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Chapter 10: Problem 59

A thin rod of length \(l\) lies on the \(+x\) -axis with its left end at the origin. A string pulls on the rod with a force \(\overrightarrow{\boldsymbol{F}}\) directed toward a point \(P\) a distance \(h\) above the rod. Where along the rod should you attach the string to get the greatest torque about the origin if point \(P\) is (a) above the right end of the rod? (b) Above the left end of the rod? (c) Above the center of the rod?

Short Answer

Expert verified
In all cases, attach the string at the right end of the rod for maximum torque.

Step by step solution

01

Understanding Torque

Torque is the measure of the force that can cause an object to rotate about an axis. The torque \( \tau \) is given by the formula \( \tau = r \cdot F \cdot \sin(\theta) \), where \( r \) is the distance from the point of rotation, \( F \) is the magnitude of the force, and \( \theta \) is the angle between \( r \) and the line of action of the force.
02

Finding Torque Direction for Maximum Torque

For the torque to be maximum, \( \sin(\theta) \) should be maximized. This occurs when \( \theta = 90^\circ \), meaning the force vector \( \overrightarrow{F} \) needs to be perpendicular to the position vector from the origin to the point of force application on the rod.
03

Case (a): Point P Above Right End of Rod

The point \( P \) is directly above the right end of the rod at \( (l, h) \). The position vector from the origin to point where force causes maximum torque is \( r = l \). The force vector is directed towards \( (l, h) \). The position on the rod at which the string should be attached for maximum torque is at its right end.
04

Case (b): Point P Above Left End of Rod

The point \( P \) is directly above the left end of the rod at \( (0, h) \). For maximum torque, the string should be attached at the farthest point from origin which is the right end of the rod, \( r = l \).
05

Case (c): Point P Above Center of Rod

The point \( P \) is directly above the center of the rod at \( \left( \frac{l}{2}, h \right) \). The maximum torque again requires the string to be attached at a point that maximizes \( r \). Hence, attaching the string at the right end of the rod, where \( r = l \), gives the maximum torque.

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Key Concepts

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

Rotational Motion
Rotational motion is when an object spins around an axis. This contrasts with linear motion, where an object moves along a straight path. In rotational motion, every particle of the object moves in a circular path around the axis. For a rod lying on the x-axis, applying force at different points along the rod will yield different rotational effects due to varying distances from the axis of rotation, which is at the origin. These effects are dictated by the concept of torque. Understanding rotational motion requires recognizing that the effectiveness of a force in causing rotation depends on where and at what angle the force is applied relative to the axis.
Physics Problem Solving
Physics problems often require a strategic approach. Here, solving the problem of where to attach a string on a rod to achieve maximum torque involves a clear understanding of the components involved:
  • Identify the forces: Recognize the force vector and the point of application.
  • Analyze the geometry: Consider the layout of the rod and point of force application.
  • Calculate using relevant equations: Here, torque equations help ascertain effects.
By breaking down the problem into smaller steps:
Determine the distances involved.
Calculate the angle between forces and lever arms.
Make sure to accurately determine the point on the rod"s length that maximizes the leverage.
Force and Motion
In mechanics, force is any interaction that, if unopposed, will change the motion of an object. Forces can cause objects to accelerate, remain in place, or rotate, depending on how and where they are applied. For the rod exercise, a force applied differently along the rod will vary the torque produced.
This principle shows why understanding how forces interact with components of a system is crucial. For maximum torque, the force must be applied perpendicularly to the rod's position vector from the rotation axis. This leads to efficient conversion of force into rotational motion, leveraging both the magnitude and position of the applied force.
Vector Analysis
Vectors provide a mathematical way to describe forces and points in space. They have both magnitude and direction, making them ideal for studying complex systems involving multiple forces and movements.
In the context of this rotational problem, vector analysis helps determine the exact impact of a force on the rod by scrutinizing:
  • The direction of the force vector.
  • The position vector from the origin to the point of force application.
  • The angle formed between the two vectors.
This analysis determines the effective component of the force that contributes to torque, maximizing the rotational effect by aligning the force perpendicularly to the lever arm.

