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Zorluk: OrtaLevers and Fulcrums

An industrial sheet metal shear operates as a second-class lever. The pivot point (fulcrum) is located at the far end of the cutting arm. A mechanic applies a downward effort force of 150 N150 \text{ N} at the handle, which is located 80 cm80 \text{ cm} from the fulcrum. The metal sheet is positioned 16 cm16 \text{ cm} from the fulcrum. Assuming the lever is rigid and frictionless, what is the downward cutting force, in newtons, applied to the metal sheet?

Cevap: 750 N

Cevap

750
A second-class lever always places the resistance between the effort and the fulcrum, providing a mechanical advantage greater than 1. The torque applied by the effort (150 N×80 cm150 \text{ N} \times 80 \text{ cm}) must equal the opposing torque exerted on the resistance (Fr×16 cmF_r \times 16 \text{ cm}). Solving this equilibrium equation yields a cutting force of 750 N750 \text{ N}.

Adım Adım Çözüm

1
Identify the lever parameters and equation
Fe=150 NF_e = 150 \text{ N}, de=80 cmd_e = 80 \text{ cm}, dr=16 cmd_r = 16 \text{ cm}
Establishing the values needed for the principle of moments (torque balance).
2
Apply the law of the lever
150×80=Fr×16150 \times 80 = F_r \times 16
In equilibrium, the input torque equals the output torque.
3
Calculate the unknown resistance force
Fr=1200016=750 NF_r = \frac{12000}{16} = 750 \text{ N}
Isolating FrF_r gives the mechanical force applied to the metal.

Anahtar Kavram

Mechanical Advantage and Torque Balance in Second-Class Levers
Tahmini Süre:1m 0s
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Konular

Levers and Fulcrums

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