A specialized clamping arm on a manufacturing machine operates as a straight lever, hinged… — Armed Services Vocational Aptitude Battery (ASVAB)

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 leverAnswer
B
$2000 \text{ N}$ ; Second-class lever
C
$1600 \text{ N}$ ; Third-class lever
D
$80 \text{ N}$ ; 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

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.

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.

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

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