/*! 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 9 A rod is lying on the top of a t... [FREE SOLUTION] | 91Ó°ÊÓ

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A rod is lying on the top of a table. One end of the rod is hinged to the table so that the rod can rotate freely on the tabletop. Two forces, both parallel to the tabletop, act on the rod at the same place. One force is directed perpendicular to the rod and has a magnitude of \(38.0 \mathrm{N}\). The second force has a magnitude of \(55.0 \mathrm{N}\) and is directed at an angle \(\theta\) with respect to the rod. If the sum of the torques due to the two forces is zero, what must be the angle \(\theta ?\)

Short Answer

Expert verified
The angle \( \theta \) is approximately \( 44.4^\circ \).

Step by step solution

01

Understanding the Problem

The task is to find the angle \( \theta \) at which the second force must act such that the sum of the torques exerted by both forces on the rod is zero. The rod is fixed at one end, allowing it to pivot around this point.
02

Identify Torque Expressions

Torque (\( \tau \)) is calculated as \( \tau = F \cdot r \cdot \sin(\phi) \), where \( F \) is the force, \( r \) is the distance from the pivot, and \( \phi \) is the angle between the force and the lever arm.
03

Apply Torque Condition

The condition given is the sum of the torques must be zero: \( \tau_1 + \tau_2 = 0 \). This means the torque produced by the first force must be equal and opposite to the torque produced by the second force.
04

Calculate Torque from First Force

The first force is perpendicular to the rod, meaning the angle is 90 degrees, and \( \sin(90^\circ) = 1 \). Thus, \( \tau_1 = 38.0 \cdot r \).
05

Calculate Torque from Second Force

For the second force, the torque is \( \tau_2 = 55.0 \cdot r \cdot \sin(\theta) \). Here, \( \theta \) is the angle to be found.
06

Set Torques Equal

Set the torques equal as per the condition: \( 38.0 \cdot r = 55.0 \cdot r \cdot \sin(\theta) \). The distance \( r \) cancels out.
07

Solve for \( \theta \)

The equation simplifies to \( 38.0 = 55.0 \cdot \sin(\theta) \). Solve for \( \sin(\theta) \): \( \sin(\theta) = \frac{38.0}{55.0} \).
08

Calculate the Angle \( \theta \)

Determine \( \theta \) using the inverse sine function: \( \theta = \arcsin\left(\frac{38.0}{55.0}\right) \). Calculate to find \( \theta \).

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Key Concepts

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

Force and Motion
Force and motion are interconnected concepts in physics that help us understand how objects move and interact. A force is any interaction that, when unopposed, changes the motion of an object. Forces can be contact forces, like friction or tension, or non-contact forces, such as gravity and magnetism.
In the context of our exercise, we have two forces acting on a rod that is free to rotate about a hinge. These forces influence the rod's motion by exerting torque, which can cause the rod to rotate around the pivot point. Each force acts parallel to the tabletop with different directions and magnitudes, impacting the rotational motion of the rod differently.
  • The first force, which is perpendicular to the rod, provides the maximum possible torque for its magnitude because the angle of action is 90 degrees, where the sine function reaches its peak value of 1.
  • The second force acts at an angle \( \theta \) with respect to the rod. Its effectiveness in causing rotation depends on the sine of \( \theta \), which ranges from 0 (no rotation) to 1 (maximum rotation) depending on \( \theta \).
Understanding how forces cause motion is key to solving problems involving dynamics and torque.
Angular Mechanics
Angular mechanics, the study of rotation, provides the tools needed to analyze situations where objects pivot around a point. Key to angular mechanics is the concept of torque, which is the rotational equivalent of force in linear motion.
Torque \( (\tau) \) is defined as the product of a force and the lever-arm distance from the axis of rotation, multiplied by the sine of the angle between the force and the lever arm. Mathematically expressed as:
\[\tau = F \cdot r \cdot \sin(\phi)\]
  • \( F \) is the force applied.
  • \( r \) is the distance from the pivot to the point of force application.
  • \( \phi \) is the angle between the force applied and the lever arm.
In our problem, we're interested in ensuring the sum of torques is zero so the rod remains balanced without rotating. This involves setting the torque from force one equal to the negative torque from force two, as their effects cancel each other out. The balance equation highlights how angular mechanics govern the equilibrium conditions in rotational systems.
Physics Problem Solving
Physics problem-solving involves breaking down complex concepts into manageable parts. For situations involving forces and torques, start by understanding the problem's parameters and desired outcomes.
In this exercise, the goal was to find the angle \( \theta \) where the rod experiences no net torque. To achieve this:
  • First, recognize the setup: a rod that can pivot with two forces applied.
  • Identify relevant formulas, such as the torque equation \( \tau = F \cdot r \cdot \sin(\phi) \).
  • Apply given conditions: here, the sum of torques equaling zero.
  • Simplify the equations step-by-step, cancel common factors where possible (such as distance \( r \)), and solve for the unknown (\( \theta \)).
Using mathematical tools like inverse trigonometric functions helps find the precise value of angles when working with sine, cosine, and other trigonometric elements. The process of physics problem-solving sharpens logical thinking and cultivates an ability to work through complex scenarios by systematically applying relevant physics principles.

