QuestionAugust 13, 2025

An 8 m long uniform plank of mass 140 kg rests on the side of a pirate ship with 3 meters extended over the edge. How far can a 60 kg prisoner venture past the odge of the ship on the plank before the plank just begins to tip? Please select the correct answer from the choices shown below: A. 1,5 m B. 1.67 m C. 2 m D. 2.33 m Reset Selection

An 8 m long uniform plank of mass 140 kg rests on the side of a pirate ship with 3 meters extended over the edge. How far can a 60 kg prisoner venture past the odge of the ship on the plank before the plank just begins to tip? Please select the correct answer from the choices shown below: A. 1,5 m B. 1.67 m C. 2 m D. 2.33 m Reset Selection
An 8 m long uniform plank of mass 140 kg rests on the side of a pirate ship with 3 meters extended over the edge. How far can a 60 kg prisoner venture past the odge of the ship on the plank before the plank just begins to tip? Please select the correct answer from the choices shown below: A. 1,5 m B. 1.67 m C. 2 m D. 2.33 m Reset Selection

Solution
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Answer

D. 2.33 m Explanation 1. Calculate the plank's center of mass The plank is uniform, so its center of mass is at its midpoint. Since 3 m extends over the edge, the center of mass is 4 - 3 = 1 m from the edge. 2. Determine torque balance condition For tipping, the torque due to the prisoner must equal the torque due to the plank's weight about the pivot point (edge). Use \text{Torque} = \text{Force} \times \text{Distance}. 3. Set up the equation for torque balance Torque by plank: 140 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 1 \, \text{m}. Torque by prisoner: 60 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times x, where x is the distance past the edge. 4. Solve for x Equate torques: 140 \times 9.8 \times 1 = 60 \times 9.8 \times x. Simplify to find x: 140 = 60x. Thus, x = \frac{140}{60} = 2.33 m.

Explanation

1. Calculate the plank's center of mass<br /> The plank is uniform, so its center of mass is at its midpoint. Since 3 m extends over the edge, the center of mass is $4 - 3 = 1$ m from the edge.<br />2. Determine torque balance condition<br /> For tipping, the torque due to the prisoner must equal the torque due to the plank's weight about the pivot point (edge). Use $\text{Torque} = \text{Force} \times \text{Distance}$.<br />3. Set up the equation for torque balance<br /> Torque by plank: $140 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 1 \, \text{m}$. Torque by prisoner: $60 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times x$, where $x$ is the distance past the edge.<br />4. Solve for $x$<br /> Equate torques: $140 \times 9.8 \times 1 = 60 \times 9.8 \times x$. Simplify to find $x$: $140 = 60x$. Thus, $x = \frac{140}{60} = 2.33$ m.
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