ASVAB Mechanical Comprehension Practice Test 129301 Results

Your Results Global Average
Questions 5 5
Correct 0 2.93
Score 0% 59%

Review

1 If A = 3 ft. and the green box weighs 25 lbs. what is the torque acting on the A side of this lever?
74% Answer Correctly
75 ft⋅lb
300 ft⋅lb
150 ft⋅lb
25 ft⋅lb
Solution
For a lever, torque is weight x distance from the fulcrum which, in this case, is: 25 ft. x 3 lbs. = 75 ft⋅lb
2

A box is resting on a smooth floor. Static friction is present:

60% Answer Correctly

only if normal force is present

at all times

when an attempt is made to move the box

if the coefficient of friction is greater than one
Solution

For any given surface, the coefficient of static friction is higher than the coefficient of kinetic friction. More force is required to initally get an object moving than is required to keep it moving. Additionally, static friction only arises in response to an attempt to move an object (overcome the normal force between it and the surface).

3

Assuming force applied remains constant, which of the following will result in more work being done?

53% Answer Correctly

moving the object farther

increasing the coefficient of friction

moving the object with more speed

moving the object with more acceleration
Solution

Work is accomplished when force is applied to an object: W = Fd where F is force in newtons (N) and d is distance in meters (m). Thus, the more force that must be applied to move an object, the more work is done and the farther an object is moved by exerting force, the more work is done.

4 If the handles of a wheelbarrow are 2.0 ft. from the wheel axle, how many pounds of force must you exert to lift the handles if it's carrying a 210 lbs. load concentrated at a point 0.5 ft. from the axle?
52% Answer Correctly
0
420
52.5
840
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:
\( F_e = \frac{F_r d_r}{d_e} \)
\( F_e = \frac{210 \times 0.5}{2.0} \)
\( F_e = \frac{105.0}{2.0} \)
\( F_e = 52.5 \)
5 If the radius of the axle is 5 and the radius of the wheel is 10, what is the mechanical advantage of this wheel and axle configuration?
52% Answer Correctly
2.0
-5
0.5
10
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 10 and the output radius (where the resistance is being applied) is 5 for a mechanical advantage of \( \frac{10}{5} \) = 2.0