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Most popular questions from this chapter

The drawing shows a rectangular piece of wood. The forces applied to corners \(\mathrm{B}\) and \(\mathrm{D}\) have the same magnitude of \(12 \mathrm{N}\) and are directed parallel to the long and short sides of the rectangle. The long side of the rectangle is twice as long as the short side. An axis of rotation is shown perpendicular to the plane of the rectangle at its center. A third force (not shown in the drawing) is applied to corner A, directed along the short side of the rectangle (either toward \(\mathrm{B}\) or away from \(\mathrm{B}\) ), such that the piece of wood is at equilibrium. Find the magnitude and direction of the force applied to corner A.

Two disks are rotating about the same axis. Disk A has a moment of inertia of \(3.4 \mathrm{kg} \cdot \mathrm{m}^{2}\) and an angular velocity of \(+7.2 \mathrm{rad} / \mathrm{s} .\) Disk \(\mathrm{B}\) is rotating with an angular velocity of \(-9.8 \mathrm{rad} / \mathrm{s} .\) The two disks are then linked together without the aid of any external torques, so that they rotate as a single unit with an angular velocity of -2.4 rad/s. The axis of rotation for this unit is the same as that for the separate disks. What is the moment of inertia of disk B?

A Ride Inside a Tractor Tire. You and your friends plan to roll down a hill on the inside of 600 -pound tractor tire (diameter \(D=1.80 \mathrm{m}\) ). The hill is inclined at an angle of \(25.0^{\circ}\) and you initially plan to start from a distance \(L=100 \mathrm{m}\) up the hill, but decide to first check whether it will be safe. (a) Assuming the masses of the tire and your 105 -pound body are concentrated at the outer rim of a thin-walled cylinder/hoop, what is the effective acceleration your body experiences at the bottom of the hill where your angular speed is greatest, i.e., how many "g's" will you experience? Assuming the human body can withstand a g-force of \(8.00 \mathrm{g}\) 's \(\left(1 \mathrm{g}=9.80 \mathrm{m} / \mathrm{s}^{2}\right),\) is it safe to make the ride from \(100 \mathrm{m}\) up the hill? (b) What is the maximum starting distance \(\left(L_{\text {max }}\right)\) up the hill that is safe?

A block (mass \(=2.0 \mathrm{kg}\) ) is hanging from a massless cord that is wrapped around a pulley (moment of inertia \(=1.1 \times 10^{-3} \mathrm{kg} \cdot \mathrm{m}^{2}\) ), as the drawing shows. Initially the pulley is prevented from rotating and the block is stationary. Then, the pulley is allowed to rotate as the block falls. The cord does not slip relative to the pulley as the block falls. Assume that the radius of the cord around the pulley remains constant at a value of \(0.040 \mathrm{m}\) during the block's descent. Find the angular acceleration of the pulley and the tension in the cord.

A 15.0 -m length of hose is wound around a reel, which is initially at rest. The moment of inertia of the reel is \(0.44 \mathrm{kg} \cdot \mathrm{m}^{2},\) and its radius is \(0.160 \mathrm{m} .\) When the reel is turning, friction at the axle exerts a torque of magnitude \(3.40 \mathrm{N} \cdot \mathrm{m}\) on the reel. If the hose is pulled so that the tension in it remains a constant \(25.0 \mathrm{N},\) how long does it take to completely unwind the hose from the reel? Neglect the mass and thickness of the hose on the reel, and assume that the hose unwinds without slipping.

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