There is a 100g mass sitting on the 10cm mark, and a 50g mass sitting on the 60cm mark. Force F acts at the 30cm mark.\

Question

There is a  100g  mass sitting on the  10cm  mark, and a  50g  mass sitting on the  60cm  mark. Force  F  acts at the  30cm  mark.\

Accepted Answer

∻Solution∻ ‖ a ⋮ To find the force $ F $ needed to balance the meter rule, we first identify the forces acting on it. The 100 g mass at the 10 cm mark exerts a downward force due to gravity, calculated as: $$ F_{100g} = 0.1 \, \text{kg} \times 9.8 \, \text{N/kg} = 0.98 \, \text{N} $$ and the 50 g mass at the 60 cm mark exerts a downward force of: $$ F_{50g} = 0.05 \, \text{kg} \times 9.8 \, \text{N/kg} = 0.49 \, \text{N} $$ Thus, the total downward force is: $$ F_{total} = 0.98 \, \text{N} + 0.49 \, \text{N} = 1.47 \, \text{N} $$ ‖ ‖ b ⋮ Next, we apply the principle of static equilibrium, which states that the sum of the torques about the pivot point (50 cm mark) must also be zero. The torque due to the 100 g mass at the 10 cm mark is: $$ \tau_{100g} = F_{100g} \times d_{100g} = 0.98 \, \text{N} \times (50 \, \text{cm} - 10 \, \text{cm}) = 0.98 \, \text{N} \times 40 \, \text{cm} = 39.2 \, \text{N cm} $$ The torque due to the 50 g mass at the 60 cm mark is: $$ \tau_{50g} = F_{50g} \times d_{50g} = 0.49 \, \text{N} \times (60 \, \text{cm} - 50 \, \text{cm}) = 0.49 \, \text{N} \times 10 \, \text{cm} = 4.9 \, \text{N cm} $$ ‖ ‖ c ⋮ To balance the torques, the torque due to force $ F $ at the 30 cm mark must equal the total torque from the weights. The distance from the pivot (50 cm) to the 30 cm mark is 20 cm, so we have: $$ F \times 20 \, \text{cm} = 39.2 \, \text{N cm} + 4.9 \, \text{N cm} $$ This simplifies to: $$ F \times 20 = 44.1 $$ Therefore, solving for $ F $: $$ F = \frac{44.1}{20} = 2.205 \, \text{N} $$ Since $ F $ acts upwards to balance the downward forces, the direction of $ F $ is upwards. ‖ ∻Answer∻ ⚹ 2.205 N acting upwards ⚹ ∻Key Concept∻ ⚹ Static Equilibrium: The sum of forces and the sum of torques acting on an object must be zero for it to be in equilibrium. Equations: $$\sum F = 0$$ and $$\sum \tau = 0$$ ⚹ ∻Explanation∻ ⚹ The forces and torques were calculated to find the necessary force $ F $ to maintain equilibrium, ensuring that the meter rule does not rotate or translate. ⚹

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