Log out Go to App
Go to App

Learning Materials

Features

Discover

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: \(F_A = 1694.29 \mathrm{~N}\) \(F_B = 205.71 \mathrm{~N}\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In a butcher shop, a horizontal steel bar of mass \(4.00 \mathrm{~kg}\) and length \(1.20 \mathrm{~m}\) is supported by two vertical wires attached to its ends. The butcher hangs a sausage of mass \(2.40 \mathrm{~kg}\) from a hook that is at a distance of \(0.20 \mathrm{~m}\) from the left end of the bar. What are the tensions in the two wires?

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.

A \(100 .-\mathrm{kg}\) uniform bar of length \(L=5.00 \mathrm{~m}\) is attached to a wall by a hinge at point \(A\) and supported in a horizontal position by a light cable attached to its other end. The cable is attached to the wall at point \(B\), at a distance \(D=2.00 \mathrm{~m}\) above point \(A\). Find: a) the tension, \(T,\) on the cable and b) the horizontal and vertical components of the force acting on the bar at point \(A\).

Why didn't the ancient Egyptians build their pyramids upside-down? In other words, use force and center-of-mass principles to explain why it is more advantageous to construct buildings with broad bases and narrow tops than the other way around.

Two uniform planks, each of mass \(m\) and length \(L,\) are connected by a hinge at the top and by a chain of negligible mass attached at their centers, as shown in the figure. The assembly will stand upright, in the shape of an \(A,\) on a frictionless surface without collapsing. As a function of the length of the chain, find each of the following: a) the tension in the chain, b) the force on the hinge of each plank, and c) the force of the ground on each plank.
See all solutions

Recommended explanations on Physics Textbooks

Oscillations

Read Explanation

Fluids

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Measurements

Read Explanation

Modern Physics

Read Explanation

Atoms and Radioactivity

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free