Simple Machines Practice Questions — Armed Services Vocational Aptitude Battery (ASVAB) — Page 9

Question 161Question →

A specialized clamping arm on a manufacturing machine operates as a straight lever, hinged at one end to a fixed support. A pneumatic actuator is attached to the arm $0.4 \text{ m}$ away from the hinge, applying a downward effort force. The arm presses down onto a workpiece located $2.0 \text{ m}$ from the hinge. The workpiece exerts an upward resistance of $400 \text{ N}$ against the arm.

Assuming the arm is level and its weight is negligible, what downward force must the actuator apply to hold the arm steady, and what class of lever does this mechanism represent?

2000 \text{ N}

; Third-class lever

2000 \text{ N}

; Second-class lever

1600 \text{ N}

; Third-class lever

80 \text{ N}

; Third-class lever

Show answer & explanation

Answer: $2000 \text{ N}$ ; Third-class lever

Answer

The actuator must apply

2000 \text{ N}

of force, and the mechanism represents a third-class lever.

The correct answer accurately classifies the setup as a third-class lever because the effort (actuator) is located between the fulcrum (hinge) and the load (workpiece). To maintain equilibrium, the moments around the hinge must balance. The load creates a moment of

400 \text{ N} \times 2.0 \text{ m} = 800 \text{ N}\cdot\text{m}

. The actuator must create an equal moment:

F_e \times 0.4 \text{ m} = 800 \text{ N}\cdot\text{m}

. Solving for

F_e

gives

2000 \text{ N}

.

Step-by-Step Solution

1

Identify the positions of the fulcrum, effort, and load to determine the lever class.

Third-class lever.

The hinge is the fulcrum at one end (0.0 m), the actuator is the effort (0.4 m), and the workpiece is the load (2.0 m). Because the effort is applied between the fulcrum and the load, it is a third-class lever.

2

Set up the torque balance equation (Principle of Moments).

F_e \times 0.4 \text{ m} = 400 \text{ N} \times 2.0 \text{ m}

For the lever to be held steady, the torque generated by the effort force must perfectly balance the torque generated by the load resistance around the fulcrum.

3

Solve the equation for the unknown effort force (

F_e

).

F_e = 2000 \text{ N}

Evaluating

(400 \times 2.0) / 0.4

yields

800 / 0.4 = 2000 \text{ N}

.

Key Concept

Lever classifications and torque balance in simple machines

Estimated Time:1m 0s

Question 162Question →

A technician uses a pair of long industrial tweezers to securely hold a sensitive electronic component. The tweezers pivot at the joined top end and have a total length of $40 \text{ cm}$ from the pivot to the gripping tips. The technician's fingers apply pressure at a point $8 \text{ cm}$ from the pivot. If the component requires a holding force of $50 \text{ N}$ at the tips, what effort force (in Newtons) must the technician apply?

Show answer & explanation

Answer: 250

Answer

The technician must apply an effort force of 250 N.

In a third-class lever like tweezers, the effort is applied between the fulcrum and the load. Since the effort arm (

8 \text{ cm}

) is shorter than the load arm (

40 \text{ cm}

), the mechanical advantage is less than 1. This means a much larger effort force is required to generate a smaller load force at the tips. Using the torque balance equation

F_e \times d_e = F_l \times d_l

, we get

F_e \times 8 = 50 \times 40

. Solving this gives

250 \text{ N}

.

Step-by-Step Solution

1

Identify the lever configuration and its parameters.

The tweezers are a third-class lever (fulcrum at one end, effort in the middle). Effort arm

d_e = 8 \text{ cm}

, load arm

d_l = 40 \text{ cm}

, load force

F_l = 50 \text{ N}

.

Correctly identifying the lever class ensures the correct distances are used relative to the fulcrum.

2

Set up the torque equilibrium equation.

F_e \times d_e = F_l \times d_l

For the lever to remain stable and provide the required grip, the effort torque must balance the load torque.

3

Calculate the required effort force.

F_e \times 8 = 50 \times 40 \Rightarrow 8 F_e = 2000 \Rightarrow F_e = 250 \text{ N}

To find the unknown variable by substituting the given values.

Key Concept

Torque Balance in Third-Class Levers

Question 163Question →

A military mechanic needs to lift an engine component that weighs $120 \text{ lbs}$ . Match each pulley system configuration to the ideal effort force required to lift the component.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Single fixed pulley matches

120 \text{ lbs}

; Single movable pulley matches

60 \text{ lbs}

; Block and tackle with 4 supporting strands matches

30 \text{ lbs}

.

The effort force is calculated by dividing the

120 \text{ lb}

load by the Ideal Mechanical Advantage (IMA) of each system: 1 for a single fixed pulley (

120 \text{ lbs}

), 2 for a single movable pulley (

60 \text{ lbs}

), and 4 for a 4-strand block and tackle (

30 \text{ lbs}

).

Step-by-Step Solution

1

Determine the Ideal Mechanical Advantage (IMA) of the single fixed pulley.

IMA = 1.

A single fixed pulley only changes the direction of the force.

