/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 A uniform \(1.4-\mathrm{kg}\) ro... [FREE SOLUTION] | 91Ó°ÊÓ

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A uniform \(1.4-\mathrm{kg}\) rod that is \(0.75 \mathrm{m}\) long is suspended at rest from the ceiling by two springs, one at each end of the rod. Both springs hang straight down from the ceiling. The springs have identical lengths when they are unstretched. Their spring constants are \(59 \mathrm{N} / \mathrm{m}\) and \(33 \mathrm{N} / \mathrm{m}\). Find the angle that the rod makes with the horizontal.

Short Answer

Expert verified
The rod makes an angle with the horizontal due to unequal spring constants, but exact calculations need specific extensions to get the precise angle.

Step by step solution

01

Understand the Problem

The rod is in equilibrium, meaning that the net force and the net torque (rotational force) acting on it are zero. The two springs at either end of the rod have different spring constants, causing the rod to tilt. Our goal is to find the angle that the rod tilts from the horizontal.
02

Define Forces Acting on the Rod

Each spring exerts a force equal to the spring constant multiplied by its extension. Let the extensions of the two springs be \( x_1 \) and \( x_2 \) such that the force from the spring with constant 59 N/m is \( F_1 = 59 x_1 \) and from the spring with constant 33 N/m is \( F_2 = 33 x_2 \).
03

Balance the Forces for Equilibrium

The net force on the rod needs to be zero. Vertically, this is expressed as: \[ m g = F_1 + F_2 \]where \( m = 1.4 \, \text{kg} \) is the mass of the rod and \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity. Calculate \( m g \) and solve for \( x_1 \) and \( x_2 \) using this equation.
04

Balance the Torques for Equilibrium

Take the pivot point at the center of the rod. The torques due to the springs need to be equal and opposite for the rod to not rotate: \[ F_1 \cdot \frac{L}{2} = F_2 \cdot \frac{L}{2} + \tau_{mg} \]where \( L = 0.75 \, \text{m} \) is the length of the rod, and \( \tau_{mg} \) is the torque due to the weight of the rod acting at the center. Since we take torques about the center, the torques from the weight of the rod are balanced by the torques from the springs.
05

Solve for the Angle

Since the rod is not moving, the extensions \( x_1 \) and \( x_2 \) will cause a slight tilt. Use the trigonometric relation that gives the difference in extensions in terms of the angle \( \theta \) the rod makes with the horizontal:\[ \tan \theta = \frac{x_1 - x_2}{L} \] Finally, solve for \( \theta \) considering \( F_1 > F_2 \) which means \( x_1 > x_2 \).

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Key Concepts

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

Torque
Torque is a fundamental concept in physics that describes how a force causes an object to rotate. Imagine trying to open a door – pushing at the handle makes it swing open, whereas pushing near the hinges requires much more effort. That's torque in action! It depends on three factors:
  • The magnitude of the force applied
  • The distance from the pivot point (also known as the lever arm)
  • The angle at which the force is applied
Mathematically, torque \( (\tau) \) is expressed as \( \tau = F \times r \times \sin \theta \), where \( F \) is the force applied, \( r \) is the length of the lever arm, and \( \theta \) is the angle between the force vector and the lever arm. In our problem, the pivot is at the center of the rod, and the forces are due to the springs pulling at either end, creating a rotational effect or torque. For equilibrium, the torques from the weights and the springs must balance each other, ensuring that the net torque is zero.
Spring Constant
The spring constant, denoted as \( k \), is characteristic of a spring's stiffness. It tells us how much force is needed to stretch (or compress) the spring by a unit distance. Think of a spring as a like a bouncy rubber band; the stiffer it is, the higher the spring constant.In the exercise, there are two springs with different spring constants, 59 N/m and 33 N/m. This difference is crucial because it causes an imbalance in the forces, influencing the tilt of the rod. Hooke's Law relates the spring constant to force and extension: \( F = k \times x \), where \( F \) is the force exerted by the spring, \( x \) is the extension, and \( k \) is the spring constant. When equilibrium is achieved, the forces from the springs match the gravitational force on the rod, playing a key role in determining the system's behavior.
Rotational Equilibrium
Rotational equilibrium occurs when an object is not experiencing any net torque, meaning it is either at rest or rotating at a constant angular velocity. For the rod in our problem, being in rotational equilibrium implies that the torque forces generated by the springs and the weight of the rod are balanced.To find the condition of rotational equilibrium, we use the principle that the sum of clockwise torques must equal the sum of counterclockwise torques. This can be represented equation-wise as:\[ \tau_\text{clockwise} = \tau_\text{counterclockwise} \]For the rod, this took into account the weights pulling down at its center and the forces exerting upward torque at each end. Achieving rotational equilibrium is essential for determining the tilt angle of the rod because any unbalance would cause it to rotate until equilibrium is restored.
Force Balance
In equilibrium scenarios, understanding force balance helps us ensure that the sum of forces acting in any direction is zero. The rod hangs at rest because all forces acting vertically are perfectly balanced.Considering forces on the rod, the upward forces from the two springs (\( F_1 \) and \( F_2 \)) must counterbalance the downward gravitational force (\( m\cdot g \)). Mathematically, this is represented as:\[ m\cdot g = F_1 + F_2 \]Breaking it down:- \( m \) is the mass of the rod (1.4 kg)- \( g \) is the acceleration due to gravity (9.8 m/s²)- \( F_1 = 59x_1 \) for the spring with 59 N/m- \( F_2 = 33x_2 \) for the spring with 33 N/mBalancing these forces ensures the rod remains in place without any vertical motion, setting the stage to address rotational aspects, which ultimately define the angle of tilt.

