ASVAB Mechanical Comprehension Practice Test 103070 Results

Your Results Global Average
Questions 5 5
Correct 0 2.90
Score 0% 58%

Review

1

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

53% Answer Correctly

first

all of these

second

third


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.


2 If the green box is 5 ft. from the fulcrum and a certain force applied 9 ft. from the fulcrum at the blue arrow balances the lever, what is the mechanical advantage?
61% Answer Correctly
5.4
2.7
1.8
0.9

Solution

Because this lever is in equilibrium, we know that the effort force at the blue arrow is equal to the resistance weight of the green box. For a lever that's in equilibrium, one method of calculating mechanical advantage (MA) is to divide the length of the effort arm (Ea) by the length of the resistance arm (Ra):

MA = \( \frac{E_a}{R_a} \) = \( \frac{9 ft.}{5 ft.} \) = 1.8

When a lever is in equilibrium, the torque from the effort and the resistance are equal. The equation for equilibrium is Rada = Rbdb where a and b are the two points at which effort/resistance is being applied to the lever.

In this problem, Ra and Rb are such that the lever is in equilibrium meaning that some multiple of the weight of the green box is being applied at the blue arrow. For a lever, this multiple is a function of the ratio of the distances of the box and the arrow from the fulcrum. That's why, for a lever in equilibrium, only the distances from the fulcrum are necessary to calculate mechanical advantage.

If the lever were not in equilibrium, you would first have to calculate the forces and distances necessary to put it in equilibrium and then divide Ea by Ra to get the mechanical advantage.


3

The principle of moments defines equilibrium in terms of:

53% Answer Correctly

power

speed

torque

energy


Solution

According to the principle of moments, you can maintain equilibrium if the moments (forces) tending to clockwise rotation are equal to the moments tending to counterclockwise rotation. Another name for these moments of force is torque.


4

A a seesaw / teeter-totter is an example of which of the following?

69% Answer Correctly

third-class lever

inclined plane

first-class lever

second-class lever


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.


5 If the green box weighs 65 lbs. and 45 lbs. of force is applied 7 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?
55% Answer Correctly
1.21 ft.
4.85 ft.
2.42 ft.
9.69 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 da, our missing value, and plugging in our variables yields:

da = \( \frac{R_bd_b}{R_a} \) = \( \frac{45 lbs. \times 7 ft.}{65 lbs.} \) = \( \frac{315 ft⋅lb}{65 lbs.} \) = 4.85 ft.