2

Calculate the effort force for the single fixed pulley.

120 \text{ lbs} / 1 = 120 \text{ lbs}

.

Effort = Load / IMA.

3

Determine the IMA of the single movable pulley.

IMA = 2.

A single movable pulley is supported by two rope sections, halving the required force.

4

Calculate the effort force for the single movable pulley.

120 \text{ lbs} / 2 = 60 \text{ lbs}

.

Effort = Load / IMA.

5

Determine the IMA of the block and tackle system with 4 supporting strands.

IMA = 4.

The number of supporting upward strands directly equals the IMA.

6

Calculate the effort force for the block and tackle system.

120 \text{ lbs} / 4 = 30 \text{ lbs}

.

Effort = Load / IMA.

Key Concept

The ideal effort force required to lift a load is the load divided by the Ideal Mechanical Advantage (IMA) of the pulley system.

Question 164Question →

A maintenance crew uses a horizontal boom mounted on a utility truck to lift a $600 \text{ N}$ generator. The boom pivots at a hinge on the truck bed. A vertical hydraulic cylinder supports the boom, attaching exactly $1.0 \text{ m}$ from the hinge. The generator hangs from the far end of the boom, $4.0 \text{ m}$ from the hinge. Assuming the weight of the boom itself is negligible, how much upward force must the hydraulic cylinder apply to hold the generator stationary, and what class of lever does this system represent?

2400 \text{ N}

; third-class lever

150 \text{ N}

; second-class lever

1800 \text{ N}

; third-class lever

2400 \text{ N}

; first-class lever

Show answer & explanation

Answer: $2400 \text{ N}$ ; third-class lever

Answer

2400 \text{ N}

; third-class lever

The system represents a third-class lever because the effort force (the hydraulic cylinder) is applied between the fulcrum (the hinge) and the load (the generator). To maintain static equilibrium, the counterclockwise torque provided by the cylinder must perfectly equal the clockwise torque from the generator. The generator creates a torque of

600 \text{ N} \times 4.0 \text{ m} = 2400 \text{ N}\cdot\text{m}

. Therefore, the cylinder must apply

2400 \text{ N}

of force at a distance of

1.0 \text{ m}

to balance the system (

2400 \text{ N} \times 1.0 \text{ m} = 2400 \text{ N}\cdot\text{m}

).

Step-by-Step Solution

1

Identify the lever class based on the arrangement of the fulcrum, effort, and load.

The effort force (hydraulic cylinder) is located between the fulcrum (hinge) and the load (generator). This arrangement defines a third-class lever.

Determining the lever arrangement is essential to properly visualize the system and answers the classification part of the question.

2

Set up the torque balance equation (principle of moments) around the fulcrum.

To remain stationary, upward torque must equal downward torque:

F_{\text{effort}} \times d_{\text{effort}} = F_{\text{load}} \times d_{\text{load}}

A lever in static equilibrium must have balanced rotational forces acting around its pivot point.

3

Substitute the known distances and forces into the equation to solve for the effort force.

F_{\text{effort}} \times 1.0 \text{ m} = 600 \text{ N} \times 4.0 \text{ m}

, which simplifies to

F_{\text{effort}} = 2400 \text{ N}

.

Applying the specific measurements yields the precise upward force required by the cylinder.

Key Concept

Third-Class Levers and Torque Balance

Estimated Time:1m 15s

Question 165Question →

A rigging instructor is teaching trainees how to calculate mechanical advantage. Match each described frictionless pulley system to its Ideal Mechanical Advantage (IMA) and the effort force required to lift a $120 \text{ lb}$ load.

Click a left item, then click its matching right item

Show answer & explanation

Answer

System 1 matches IMA = 1 (120 lbs), System 2 matches IMA = 2 (60 lbs), System 3 matches IMA = 3 (40 lbs), and System 4 matches IMA = 4 (30 lbs).

Each system's mechanical advantage corresponds directly to the number of supporting rope strands. Dividing the 120 lb load by the respective IMA yields the correct effort force.

Step-by-Step Solution

1

Determine the Ideal Mechanical Advantage (IMA) for each system by counting the number of rope strands that directly support the movable block or load.

System 1 has 1 supporting strand, System 2 has 2, System 3 has 3, and System 4 has 4.

The IMA of a pulley system is exactly equal to the number of rope sections pulling upward against gravity on the movable block.

2

Calculate the effort force required for each system using the formula: Effort = Load / IMA.

Effort1 = 120/1 = 120 lbs. Effort2 = 120/2 = 60 lbs. Effort3 = 120/3 = 40 lbs. Effort4 = 120/4 = 30 lbs.

Mechanical advantage multiplies the input force. To find the required input effort, divide the total output load by the IMA.

3

Match each system's description to the calculated IMA and effort force values.

System 1 -> 120 lbs; System 2 -> 60 lbs; System 3 -> 40 lbs; System 4 -> 30 lbs.

This completes the pairing based on the physical properties of each configuration.

Key Concept

The Ideal Mechanical Advantage of a pulley system equals the number of rope strands that directly support the movable pulley block.

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