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Most popular questions from this chapter

The carbide tips of the cutting teeth of a circular saw are 8.6 \(\mathrm{cm}\) from the axis of rotation. (a) The no-toad speed of the saw, when it is not cutting anything, is 4800 rev/min. Why is no-load power output negligible? (b) While the saw is cutting lumber, its angular speed slows to 2400 rev/min and the power outputis 1.9 \(\mathrm{hp}\) . What is the tangential force that the wood exerts on the carbidetips?

A playground merry-go-round has radius 2.40 \(\mathrm{m}\) and moment of inertia 2100 \(\mathrm{kg} \cdot \mathrm{m}^{2}\) about a vertical axle through its center, and it turns with negligible friction. (a) A child applies an 18.0 \(\mathrm{N}\) force tangentially to the edge of the merry-go-round for 15.0 \(\mathrm{s}\) . If the merry-go-round is initially at rest, what is its angular speed after this 15.0 -s interval? (b) How much work did the child do on the merry-go-round? (c) What is the average power supplied by the child?

A small 10.0-g bug stands at one end of a thin uniform bar that is initially at rest on a smooth horizontal table. The other end of the bar pivots about a nail driven into the table and can rotate freely, without friction. The bar has mass 50.0 \(\mathrm{g}\) and is 100 \(\mathrm{cm}\) in length. The bug jumps off in the horizontal direction, perpendicular to the bar, with a speed of 20.0 \(\mathrm{cm} / \mathrm{s}\) relative to the table. (a) What is the angular speed of the bar just after the frisky insect leaps? (b) What is the total kinetic energy of the system just after the bug leaps? (c) Where does this energy come from?

When an object is rolling without slipping, the rolling friction force is much less than the friction force when the object is sliding; a silver dollar will roll on its edge much farther than it will slide on its flat side (see Section 5.3\() .\) When an object is rolling without slipping on a horizontal surface, we can approximate the friction force to be zero, so that \(a_{x}\) and \(\alpha_{z}\) are approximately zero and \(v_{x}\) and \(\omega_{z}\) are approximately constant. Rolling without slipping means \(v_{x}=r \omega_{z}\) and \(a_{x}=r \alpha_{x} .\) If an object is set in motion on a surface without these equalities, sliding (kinetic) friction will act on the object as it slips until rolling without slipping is established, A solid cylinder with mass \(M\) and radius \(R\) , rotating with angular speed \(\omega_{0}\) about an axis through its center, is set on a horizontal surface for which the kinetic friction coefficient is \(\mu_{k}\) , (a) Draw a free-body diagram for the cylinder on the surface. Think carefully about the direction of the kinetic friction force on the cylinder. Calculate the accelerations \(a_{x}\) of the center of mass and \(\alpha_{z}\) of rotation about the center of mass. (b) The cylinder is initially slipping completely, so initially \(\omega_{z}=\omega_{0}\) but \(v_{x}=0 .\) Rolling without slipping sets in when \(v_{x}=R \omega_{z}\) . Calculate the distance the cylinder rolls before slipping stops. (c) Calculate the work done by the friction force on the cylinder as it moves from where it was set down to where it begins to roll without slipping.

A child rolls a 0.600 -kg basketball up a long ramp. The bas- ketball can be considered a thin-walled, hollow sphere. When the child releases the basketball at the bottom of the ramp, it has a speed of 8.0 \(\mathrm{m} / \mathrm{s}\) . When the ball returns to her after rolling up the ramp and then rolling back down, it has a speed of 4.0 \(\mathrm{m} / \mathrm{s}\). Assume the work done by friction on the basketball is the same when the ball moves up or down the ramp and that the basketball rolls without slipping. Find the maximum vertical height increase of the ball as it rolls up the ramp.
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