ASVAB Mechanical Comprehension Practice Test 330205 Results

	Your Results	Global Average
Questions	5	5
Correct	0	2.68
Score	0%	54%

Review

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
	third
	all of these
	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.

2 A = 9 ft., the green box weighs 20 lbs., and the blue box weighs 35 lbs. What does distance B need to be for this lever to balance?

65% Answer Correctly

	0 ft.
	5.14 ft.
	180 ft.
	20.57 ft.

Solution

In order for this lever to balance, the torque acting on side A must equal the torque acting on side B. Torque is weight x distance from the fulcrum which means that the following must be true for the lever to balance:

f_Ad_A = f_Bd_B

For this problem, the equation becomes:

20 lbs. x 9 ft. = 35 lbs. x d_B

d_B = \( \frac{20 \times 9 ft⋅lb}{35 lbs.} \) = \( \frac{180 ft⋅lb}{35 lbs.} \) = 5.14 ft.

Force of friction due to kinetic friction is __________ the force of friction due to static friction.

41% Answer Correctly

	the same as
	higher than
	opposite
	lower than

Solution

The formula for force of friction (F_f) is the same whether kinetic or static friction applies: F_{f =}μF_N. To distinguish between kinetic and static friction, μ_k and μ_s are often used in place of μ.

Which of the following is not true of a first-class lever?

51% Answer Correctly

	increases force
	changes the direction of force
	increases distance
	decreases distance

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 is 8 ft. from the fulcrum and a certain force applied 6 ft. from the fulcrum at the blue arrow balances the lever, what is the mechanical advantage?

61% Answer Correctly

	0.38
	2.25
	0.75
	3.75

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 (E_a) by the length of the resistance arm (R_a):

MA = \( \frac{E_a}{R_a} \) = \( \frac{6 ft.}{8 ft.} \) = 0.75

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

In this problem, R_a and R_b 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 E_a by R_a to get the mechanical advantage.