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Chapter 11: Problem 65

A \(5.00-\mathrm{m}\) -long board of mass \(50.0 \mathrm{~kg}\) is used as a seesaw. On the left end of the seesaw sits a 45.0 -kg girl, and on the right end sits a 60.0 -kg boy. Determine the position of the pivot point for static equilibrium.

Short Answer

Expert verified
Answer: The pivot point should be placed approximately 1.67 meters from the left end of the seesaw for static equilibrium.

Step by step solution

01

Define variables and write the formula for static equilibrium

Let's define the variables: The weight of the board (W_board) = 50 kg The weight of the girl (W_girl) = 45 kg The weight of the boy (W_boy) = 60 kg The length of the seesaw (L) = 5 m Distance of the pivot point to the left end (x_pivot) In static equilibrium, the total clockwise torque equals the total counterclockwise torque. Thus, the formula for the equilibrium is: torque_girl + torque_board = torque_boy
02

Calculate the torques

To calculate the individual torques, we will multiply the weights by their respective distances from the pivot point. For the girl, the distance is equal to the distance of the pivot point to the left end (x_pivot). For the board, we account for the uniform distribution of weight across the seesaw and use half of the length of the seesaw as the distance from the pivot (0.5 * L). For the boy, the distance is the total length of the seesaw minus the distance of the pivot from the left end (L - x_pivot). torque_girl = W_girl * x_pivot torque_board = W_board * (0.5 * L) torque_boy = W_boy * (L - x_pivot)
03

Solve for x_pivot

Now, plug the torques into the static equilibrium formula and solve for x_pivot: W_girl * x_pivot + W_board * (0.5 * L) = W_boy * (L - x_pivot) 45 * x_pivot + 50 * (0.5 * 5) = 60 * (5 - x_pivot)
04

Simplify the equation

Simplify the equation: 45 * x_pivot + 125 = 300 - 60 * x_pivot Now, combine the x_pivot terms: 105 * x_pivot = 175 Now, divide by 105 to find the value of x_pivot: x_pivot = 175 / 105 x_pivot ≈ 1.67 m The position of the pivot point for static equilibrium is approximately 1.67 meters from the left end of the seesaw.

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Most popular questions from this chapter

A ladder of mass \(M\) and length \(L=4.00 \mathrm{~m}\) is on a level floor leaning against a vertical wall. The coefficient of static friction between the ladder and the floor is \(\mu_{\mathrm{s}}=\) 0.600 , while the friction between the ladder and the wall is negligible. The ladder is at an angle of \(\theta=50.0^{\circ}\) above the horizontal. A man of mass \(3 M\) starts to climb the ladder. To what distance up the ladder can the man climb before the ladder starts to slip on the floor?

You have a meter stick that balances at the \(55-\mathrm{cm}\) mark. Is your meter stick homogeneous?

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