class lever is a broom. The fulcrum is at your hand on the end of the broom, the effort force is your other hand in the middle, and the resistance is at the bottom bristles. The effort force of your hand in the middle multiplies the distance and speed of the bristles at the bottom but at the expense of producing a brushing force that's less than the force you're applying with your hand.

2

Which of the following is the formula for gravitational potential energy?

61% Answer Correctly

\(PE = mgh\)

\(PE = mg^2h\)

\(PE = { 1 \over 2} mg^2\)

\(PE = { 1 \over 2} mv^2\)
Solution

Gravitational potential energy is energy by virtue of gravity. The higher an object is raised above a surface the greater the distance it must fall to reach that surface and the more velocity it will build as it falls. For gravitational potential energy, PE = mgh where m is mass (kilograms), h is height (meters), and g is acceleration due to gravity which is a constant (9.8 m/s2).

3 If the handles of a wheelbarrow are 2.5 ft. from the wheel axle, how many pounds of force must you exert to lift the handles if it's carrying a 170 lbs. load concentrated at a point 1.0 ft. from the axle?
52% Answer Correctly
68
122.6
9
8.5
Solution
This problem describes a second-class lever and, for a second class lever, the effort force multiplied by the effort distance equals the resistance force multipied by the resistance distance: Fede = Frdr. In this problem we're looking for effort force:
\( F_e = \frac{F_r d_r}{d_e} \)
\( F_e = \frac{170 \times 1.0}{2.5} \)
\( F_e = \frac{170.0}{2.5} \)
\( F_e = 68 \)
4 If the radius of the axle is 3 and the radius of the wheel is 8, what is the mechanical advantage of this wheel and axle configuration?
41% Answer Correctly
0.38
8
2.67
3
Solution

The mechanical advantage of a wheel and axle is the input radius divided by the output radius:

MA = \( \frac{r_i}{r_o} \)

In this case, the input radius (where the effort force is being applied) is 3 and the output radius (where the resistance is being applied) is 8 for a mechanical advantage of \( \frac{3}{8} \) = 0.38

5 If the green box weighs 45 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
11 lbs.
78.75 lbs.
236.25 lbs.
180 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 = \( \frac{R_ad_a}{d_b} \) = \( \frac{45 lbs. \times 7 ft.}{4 ft.} \) = \( \frac{315 ft⋅lb}{4 ft.} \) = 78.75 lbs.