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Chapter 11: Problem 33

A 600.0-N bricklayer is \(1.5 \mathrm{~m}\) from one end of a uniform scaffold that is \(7.0 \mathrm{~m}\) long and weighs \(800.0 \mathrm{~N}\). A pile of bricks weighing \(500.0 \mathrm{~N}\) is \(3.0 \mathrm{~m}\) from the same end of the scaffold. If the scaffold is supported at both ends, calculate the force on each end.

Short Answer

Expert verified
Answer: The force at point A on the scaffold is 1694.29 N, and the force at point B is 205.71 N.

Step by step solution

01

Define the variables

Let's define the given variables, and set the end closer to the bricklayer as point A and the other end as point B: - The mass of the bricklayer (\(m_1\)) = \(600.0 \mathrm{~N}\) - Distance of the bricklayer from point A (\(d_1\)) = \(1.5 \mathrm{~m}\) - The mass of the pile of bricks (\(m_2\)) = \(500.0 \mathrm{~N}\) - Distance of the bricks pile from point A (\(d_2\)) = \(3.0 \mathrm{~m}\) - The mass of the scaffold (\(m_3\)) = \(800.0 \mathrm{~N}\) - Distance of the center of mass of the scaffold (\(d_3\)) = \(\frac{7.0 \mathrm{~m}}{2}\) (Uniform scaffold)
02

Define the forces on point A and point B

Let's assume the force on point A is \(F_A\) and the force on point B is \(F_B\). On a balanced scaffold, the sum of all the forces must add up to zero. We can write that down as: \(F_A + F_B = m_1 + m_2 + m_3\)
03

Calculate the torque

Torque is the measure of how much force acting on an object causes that object to rotate. Since the scaffold is in equilibrium, the sum of the torques around any point must also be equal to zero. Let's calculate the torque around point A due to the bricklayer, bricks, and the scaffold itself: \(\tau_1 = m_1 \cdot d_1\) \(\tau_2 = m_2 \cdot d_2\) \(\tau_3 = m_3 \cdot d_3\) Since, the sum of all torques around point A must be equal to zero, we have: \(-(\tau_1 + \tau_2 + \tau_3) + F_B \cdot 7.0 \mathrm{~m} = 0\)
04

Solve for the force at point B

Plugging in the values and solving for \(F_B\), we get: \(-(600.0 \cdot 1.5 + 800.0 \cdot 3.5 + 500.0 \cdot 3) + F_B \cdot 7.0 = 0\) After calculating, we get the force at point B: \(F_B = 205.71 \mathrm{~N}\)
05

Solve for the force at point A

Now, we can plug the value of \(F_B\) into our force balance equation to get the force at point A: \(F_A = 600 + 800 + 500 - 205.71\) \(F_A = 1694.29 \mathrm{~N}\) So, the force on each end of the scaffold is: \(F_A = 1694.29 \mathrm{~N}\) \(F_B = 205.71 \mathrm{~N}\)

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Most popular questions from this chapter

In the absence of any symmetry or other constraints on the forces involved, how many unknown force components can be determined in a situation of static equilibrium in each of the following cases? a) All forces and objects lie in a plane. b) Forces and objects are in three dimensions. c) Forces act in \(n\) spatial dimensions.

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