ASVAB Mechanical Comprehension Practice Test 594516 Results

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

Review

1

Which of the following is the formula for torque?

61% Answer Correctly

τ = F/r2

τ = r/F

τ = rF

τ = F/r
Solution

Torque measures force applied during rotation: τ = rF.  Torque (τ, the Greek letter tau) = the radius of the lever arm (r) multiplied by the force (F) applied. Radius is measured from the center of rotation or fulcrum to the point at which the perpendicular force is being applied. The resulting unit for torque is newton-meter (N-m) or foot-pound (ft-lb).

2

Which class of lever is used to increase force on an object in the same direction as the force is applied?

52% Answer Correctly

first

all of these

third

second
Solution

A second-class lever is used to increase force on an object in the same direction as the force is applied. This lever requires a smaller force to lift a larger load but the force must be applied over a greater distance. The fulcrum is placed at one end of the lever and mechanical advantage increases as the object being lifted is moved closer to the fulcrum or the length of the lever is increased. An example of a second-class lever is a wheelbarrow.

3

An inclined plane increases ___________ to reduce ____________.

58% Answer Correctly

force, power

distance, force

distance, power

force, distance
Solution

An inclined plane is a simple machine that reduces the force needed to raise an object to a certain height. Work equals force x distance and, by increasing the distance that the object travels, an inclined plane reduces the force necessary to raise it to a particular height. In this case, the mechanical advantage is to make the task easier. An example of an inclined plane is a ramp.

4

If the handles of a wheelbarrow are 3 ft. from the wheel axle, what force must you exert to lift the handles if it's carrying a 270 lb. load concentrated at a point 0.5 ft. from the axle?

56% Answer Correctly

810 lbs

0.83 lbs

45 lbs

90 lbs
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. Plugging in the variables from this problem yields:

Fe x 3 ft. = 270 lbs x 0.5 ft
Fe = 135 ft-lb. / 3 ft 
Fe = 45 lbs

5 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 200 lbs. load concentrated at a point 0.5 ft. from the axle?
52% Answer Correctly
50
61
200
-88.1
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:
\( F_e = \frac{F_r d_r}{d_e} \)
\( F_e = \frac{200 \times 0.5}{2.0} \)
\( F_e = \frac{100.0}{2.0} \)
\( F_e = 50 \)