ASVAB Mechanical Comprehension Practice Test 200288 Results

Your Results Global Average
Questions 5 5
Correct 0 2.94
Score 0% 59%

Review

1 The green box weighs 75 lbs. and a 20 lbs. weight is placed 1 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?
57% Answer Correctly
0 ft.
0.09 ft.
0.53 ft.
0.27 ft.
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 da, our missing value, and plugging in our variables yields:

da = \( \frac{R_bd_b}{R_a} \) = \( \frac{20 lbs. \times 1 ft.}{75 lbs.} \) = \( \frac{20 ft⋅lb}{75 lbs.} \) = 0.27 ft.

2 What is the efficiency of a machine has work input of 195 ft⋅lb and work output of 126 ft⋅lb?
67% Answer Correctly
260%
16%
65%
0%
Solution
Due to friction, a machine will never be able to utilize 100% of its work input. A certain percentage of that input will be lost in overcoming friction within the machine. Effeciency is a measure of how much of a machine's work input can be turned into useful work output and is calculated by dividing work output by work input and multiplying the result by 100:
\( Efficiency = \frac{Work_{out}}{Work_{in}} \times 100 \) \( = \frac{126 ft⋅lb}{195 ft⋅lb} \times 100 \) \( = 65% \) %
3 If the green box weighs 75 lbs. and is 5 ft. from the fulcrum, how far from the fulcrum would a 50 lbs. force need to be applied to balance the lever?
58% Answer Correctly
22.5 ft.
0 ft.
15 ft.
7.5 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 db, our missing value, and plugging in our variables yields:

db = \( \frac{R_ad_a}{R_b} \) = \( \frac{75 lbs. \times 5 ft.}{50 lbs.} \) = \( \frac{375 ft⋅lb}{50 lbs.} \) = 7.5 ft.

4 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 120 lbs. load concentrated at a point 1.0 ft. from the axle?
52% Answer Correctly
0
80
-25.8
180
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:
\( F_e = \frac{F_r d_r}{d_e} \)
\( F_e = \frac{120 \times 1.0}{1.5} \)
\( F_e = \frac{120.0}{1.5} \)
\( F_e = 80 \)
5

The work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle. This defines which of the following?

60% Answer Correctly

conservation of mechanical energy

mechanical advantage

work-energy theorem

Pascal's law
Solution

The work-energy theorem states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle. Simply put, work imparts kinetic energy to the matter upon which the work is being done.