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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 3.00-m-long, \(190 \mathrm{~N}\), uniform rod at the \(\mathrm{zoo}\) is held in a horizontal position by two ropes at its ends (Fig. E11.21). The left rope makes an angle of \(150^{\circ}\) with the rod, and the right rope makes an angle \(\theta\) with the horizontal. A \(90 \mathrm{~N}\) howler monkey (Alouatta seniculus) hangs motionless \(0.50 \mathrm{~m}\) from the right end of the rod as he carefully studies you. Calculate the tensions in the two ropes and the angle \(\theta\). First make a free-body diagram of the rod.

A claw hammer is used to pull a nail out of a board (see Fig. \(\mathrm{P} 11.45\) ). The nail is at an angle of \(60^{\circ}\) to the board, and a force \(\overrightarrow{\boldsymbol{F}}_{1}\) of magnitude \(400 \mathrm{~N}\) applied to the nail is required to pull it from the board. The hammer head contacts the board at point \(A,\) which is \(0.080 \mathrm{~m}\) from where the nail enters the board. A horizontal force \(\vec{F}_{2}\) is applied to the hammer handle at a distance of \(0.300 \mathrm{~m}\) above the board. What magnitude of force \(\overrightarrow{\boldsymbol{F}}_{2}\) is required to apply the required \(400 \mathrm{~N}\) force \(\left(F_{1}\right)\) to the nail? (Ignore the weight of the hammer.)

The left-hand end of a uniform rod of mass \(2.00 \mathrm{~kg}\) andlength \(1.20 \mathrm{~m}\) is attached to a vertical wall by a frictionless hinge. The rod is held in a horizontal position by an aluminum wire that runs between the right-hand end of the rod and a point on the wall that is above the hinge. The cross-sectional radius of the wire is \(2.50 \mathrm{~mm}\) and the wire makes an angle of \(30.0^{\circ}\) with the rod. (a) What is the length of the wire? (b) An object of mass \(90.0 \mathrm{~kg}\) is suspended from the right-hand end of the rod. What is the increase in the length of the wire when this object is added? In your analysis do you need to be concerned that the lengthening of the wire means that the rod is no longer horizontal?

Mountaineers often use a rope to lower themselves down the face of a cliff (this is called rappelling). They do this with their body nearly horizontal and their feet pushing against the cliff (Fig. \(\mathbf{P 1 1 . 4 9}\) ). Suppose that an \(82.0 \mathrm{~kg}\) climber, who is \(1.90 \mathrm{~m}\) tall and has a center of gravity \(1.1 \mathrm{~m}\) from his feet, rappels down a vertical cliff with his body raised \(35.0^{\circ}\) above the horizontal. He holds the rope \(1.40 \mathrm{~m}\) from his feet, and it makes a \(25.0^{\circ}\) angle with the cliff face. (a) What tension does his rope need to support? (b) Find the horizontal and vertical components of the force that the cliff face exerts on the climber's feet. (c) What minimum coefficient of static friction is needed to prevent the climber's feet from slipping on the cliff face if he has one foot at a time against the cliff?

In a materials testing laboratory, a metal wire made from a new alloy is found to break when a tensile force of \(90.8 \mathrm{~N}\) is applied perpendicular to each end. If the diameter of the wire is \(1.84 \mathrm{~mm},\) what is the breaking stress of the alloy?
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