ASVAB Mechanical Comprehension Practice Test 937453 Results

Your Results Global Average
Questions 5 5
Correct 0 3.15
Score 0% 63%

Review

1

What defines the mechanical advantage of a first class lever?

65% Answer Correctly

position of the fulcrum

input force

output distance

output force 
Solution

A first-class lever is used to increase force or distance while changing the direction of the force. The lever pivots on a fulcrum and, when a force is applied to the lever at one side of the fulcrum, the other end moves in the opposite direction. The position of the fulcrum also defines the mechanical advantage of the lever. If the fulcrum is closer to the force being applied, the load can be moved a greater distance at the expense of requiring a greater input force. If the fulcrum is closer to the load, less force is required but the force must be applied over a longer distance. An example of a first-class lever is a seesaw / teeter-totter.

2 If the green box weighs 35 lbs. and is 4 ft. from the fulcrum, how much force would need to be applied at the blue arrow to balance the lever if the arrow's distance from the fulcrum is 7 ft.?
62% Answer Correctly
60 lbs.
10 lbs.
20 lbs.
0 lbs.
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 Rb, our missing value, and plugging in our variables yields:

Rb = \( \frac{R_ad_a}{d_b} \) = \( \frac{35 lbs. \times 4 ft.}{7 ft.} \) = \( \frac{140 ft⋅lb}{7 ft.} \) = 20 lbs.

3 If the handles of a wheelbarrow are 1.5 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 1.0 ft. from the axle?
52% Answer Correctly
140
572.6
0
98.7
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:
\( F_e = \frac{F_r d_r}{d_e} \)
\( F_e = \frac{210 \times 1.0}{1.5} \)
\( F_e = \frac{210.0}{1.5} \)
\( F_e = 140 \)
4 What's the mechanical advantage of a wedge that's 3 inches wide and 18 inches long?
83% Answer Correctly
6
3
6.6
9
Solution

The mechanical advantage (MA) of a wedge is its length divided by its thickness:

MA = \( \frac{l}{t} \) = \( \frac{18 in.}{3 in.} \) = 6

5 The radius of the axle is 5, the radius of the wheel is 10, and the blue box weighs 45 lbs. What is the effort force necessary to balance the load?
53% Answer Correctly
22.5 lbs.
5 lbs.
20 lbs.
50 lbs.
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

MA = \( \frac{load}{effort} \) so effort = \( \frac{load}{MA} \) = \( \frac{45 lbs.}{2.0} \) = 22.5 lbs.