Question

A 600.0 - N bricklayer is \(1.50 \mathrm{~m}\) from one end of a uniform scaffold that is \(7.00 \mathrm{~m}\) long and weighs \(800.0 \mathrm{~N}\). A pile of bricks weighing \(500.0 \mathrm{~N}\) is \(3.00 \mathrm{~m}\) from the same end of the scaffold. If the scaffold is supported at both ends, calculate the force on each end.

Short answer

In this static equilibrium problem, we determined the supporting forces at each end of a uniform scaffold with two point loads (a bricklayer and a pile of bricks). After identifying the forces and moments acting on the scaffold, we wrote equilibrium equations for these forces and moments. By solving these equations, we found that the force on the left-end support is \(F_A = 886.43N\), and the force on the right-end support is \(F_B = 1013.57N\).

Step-by-step solution

Identify the forces and moments acting on the scaffold

Let's first label the forces acting on the scaffold: - \(F_A\): Force at left-end support (unknown) - \(F_B\): Force at right-end support (unknown) - \(W_s\): Weight of the scaffold at its center=\(800.0N\) - \(W_b\): Weight of the bricks at a distance of \(3.00m\) from the left-end of the scaffold=\(500.0N\) - \(W_l\): Weight of the bricklayer at a distance of \(1.50m\) from the left-end of the scaffold=\(600.0N\) In this static situation, vertical forces will be balanced.

Write the equilibrium equations for the forces and moments

Write the force equilibrium equation in the vertical direction: \(F_A + F_B = W_s + W_b + W_l\) Next, write the moment equilibrium equation about point A (left-end support), where the distances are in meters: \(7.00 F_B - 3.50 W_s - 3.00 W_b - 1.50 W_l = 0\)

Solve the equilibrium equations

Solve the force equilibrium equation for \(F_B\): \(F_B = W_s + W_b + W_l - F_A\) Substitute the expression for \(F_B\) in the moment equilibrium equation, in order to have an equation only with \(F_A\): \(7.00 (W_s + W_b + W_l - F_A) - 3.50 W_s - 3.00 W_b - 1.50 W_l = 0\) Now, solve for \(F_A\): \(F_A = \frac{7.00(W_s + W_b + W_l) - 3.50 W_s - 3.00 W_b - 1.50 W_l}{7.00}\) Plug in the given values: \(F_A = \frac{7.00(800.0 + 500.0 + 600.0) - 3.50(800.0) - 3.00(500.0) - 1.50(600.0)}{7.00}\) \(F_A = 886.43N\) Now, use the force equilibrium equation to find \(F_B\): \(F_B = (800.0 + 500.0 + 600.0) - 886.43\) \(F_B = 1013.57N\) The force on the left-end support is \(F_A = 886.43N\), and the force on the right-end support is \(F_B = 1013.57N\).

Key concepts

Force Equilibrium

Understanding force equilibrium is essential when dealing with statics problems. This principle states that an object is in a state of force equilibrium if the sum of all forces acting upon it is zero, resulting in no acceleration.

In the example given, we consider a scaffold in static equilibrium. The force equilibrium equation, \(F_A + F_B = W_s + W_b + W_l\), represents the balance of the vertical forces. In this situation, we have the weights of the scaffold, the bricks, and the bricklayer balanced by the forces at the supports named \(F_A\) and \(F_B\). This balance of forces ensures the scaffold remains at rest, not moving up or down.

When solving static equilibrium problems, always start by identifying all forces acting on the object and express their sum as an equation. By doing so, you set up a crucial part of figuring out the unknown forces that keep the object at rest.

Moment Equilibrium

Moment equilibrium is another fundamental concept in solving statics problems. A moment, often equated with torque in this context, is the product of a force and the perpendicular distance from that force to a pivot point.

For an object to be in moment equilibrium, the sum of the clockwise moments about any point must equal the sum of the counterclockwise moments about that same point. The equation \(7.00 F_B - 3.50 W_s - 3.00 W_b - 1.50 W_l = 0\) represents this balance about point A for our scaffold. The distances multiplied by the corresponding forces represent the moments. Here, the distances are crucial because a force applied further from the pivot point will generate a larger moment, potentially disrupting the equilibrium if not properly balanced by other moments.

To solve a moment equilibrium problem, choose a pivot point, identify all the moments, and set up an equation ensuring that the sum of the clockwise moments equals the sum of the counterclockwise moments. Solving this equation will help determine unknown forces or torques ensuring stability.

Torque and Rotational Statics

Torque is a measure of how much a force acting on an object causes that object to rotate. The concept of torque is closely related to moment equilibrium in rotational statics, which explores objects in a stable rotational state.

In the given problem, understanding torque is critical when computing the moments generated by the various weights about a point of rotation. The formula for a moment, or torque, is given by \( \tau = r \times F \), where \(r\) is the distance from the pivot point to the point of application of the force, and \(F\) is the magnitude of that force.

By analyzing torques and ensuring they are balanced (clockwise torques equal counterclockwise torques), you can predict the rotational behavior of static objects. This concept is vital in designing and analyzing structures or mechanical systems to ensure they won't start spinning or tipping unexpectedly.

Mechanical Equilibrium

Mechanical equilibrium is achieved when an object is in both force and moment equilibrium, meaning there are no net forces or moments causing acceleration or rotation.

The scaffold with weights problem perfectly embodies mechanical equilibrium: not only are the up and down forces balanced, but also the rotational effects around any pivot points are negated. When dealing with static structures, such as bridges, buildings, or scaffolds, engineers use the principles of mechanical equilibrium to design systems that will remain stationary and stable under various loads.

If the body is in mechanical equilibrium, you will find that resolving both force equilibrium equations and moment equilibrium equations is necessary to ensure the safety and stability of the structure. In our exercise, calculating the precise forces each end must bear, considering both the scaffold's weight and the added load from the bricklayer and bricks, illustrates how an object remains static only when all aspects of mechanical equilibrium are satisfied.