12. A beam pivoted at one end has a force of 5.0 N acting vertically upwards on it as shown in the Figure below. What is the weight of the beam?

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12. A beam pivoted at one end has a force of 5.0 N acting vertically upwards on it as shown in the Figure below. What is the weight of the beam?

Answer 1

To solve this problem, we will use the principle of moments (also known as the principle of torque). The principle of moments states that for an object to be in equilibrium, the sum of the clockwise moments about any pivot must be equal to the sum of the anticlockwise moments about that pivot.The moment (torque) is calculated by multiplying the force by the perpendicular distance from the pivot point to the line of action of the force. The formula for the moment is:\[\text{Moment} = \text{Force} \times \text{Distance from pivot}\]Given that the beam is in equilibrium, the anticlockwise moment due to the weight of the beam (W) must be equal to the clockwise moment due to the upward force of 5.0 N.Let's denote the weight of the beam as \(W\) (in newtons) and the distance from the pivot to the point where the weight of the beam acts as \(d\) (in meters). The weight of the beam will act at its center of gravity, which, assuming the beam is uniform, will be at the midpoint of the beam.Since the beam is 5.0 cm in total length (2.0 cm + 3.0 cm), the center of gravity will be at 2.5 cm from either end. However, since the pivot is 2.0 cm from the left end, the distance from the pivot to the center of gravity is \(2.5 \text{ cm} - 2.0 \text{ cm} = 0.5 \text{ cm}\).Now, we convert this distance to meters for consistency in units:\[d = 0.5 \text{ cm} \times \frac{1 \text{ m}}{100 \text{ cm}} = 0.005 \text{ m}\]Now we can set up the equation for the moments about the pivot:\[\text{Anticlockwise moment} = \text{Clockwise moment}\]\[W \times d = 5.0 \text{ N} \times 0.03 \text{ m}\]Now we solve for \(W\):\[W \times 0.005 \text{ m} = 5.0 \text{ N} \times 0.03 \text{ m}\]\[W = \frac{5.0 \text{ N} \times 0.03 \text{ m}}{0.005 \text{ m}}\]\[W = \frac{0.15 \text{ Nm}}{0.005 \text{ m}}\]\[W = 30.0 \text{ N}\]Therefore, the weight of the beam is 30.0 N.

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12. A beam pivoted at one end has a force of 5.0 N acting vertically upwards on it as shown in the Figure below. What is the weight of the beam?

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