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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

A relaxed biceps muscle requires a force of 25.0 N for an elongation of 3.0 cm; the same muscle under maximum tension requires a force of 500 N for the same elongation. Find Young's modulus for the muscle tissue under each of these conditions if the muscle is assumed to be a uniform cylinder with length 0.200 m and cross-sectional area 50.0 cm\(^2\).

A holiday decoration consists of two shiny glass spheres with masses 0.0240 kg and 0.0360 kg suspended from a uniform rod with mass 0.120 kg and length 1.00 m (\(\textbf{Fig. P11.62}\)). The rod is suspended from the ceiling by a vertical cord at each end, so that it is horizontal. Calculate the tension in each of the cords \(A\) through \(F\).

A steel cable with cross-sectional area 3.00 cm\(^2\) has an elastic limit of 2.40 \(\times\) 10\(^8\) Pa. Find the maximum upward acceleration that can be given a 1200-kg elevator supported by the cable if the stress is not to exceed one-third of the elastic limit.

A nonuniform beam 4.50 m long and weighing 1.40 kN makes an angle of 25.0\(^\circ\) below the horizontal. It is held in position by a frictionless pivot at its upper right end and by a cable 3.00 m farther down the beam and perpendicular to it (\(\textbf{Fig. E11.20}\)). The center of gravity of the beam is 2.00 m down the beam from the pivot. Lighting equipment exerts a 5.00-kN downward force on the lower left end of the beam. Find the tension \(T\) in the cable and the horizontal and vertical components of the force exerted on the beam by the pivot. Start by sketching a free-body diagram of the beam.

You are a summer intern for an architectural firm. An 8.00-m-long uniform steel rod is to be attached to a wall by a frictionless hinge at one end. The rod is to be held at 22.0\(^\circ\) below the horizontal by a light cable that is attached to the end of the rod opposite the hinge. The cable makes an angle of 30.0\(^\circ\) with the rod and is attached to the wall at a point above the hinge. The cable will break if its tension exceeds 650 N. (a) For what mass of the rod will the cable break? (b) If the rod has a mass that is 10.0 kg less than the value calculated in part (a), what are the magnitude and direction of the force that the hinge exerts on the rod?

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