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A uniform rod is 2.00 m long and has mass 1.80 kg. A 2.40-kg clamp is attached to the rod. How far should the center of gravity of the clamp be from the left-hand end of the rod in order for the center of gravity of the composite object to be 1.20 m from the left-hand end of the rod?

Short Answer

Expert verified
The center of gravity of the clamp should be 1.35 m from the left-hand end of the rod.

Step by step solution

01

Identify the known values

The rod is 2.00 m long and has a mass of 1.80 kg. The clamp has a mass of 2.40 kg. The center of gravity of the composite object is required to be 1.20 m from the left-hand end of the rod.
02

Set up the equation for the center of gravity

The formula for the center of gravity of a system of particles is given by:\[ x_{cg} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2} \]where \(m_1\) and \(m_2\) are the masses of the rod and the clamp, respectively, and \(x_1\) and \(x_2\) are their distances from a reference point, in this case, the left-hand end of the rod.
03

Substitute known values into the equation

Substituting the given values into the center of gravity equation:\[ 1.20 = \frac{1.80 \times 1.00 + 2.40 \times x_2}{1.80 + 2.40} \]Here, \(x_1 = 1.00 \text{ m}\) because the center of gravity of the uniform rod is at its midpoint.
04

Solve for \(x_2\)

First, calculate the combined mass:\[ 1.80 + 2.40 = 4.20 \text{ kg} \]Then set up the equation:\[ 1.20 = \frac{1.80 + 2.40x_2}{4.20} \]Rearrange and solve for \(x_2\):\[ 1.20 \times 4.20 = 1.80 + 2.40x_2 \]\[ 5.04 = 1.80 + 2.40x_2 \]\[ 5.04 - 1.80 = 2.40x_2 \]\[ 3.24 = 2.40x_2 \]\[ x_2 = \frac{3.24}{2.40} \approx 1.35 \text{ m} \]
05

Verify the solution

Double-check the calculation by substituting \(x_2 = 1.35\) m back into the original equation to ensure that the center of gravity is indeed 1.20 m from the left-hand end:\[ \frac{1.80 \times 1.00 + 2.40 \times 1.35}{4.20} = 1.20 \]The left side of the equation simplifies to 1.20, confirming that the solution is correct.

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Key Concepts

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

Uniform Rod
A uniform rod is a fascinating concept in physics. It refers to a rod that has the same mass per unit length throughout its entire structure. This consistency means that the material of the rod is evenly distributed, so its mass is spread uniformly. When you think about a uniform rod's center of gravity, it's quite simple: it's right in the middle. For example, a 2-meter long uniform rod will have its center of gravity at the 1-meter mark.
This concept is important in many physics problems as it simplifies calculations involving the rod's mass and position. Knowing the location of the center of gravity allows us to predict how the rod will react in different scenarios, such as balancing or rotating. A clear grasp of how uniform rods work not only helps with solving problems but also supports understanding more complex systems.
Composite Object
The term 'composite object' is used in physics to describe an object made up of two or more different parts, especially with different masses. In our exercise, we have a uniform rod and a clamp, which together form a composite object. To find the center of gravity of this composite object, we need to consider the mass and position of each component.
When dealing with composite objects, the center of gravity is found by calculating a weighted average of the positions of the individual parts. This means taking into account both the mass and position of each part relative to a common reference point. In this case, we're looking at the left end of the rod.
Understanding how to calculate the center of gravity for composite objects allows us to solve problems where multiple objects interact, helping to predict how they will behave as a unit.
Mass Distribution
Mass distribution refers to how mass is spread across an object or system. In our exercise, understanding mass distribution is crucial when considering how both the rod and the attached clamp contribute to the overall system's center of gravity.
With uniform rods, every section of the rod contributes equally to the total mass. For non-uniform systems like a rod with an attached clamp, each part has a different mass contribution based on its mass and position. Therefore, when solving such problems, it's essential to carefully analyze how each part's mass is distributed in relation to the whole system.
By understanding mass distribution, we can predict how a system will react to forces, aiding in the design and analysis of everything from simple physics problems to complex engineering structures.
Physics Problem Solving
In physics, problem solving is a vital skill that helps us apply theoretical concepts to real-world scenarios. The step-by-step approach used in our example is key for tackling problems effectively. Begin by identifying all known variables clearly, as this sets a solid foundation.
Next, use relevant equations and substitute the known values. For instance, the center of gravity formula in our example helps calculate where the center of gravity of the composite object should be. Solving the equation step by step ensures accuracy.
Finally, double-check your solution by verifying if it satisfies the original problem's conditions. This not only confirms the correctness of your solution but also deepens your understanding of the problem. Developing a structured problem-solving approach will benefit anyone learning physics and related fields.

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Most popular questions from this chapter

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 (\({T_1}\)) 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.

An angler hangs a 4.50-kg fish from a vertical steel wire 1.50 m long and 5.00 \(\times\) 10\(^{-3}\) cm\(^2\) in cross-sectional area. The upper end of the wire is securely fastened to a support. (a) Calculate the amount the wire is stretched by the hanging fish. The angler now applies a varying force \(\overrightarrow{F}\) at the lower end of the wire, pulling it very slowly downward by 0.500 mm from its equilibrium position. For this downward motion, calculate (b) the work done by gravity; (c) the work done by the force \(\overrightarrow{F}\), (d) the work done by the force the wire exerts on the fish; and (e) the change in the elastic potential energy (the potential energy associated with the tensile stress in the wire). Compare the answers in parts (d) and (e). \(\textbf{Torques and Tug-of-War.}\) In a study of the biomechanics of the tug-of-war, a 2.0-m-tall, 80.0-kg competitor in the middle of the line is considered to be a rigid body leaning back at an angle of 30.0\(^\circ\) to the vertical. The competitor is pulling on a rope that is held horizontal a distance of 1.5 m from his feet (as measured along the line of the body). At the moment shown in the figure, the man is stationary and the tension in the rope in front of him is \({T_1} =\) 1160 N. Since there is friction between the rope and his hands, the tension in the rope behind him, \({T_2}\) is not equal to \({T_1}\). His center of mass is halfway between his feet and the top of his head. The coefficient of static friction between his feet and the ground is 0.65.

A nylon rope used by mountaineers elongates 1.10 m under the weight of a 65.0-kg climber. If the rope is 45.0 m in length and 7.0 mm in diameter, what is Young's modulus for nylon?

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

A metal rod that is 4.00 m long and 0.50 cm\(^2\) in crosssectional area is found to stretch 0.20 cm under a tension of 5000 N. What is Young's modulus for this metal?

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