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Chapter 11: Problem 1

A \(0.120 \mathrm{~kg},\) 50.0-cm-long uniform bar has a small \(0.055 \mathrm{~kg}\) mass glued to its left end and a small \(0.110 \mathrm{~kg}\) mass glued to the other end. The two small masses can each be treated as point masses. You want to balance this system horizontally on a fulcrum placed just under its center of gravity. How far from the left end should the fulcrum be placed?

Short Answer

Expert verified
The fulcrum should be placed \(26.7 \mathrm{~cm}\) from the left end.

Step by step solution

01

Identify given quantities

First identify the given quantities. The mass of the uniform bar is \(0.120 \mathrm{~kg}\), and its length is \(50.0 \mathrm{~cm} = 0.5 \mathrm{~m}\). The small mass on the left end is \(0.055 \mathrm{~kg}\) and the one on the right end is \(0.110 \mathrm{~kg}\).
02

Calculate Center of Gravity

To get the center of gravity, use its formula: \(x_{cg}=\frac{\sum \limits _{i} x_i m_i}{\sum \limits _{i} m_i}\). Here, \(x_i\) and \(m_i\) are the position and mass of ith object. The centre of the bar is at \(25 \mathrm{~cm}=0.25 \mathrm{~m}\). Thus, \(x_{cg}=\frac{(0)(0.055)+(0.25)(0.120)+(0.5)(0.110)}{0.055+0.120+0.110}\)
03

Calculate Result

Calculate, \(x_{cg}=\frac{(0)(0.055)+(0.25)(0.120)+(0.5)(0.110)}{0.055+0.120+0.110} = 0.267 \mathrm{~m}\) or \(26.7 \mathrm{~cm}\). Therefore, the fulcrum should be placed \(26.7 \mathrm{~cm}\) from the left end to balance the bar.

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Key Concepts

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

Torque and Equilibrium
When objects are subject to forces at various points, torque, also known as the moment of force, tells us how these forces cause the object to rotate. The larger the force or the longer the distance from the pivot point (fulcrum), the greater the torque. To achieve equilibrium, where an object is balanced and stationary, the clockwise torques must equal the counter-clockwise torques.

For example, with a seesaw, if two children of different weights want to balance, the heavier child must sit closer to the pivot, reducing their torque, while the lighter child sits farther away, increasing their torque. Applying this to the bar from our exercise, we calculate the torque produced by each mass about the pivot point. Equilibrium is reached when the sum of the clockwise torques equals the sum of the counter-clockwise torques, allowing the system to be perfectly balanced at the center of gravity.
Uniform Bar Center of Mass
A uniform bar, which is an object with mass distributed evenly along its length, has its center of mass at the geometric center. For such bars, this point serves as a pivot around which the bar can balance perfectly if supported. This is due to the symmetrical distribution of weight, creating equal torques on each side of the center.

In the problem provided, the bar itself, disregarding additional masses, would balance at 25.0 cm from either end. However, the added point masses change the center of mass. The mass closest to the end, even though smaller, has a greater leverage effect and impacts the balance of the bar. To calculate the new equilibrium point, we consider the added masses as part of the system.
Point Masses Balance
Point masses are small objects whose size can be ignored compared to the distances between them. For calculating the balance point, called the center of gravity, we must consider both the mass of these objects and their distance from a reference point. The center of gravity is the point where the total weight of the system is considered to act and is essential for understanding stability.

In our exercise, the small masses glued to the ends of the bar are treated as point masses. By calculating the center of gravity as shown in the solution, we find where to place the fulcrum for the system to balance. This is done by creating a proportional relationship that accounts for each mass and its distance from the pivot, resulting in a setup where all forces and torques are balanced, keeping the system in a state of equilibrium.

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Most popular questions from this chapter

A steel wire with radius \(r_{\text {steel }}\) has a fractional increase in length of \(\left(\Delta l / l_{0}\right)_{\text {steel }}\) when the tension in the wire is increased from zero to \(T_{\text {steel }}\). An aluminum wire has radius \(r_{\text {al }}\) that is twice the radius of the steel wire: \(r_{\mathrm{al}}=2 r_{\text {steel }} .\) In terms of \(T_{\text {steel }},\) what tension in the aluminum wire produces the same fractional change in length as in the steel wire?

A ladder carried by a fire truck is \(20.0 \mathrm{~m}\) long. The ladder weighs \(3400 \mathrm{~N}\) and its center of gravity is at its center. The ladder is pivoted at one end (A) about a pin (Fig. E11.7); ignore the friction torque at the pin. The ladder is raised into position by a force applied by a hydraulic piston at \(C\). Point \(C\) is \(8.0 \mathrm{~m}\) from \(A,\) and the force \(\overrightarrow{\boldsymbol{F}}\) exerted by the piston makes an angle of \(40^{\circ}\) with the ladder. What magnitude must \(\boldsymbol{F}\) have to just lift the ladder off the support bracket at \(B ?\) Start with a free-body diagram of the ladder.

His body is again leaning back at \(30.0^{\circ}\) to the vertical, but now the height at which the rope is held above-but still parallel to - the ground is varied. The tension in the rope in front of the competitor \(\left(T_{1}\right)\) is measured as a function of the shortest distance between the rope and the ground (the holding height). Tension \(T_{1}\) is found to decrease as the holding height increases. What could explain this observation? As the holding height increases, (a) the moment arm of the rope about his feet decreases due to the angle that his body makes with the vertical; (b) the moment arm of the weight about his feet decreases due to the angle that his body makes with the vertical; (c) a smaller tension in the rope is needed to produce a torque sufficient to balance the torque of the weight about his feet; (d) his center of mass moves down to compensate, so less tension in the rope is required to maintain equilibrium.

You apply a force of magnitude \(F_{\perp}\) to one end of a wire and another force \(F_{\perp}\) in the opposite direction to the other end of the wire. The cross-sectional area of the wire is \(8.00 \mathrm{~mm}^{2}\). You measure the fractional change in the length of the wire, \(\Delta l / l_{0},\) for several values of \(F_{\perp}\) When you plot your data with \(\Delta l / l_{0}\) on the vertical axis and \(F_{\perp}\) (in units of \(\mathrm{N}\) ) on the horizontal axis, the data lie close to a line that has slope \(8.0 \times 10^{-7} \mathrm{~N}^{-1} .\) What is the value of Young's modulus for this wire?

The left-hand end of a slender uniform rod of mass \(m\) is placed against a vertical wall. The rod is held in a horizontal position by friction at the wall and by a light wire that runs from the right-hand end of the rod to a point on the wall above the rod. The wire makes an angle \(\theta\) with the rod. (a) What must the magnitude of the friction force be in order for the rod to remain at rest? (b) If the coefficient of static friction between the rod and the wall is \(\mu_{\mathrm{s}},\) what is the maximum angle between the wire and the rod at which the rod doesn't slip at the wall?
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