91Ó°ÊÓ

You need to measure the mass \(M\) of a \(4.00-\mathrm{m}\) -long bar. The bar has a square cross section but has some holes drilled along its length, so you suspect that its center of gravity isn't in the middle of the bar. The bar is too long for you to weigh on your scale. So, first you balance the bar on a knife-edge pivot and determine that the bar's center of gravity is \(1.88 \mathrm{~m}\) from its left-hand end. You then place the bar on the pivot so that the point of support is \(1.50 \mathrm{~m}\) from the left-hand end of the bar. Next you suspend a \(2.00 \mathrm{~kg}\) mass \(\left(m_{1}\right)\) from the bar at a point \(0.200 \mathrm{~m}\) from the left-hand end. Finally, you suspend a mass \(m_{2}=1.00 \mathrm{~kg}\) from the bar at a distance \(x\) from the left-hand end and adjust \(x\) so that the bar is balanced. You repeat this step for other values of \(m_{2}\) and record each corresponding value of \(x\). The table gives your results. $$ \begin{array}{l|llllll} m_{2}(\mathrm{~kg}) & 1.00 & 1.50 & 2.00 & 2.50 & 3.00 & 4.00 \\ \hline x(\mathrm{~m}) & 3.50 & 2.83 & 2.50 & 2.32 & 2.16 & 2.00 \end{array} $$ (a) Draw a free-body diagram for the bar when \(m_{1}\) and \(m_{2}\) are suspended from it. (b) Apply the static equilibrium equation \(\Sigma \tau_{z}=0\) with the axis at the location of the knife-edge pivot. Solve the equation for \(x\) as a function of \(m_{2}\). (c) Plot \(x\) versus \(1 / m_{2}\). Use the slope of the best-fit straight line and the equation you derived in part (b) to calculate that bar's mass \(M .\) Use \(g=9.80 \mathrm{~m} / \mathrm{s}^{2} .\) (d) What is the \(y\) -intercept of the straight line that fits the data? Explain why it has this value.

Step by step solution

01

Draw the Free Body Diagram

In the diagram, there are four forces acting on the bar. At the pivot point there is the normal force \(N\). The weight \(m_{1}g\) is acting at a distance of \(0.200 m\) from the left-hand end. The weight of the bar \(Mg\) is acting at a distance of \(1.88 m\). The weight \(m_{2}g\) is acting at \(x\) distance from the left-hand end.

02

Apply the Static Equilibrium Equation

The static equilibrium equation is given by \(\Sigma \tau_{z}=0\). Setting up the torque equation, it should be like this: \(\tau_{N}= \tau_{Mg}+\tau_{m_{1}g}+\tau_{m_{2}g}\), where \(\tau_{N}\) is the normal force at the pivot point which is zero since the distance is zero. Solve the equation for \(x\), which will be \(x = \frac{Mg(1.88 \mathrm{m}) + m_{1}g(0.200 \mathrm{m})}{m_{2}g} - 1.50\mathrm{m}\)

03

Plot \(x\) versus \(1 / m_{2}\)

With the calculated values of \(m_{2}\) and \(x\) from the exercise, a plot of \(1/m_{2}\) versus \(x\) can be generated. By linear regression, find the slope of the line.

04

Calculate the Bar's Mass and Interpret the Plot

From the straight line plot, the slope of the line can be used to calculate the mass of the bar in the equation: \(M = \frac{slope}{g(1.88 \mathrm{m} - 1.50 \mathrm{m})}\) where \(g\) is the acceleration due to gravity. The y-intercept represents the value of \(x\) when \(1/m_{2}= 0\), in effect when \(m_{2}\) is infinite; which logically would mean the second mass is so large that the bar effectively acts as if there's no added mass.

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Key Concepts

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

Torque and Equilibrium

When we talk about static equilibrium in physics, we're focusing on objects that are not moving or rotating. To ensure this state, the net force and net torque on the object must be zero. Torque, a concept often compared to force, tells us how much a force acting on an object causes that object to rotate around an axis. It's calculated as the product of the force and the perpendicular distance from the axis to the line of action of the force, expressed as \( \tau = r \times F \), where \( \tau \) represents torque, \( r \) is the distance to the axis, and \( F \) is the force.

In the context of this exercise, you'll see torque and equilibrium at play when balancing a bar with uneven mass distribution. By using a pivot point, we simulate the axis of rotation, and the mass of the bar together with additional weights creates torque when gravity pulls them down. Static equilibrium is achieved when the sum of all torques around the pivot point is zero, which means the bar stays balanced without rotating. This happens because the clockwise and counterclockwise moments cancel each other out. Understanding these principles helps to analyze and solve real-world problems involving balance and stability.

Center of Gravity

The center of gravity is the point where the total weight of the body is considered to be concentrated for the purpose of analyzing the gravitational forces. In symmetric objects of uniform density, the center of gravity is at the geometric center. However, if the density isn't uniform – as with the bar mentioned in the exercise with drilled holes – the center of gravity shifts from that central point.

For objects on Earth, the acceleration due to gravity (\( g \)) acts directly downward, so we place the center of gravity in our analysis at the point where the object would balance perfectly on a pivot if subjected to no other forces. In the bar example, finding the true center of gravity was crucial to calculating its mass correctly, as it influences the distribution of torque along the length of the bar. Knowing how to locate the center of gravity is not only useful for solving physics problems but also for engineering and design, where the stability of structures and vehicles depends on accurately predicting how they will balance and move.

Free-Body Diagram

A free-body diagram (FBD) is a graphical illustration used to visualize the forces acting on an object. This tool is immensely helpful in solving physics problems, including those dealing with static equilibrium, because it allows us to focus on one body, ignoring the rest, while considering the interactions through forces.

In the case of the bar exercise, the FBD would include the downward force of gravity acting on the mass of the bar itself, the additional weights attached to it, and the opposing upward normal force exerted by the pivot or support. The specific distances from the pivot to where these forces act are crucial, as they determine the moments or torques that cause rotation. When you're drawing an FBD, it's imperative to accurately depict the direction and point of application of each force, as these details directly affect the calculations for torque and ultimately the equilibrium condition.