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Most popular questions from this chapter

You are running from pirates on a tropical island somewhere in the Caribbean. You have somehow become separated from the rest of your group and now find yourself on the edge of a cliff with your pursuers less than 10 minutes behind you. According to a sign posted on the guardrail at the cliff's edge, the drop to the beach below is \(h=140\) feet. Your team members (waiting for you on the beach, near your boat) have a rope, but there is no time for anyone to climb the cliff to save you. You break into a deserted cabin nearby, and rummage around for a rope. Instead, you find a brand new, still-in-package, bungee cord that must have been intended for tourists jumping from a nearby bridge. You figure you might be able to attach it to the guardrail and jump to the beach, letting go at the bottom before it reverses your motion. You read the bungee cord specifications on the package: the total length of the cord is \(L_{0}=100 \mathrm{m},\) the maximum elastic deformation is \(200 \%\) (i.e., it can safely triple its length), and the elastic constant is \(k=75.0 \mathrm{N} / \mathrm{m}\). (a) If you weigh \(170 \mathrm{lb},\) how far is the bungee cord designed to let you fall before it stops you and reverses your direction? Will this afford you a safe landing? (b) You realize that you don't have to hang from the very end of the bungee, but rather from some point in the middle. How far from the attached end should you grasp the unstretched bungee cord so that you land softly on the beach? Will you be able to perform the jump and stay under the elastic deformation limit?

Depending on how you fall, you can break a bone easily. The severity of the break depends on how much energy the bone absorbs in the accident, and to evaluate this let us treat the bone as an ideal spring. The maximum applied force of compression that one man's thighbone can endure without breaking is \(7.0 \times 10^{4} \mathrm{N} .\) The minimum effective cross-sectional area of the bone is \(4.0 \times 10^{-4} \mathrm{m}^{2},\) its length is \(0.55 \mathrm{m},\) and Young's modulus is \(Y=9.4 \times 10^{9} \mathrm{N} / \mathrm{m}^{2} .\) The mass of the man is \(65 \mathrm{kg} .\) He falls straight down without rotating, strikes the ground stiff-legged on one foot, and comes to a halt without rotating. To see that it is easy to break a thighbone when falling in this fashion, find the maximum distance through which his center of gravity can fall without his breaking a bone.

A spring lies on a horizontal table, and the left end of the spring is attached to a wall. The other end is connected to a box. The box is pulled to the right, stretching the spring. Static friction exists between the box and the table, so when the spring is stretched only by a small amount and the box is released, the box does not move. The mass of the box is \(0.80 \mathrm{kg}\), and the spring has a spring constant of \(59 \mathrm{N} / \mathrm{m}\). The coefficient of static friction between the box and the table on which it rests is \(\mu_{\mathrm{s}}=0.74 .\) How far can the spring be stretched from its unstrained position without the box moving when it is released?

A simple pendulum is made from a 0.65-m-long string and a small ball attached to its free end. The ball is pulled to one side through a small angle and then released from rest. After the ball is released, how much time elapses before it attains its greatest speed?

A spring \((k=830 \mathrm{N} / \mathrm{m})\) is hanging from the ceiling of an elevator, and a 5.0-kg object is attached to the lower end. By how much does the spring stretch (relative to its unstrained length) when the elevator is accelerating upward at \(a=0.60 \mathrm{m} / \mathrm{s}^{2} ?\)

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