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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 \(F\).

Short Answer

Expert verified
Each cord has a tension of 0.8829 N.

Step by step solution

01

Understanding the System

The system consists of a rod and two masses (spheres) attached. The masses and rod are suspended by vertical cords at each end, maintaining the rod in a horizontal position. The system involves calculating tensions in these cords due to gravitational forces acting downward.
02

Calculate Weight of Each Object

Calculate the weight of each sphere and the rod using the formula: \( W = m imes g \) where \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity. - Sphere 1: \( W_1 = 0.0240 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 0.23544 \, \text{N} \) - Sphere 2: \( W_2 = 0.0360 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 0.35316 \, \text{N} \) - Rod: \( W_r = 0.120 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 1.1772 \, \text{N} \)
03

Determine Force Equilibrium

Since the system is in equilibrium, the sum of forces and moments around any point must be zero. The combined weight of the spheres and rod must equal the tension in the cords when summed.Total weight \( W_{total} = W_1 + W_2 + W_r = 0.23544 + 0.35316 + 1.1772 = 1.7658 \, \text{N} \)
04

Calculate Tension in the Cords

Each cord contributes to balancing the total weight of the system. As the setup is symmetric and in equilibrium, the tension in the cords at each end (A and B) is equal, sharing the load equally.Tension in each cord (T_A and T_B):\[ T_A = T_B = \frac{W_{total}}{2} = \frac{1.7658}{2} = 0.8829 \, \text{N} \]

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Key Concepts

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

Tension in Cords
In our given system, the shiny glass spheres and the rod are suspended by cords, which introduces the concept of tension. Tension refers to the force transmitted through a string, rope, cable, or cord when it is pulled tight by forces acting from opposite ends. Imagine a tug-of-war rope; the more people pulling on either side, the tighter it gets. Similarly, the cords holding the system handle the weight of the spheres and the rod, pulling in opposite directions to keep them balanced in the air.

There are a couple of things to remember about tension:
  • Tension acts along the direction of the cord and away from the object it's supporting.
  • In general, tension is constant throughout a single, uninterrupted cord if only gravity acts on the system.
In this symmetrical setup, the tension in cords at each end of the rod must be equal. This is because there is uniform distribution of mass, and equal force needs to hold the rod horizontally.
Weight Calculation
Weight is the force exerted by gravity on an object. To compute the weight of each sphere and the rod, we use the weight equation: \[ W = m \times g \]where
  • \( m \) is the mass of the object,
  • \( g \) is the acceleration due to gravity \((9.81 \, \text{m/s}^2)\).
For our system:
  • The weight of the first sphere (\( W_1 \)) was calculated as 0.23544 N.
  • The weight of the second sphere (\( W_2 \)) came out to be 0.35316 N.
  • For the rod, we found \( W_r = 1.1772 \) N.
Don't forget that even though these numbers may seem small, they are essential for balancing forces in the suspended setup, contributing to the overall tension in the cords.
Symmetrical System
A symmetrical system is when both sides of it mirror one another in terms of mass distribution. This symmetry is key for many physics problems because it simplifies the calculations you need.
  • In our specific case, since the rod is uniform, and the masses are neatly suspended, the forces are evenly spread out.
  • This symmetry assures us that the tensions in the opposing cords will be equal.
Whenever you spot symmetry, it's a strong hint that some forces or conditions will mirror each other. You'll often find solutions easier when symmetry is present since forces balance naturally across the system. In this exercise, the rod's horizontal position means it's not tilting left or right, pointing to a symmetrical equilibrium.
Force Equilibrium
In a state of mechanical equilibrium, all forces acting on an object are balanced, meaning the object remains at rest or moves at a constant velocity. Mathematically, this means the sum of all forces (\( \Sigma F \)) and the sum of all torques (\( \Sigma \tau \)) must equal zero.
  • The sum of vertical forces is zero because the gravitational forces on the spheres and rod are precisely balanced by the upward tension in the cords.
  • Torque also balances around any point since the symmetric arrangement negates any rotational forces.
In our setup:
  • The combined weight of all components equals the sum of tensions, \( W_{total} = T_A + T_B \).
  • This balance assures the rod stays level, without tipping or twisting, as represented in the tension distribution between the cords.
Understanding force equilibrium is vital in predicting how systems behave under various forces, especially in construction and engineering applications.

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Most popular questions from this chapter

The compressive strength of our bones is important in everyday life. Young's modulus for bone is about 1.4 \(\times\) 10\(^{10}\) Pa. Bone can take only about a 1.0% change in its length before fracturing. (a) What is the maximum force that can be applied to a bone whose minimum cross-sectional area is 3.0 cm\(^2\)? (This is approximately the crosssectional area of a tibia, or shin bone, at its narrowest point.) (b) Estimate the maximum height from which a 70-kg man could jump and not fracture his tibia. Take the time between when he first touches the floor and when he has stopped to be 0.030 s, and assume that the stress on his two legs is distributed equally.

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