/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 A uniform plank of length \(6.00... [FREE SOLUTION] | 91Ó°ÊÓ

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A uniform plank of length \(6.00 \mathrm{m}\) and mass \(30.0 \mathrm{kg}\) rests horizontally across two horizontal bars of a scaffold. The bars are \(4.50 \mathrm{m}\) apart, and \(1.50 \mathrm{m}\) of the plank hangs over one side of the scaffold. Draw a free-body diagram of the plank. How far can a painter of mass \(70.0 \mathrm{kg}\) walk on the overhanging part of the plank before it tips?

Short Answer

Expert verified
The painter can walk on the overhanging part of the plank until he is at a distance \(x\) calculated using the formula \((m_p * (d / 2)) / m = x\).

Step by step solution

01

Identify Known and Unknown Variables

The known variables include the length of the plank \(L = 6.00m\), the mass of the plank \(m_p = 30.0 kg\), the distance between the bars \(d = 4.50m\), the length of the overhang \(O = 1.50m\), and the mass of the painter \(m = 70.0 kg\). The unknown variable is the distance \(x\) that the painter can walk on the overhanging part of the plank before it tips.
02

Draw a Free-Body Diagram and Apply the Equilibrium Condition

First, draw a free-body diagram of the plank. The weight of the plank \(W_p = m_p * g\) (where \(g\) is the acceleration due to gravity) acts downward from the center of the plank, the weight of the painter \(W = m * g\) acts downward at a distance \(x\) from the end of the overhang, and the forces exerted by the bars act upwards. Since the plank is in equilibrium, the sum of the torques about any point must equal zero. Using the point at the edge of the second bar as the pivot point generates the equation \(W_p * (d / 2) = W * x\).
03

Solve for the Unknown Variable

Substitute the known quantities into the equation and solve for \(x\). \((m_p * g * (d / 2)) / (m * g) = x\). Cancel out \(g\) to simplify the calculations. Then solve for the painter’s distance \(x\) from the end of the plank.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Free-Body Diagram
Understanding the forces acting on an object is crucial for analyzing real-world problems in physics. A free-body diagram is a graphical representation that shows all the forces acting on an object. It helps to visually simplify complex scenarios. For objects in equilibrium, these diagrams are particularly useful as they set the stage for calculating net forces and torques.

When constructing a free-body diagram, each force is represented by an arrow pointing in the direction in which it acts, with its tail at the point of application on the object. The length of the arrow reflects the magnitude of the force. In the case of the uniform plank, we identify the gravitational forces due to the weights of the plank and the painter, as well as the reactive forces from the scaffold's bars. To solve for equilibrium problems, it's also helpful to specify the pivot point, which is where we'll assess the torques.
Torque and Equilibrium
Torque is a measure of the force causing an object to rotate about an axis or pivot. It is calculated as the product of the force and the perpendicular distance from the pivot point to the line of action of the force. Mathematically, it's expressed as \( \tau = r \times F \), where \( r \) is the distance to the pivot and \( F \) is the force applied.

Torque and equilibrium are closely linked in physics problems involving rotational motion. An object is in rotational equilibrium when the sum of all torques acting on it is zero, meaning that it will not start rotating spontaneously. In our plank scenario, the torques due to the plank's weight and the applied weight of the painter about the scaffold's second bar must balance for the plank not to tip. By setting the sum of these torques to zero, we can determine the maximum distance the painter can venture out on the overhanging part of the plank.
Center of Mass
The center of mass is a point representing the mean location of the mass distribution of an object or system. For uniform, symmetrical objects, such as the plank described in the exercise, the center of mass is at the geometric center. This concept is crucial because the weight of the object acts as though it's concentrated at this point.

If we understand where the center of mass is, we can predict how an object will balance or tip. The plank will remain in equilibrium as long as its center of mass remains over the area supported by the scaffold. Once the painter's weight adds enough counter-torque to shift the plank's center of mass off the supporting scaffold, the plank will tip. Knowing the location of the center of mass allows us to use its distance from the pivot point to calculate the torques and solve equilibrium problems accurately.

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Most popular questions from this chapter

Consider the following mass distribution: \(5.00 \mathrm{kg}\) at \((0,0) \mathrm{m}, 3.00 \mathrm{kg}\) at \((0,4.00) \mathrm{m},\) and \(4.00 \mathrm{kg}\) at (3.00,0) \(\mathrm{m}\) Where should a fourth object of mass \(8.00 \mathrm{kg}\) be placed so that the center of gravity of the four-object arrangement will be at (0,0)\(?\)

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