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Zorluk: OrtaLevers and Fulcrums

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 m0.4 \text{ m} away from the hinge, applying a downward effort force. The arm presses down onto a workpiece located 2.0 m2.0 \text{ m} from the hinge. The workpiece exerts an upward resistance of 400 N400 \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?

  1. 2000 N2000 \text{ N}; Third-class leverCevap
  2. B
    2000 N2000 \text{ N}; Second-class lever
  3. C
    1600 N1600 \text{ N}; Third-class lever
  4. D
    80 N80 \text{ N}; Third-class lever

Cevap

The actuator must apply 2000 N2000 \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 N×2.0 m=800 N⋅m400 \text{ N} \times 2.0 \text{ m} = 800 \text{ N}\cdot\text{m}. The actuator must create an equal moment: Fe×0.4 m=800 N⋅mF_e \times 0.4 \text{ m} = 800 \text{ N}\cdot\text{m}. Solving for FeF_e gives 2000 N2000 \text{ N}.

Adım Adım Çözüm

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).
Fe×0.4 m=400 N×2.0 mF_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 (FeF_e).
Fe=2000 NF_e = 2000 \text{ N}
Evaluating (400×2.0)/0.4(400 \times 2.0) / 0.4 yields 800/0.4=2000 N800 / 0.4 = 2000 \text{ N}.

Anahtar Kavram

Lever classifications and torque balance in simple machines
Tahmini Süre:1m 0s
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