Lever: 6 kg on one side, find the force

A lever is balanced in a horizontal position, with a 6 kg object hanging 0.2 m to the left of the fulcrum. What force must be applied 0.4 m to the right of the fulcrum? Take g = 10 N/kg.

Step-by-step

1. Weight on the left F₁ = mg = 6×10 = 60
   The object's weight provides the effort on the left
2. F₁·L₁ = F₂·L₂ → F₂ = F₁×L₁/L₂ = 60×0.2/0.4 = 30
   Lever balance: effort × effort arm = load × load arm
3. Equivalent mass = 30/10 = 3 kg
   Convert to the everyday "how heavy"

Why it works

The condition for lever balance is that the torques on both sides are equal, not that the forces are equal. Torque equals force multiplied by the moment arm, so a smaller force can balance a larger force if it has a longer moment arm. When solving these problems, write out the torque expressions for both sides and set them equal before simplifying; this is less error-prone than directly applying a ratio formula. The moment arm is the perpendicular distance from the fulcrum to the line of action of the force, not the length of the lever.

Common mistakes

• Writing the balance condition as equal forces on both sides, ignoring the moment arm.
• Treating the moment arm as the total length of the lever or the distance from the fulcrum to the center of mass.
• Calculating weight as mass times ten but forgetting that the unit of mass is kilograms and mistakenly using grams.

Related

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