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Chapter 9: Problem 8

One end of a meter stick is pinned to a table, so the stick can rotate freely in a plane parallel to the tabletop. Two forces, both parallel to the tabletop, are applied to the stick in such a way that the net torque is zero. The first force has a magnitude of 2.00 N and is applied perpendicular to the length of the stick at the free end. The second force has a magnitude of 6.00 N and acts at a $$30.0^{\circ}$$ angle with respect to the length of the stick. Where along the stick is the 6.00-N force applied? Express this distance with respect to the end of the stick that is pinned.

Short Answer

Expert verified
6.00-N force is applied 0.667 m from pinned end.

Step by step solution

01

Identify known values and concepts

Identify that net torque is zero and that torque \( \tau \) is given by \( \tau = F \cdot r \cdot \sin\theta \), where \( F \) is force, \( r \) is the lever arm (distance from pivot), and \( \theta \) is the angle between force and lever arm.
02

Calculate torque of first force

The first force of 2.00 N is applied perpendicularly at the end of the stick with \( r_1 = 1 \text{ meter} \). Therefore, \( \tau_1 = 2.00 \text{ N} \times 1 \text{ m} \times \sin 90^{\circ} = 2.00 \text{ Nm} \).
03

Set up equation for the second force

For the second force, with magnitude 6.00 N and applied at an angle of \( 30.0^{\circ} \), the torque is \( \tau_2 = 6.00 \text{ N} \times r_2 \times \sin 30^{\circ} \). Since net torque is zero: \( 2.00 \text{ Nm} = 6.00 \text{ N} \times r_2 \times 0.5 \).
04

Solve for distance for second force

Rearrange the equation to find \( r_2 \): \( r_2 = \frac{2.00 \text{ Nm}}{6.00 \text{ N} \times 0.5} = \frac{2.00}{3.00} = 0.667 \text{ meters} \).
05

Express the position of the force

The 6.00-N force is applied 0.667 meters from the end of the stick that is pinned.

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Key Concepts

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

Lever Arm
The lever arm, also known as the moment arm, is a fundamental concept in understanding torque. In simple terms, it is the perpendicular distance from the axis of rotation (or the pivot point) to the line of action of the force being applied. This distance directly impacts the amount of rotational force, or torque, exerted by the force.

Key points to remember about lever arms:
  • The longer the lever arm, the greater the torque for the same force applied. This is why long-handled tools make it easier to apply force.
  • If the force is applied at the pivot point itself, the lever arm is zero, resulting in no torque.
In the given problem, the lever arm for the first force is the entire length of the stick (1 meter), as the force is applied at the free end. For the second force, determining the lever arm involves solving for how far along the stick from the pivot point this force acts, given that the net torque must be zero.
Angle of Force Application
The angle of force application is crucial when calculating torque, as it determines how effectively a force is rotated. Torque is calculated using the formula: \[\tau = F \cdot r \cdot \sin\theta\]where \( F \) is the magnitude of the force, \( r \) is the lever arm, and \( \theta \) is the angle between the force and the lever arm.

Important aspects of the angle of force application include:
  • If the force is perpendicular to the lever arm (\( \theta = 90^{\circ} \)), the torque is maximized because \( \sin 90^{\circ} = 1 \).
  • If the force is parallel to the lever arm (\( \theta = 0^{\circ} \) or \( 180^{\circ} \)), no torque is produced since \( \sin 0^{\circ} = 0 \).
In the exercise, the first force is applied perpendicularly, leading to effective torque. The second force, however, is applied at a 30-degree angle, which requires adjusting for this less-than-optimal angle using the sine function, \( \sin 30^{\circ} \), which equals 0.5.
Net Torque
Net torque is the total rotational effect produced by all forces acting on an object. For an object to remain in rotational equilibrium, like our pinned meter stick, the net torque must be zero.

Understanding net torque:
  • If net torque is zero, it means the clockwise torques are balanced by the counterclockwise torques, resulting in no overall rotation.
  • In problems, we often set the algebraic sum of all individual torques to zero to solve for unknowns, such as the position or magnitude of a force.
In the given problem, the solution involved balancing the torque from the first force with the torque from the second to ensure the net torque was zero. This balance allows us to calculate the exact position along the stick where the 6.00-N force must be applied, ensuring the stick does not rotate.

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Most popular questions from this chapter

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.

The parallel axis theorem provides a useful way to calculate the moment of inertia \(I\) about an arbitrary axis. The theorem states that \(I=I_{c m}+M h^{2},\) where \(I_{c m}\) is the moment of inertia of the object relative to an axis that passes through the center of mass and is parallel to the axis of interest, \(M\) is the total mass of the object, and \(h\) is the perpendicular distance between the two axes. Use this theorem and information to determine an expression for the moment of inertia of a solid cylinder of radius \(R\) relative to an axis that lies on the surface of the cylinder and is perpendicular to the circular ends.

A 9.75-m ladder with a mass of 23.2 kg lies flat on the ground. A painter grabs the top end of the ladder and pulls straight upward with a force of \(245 \mathrm{N}\). At the instant the top of the ladder leaves the ground, the ladder experiences an angular acceleration of \(1.80 \mathrm{rad} / \mathrm{s}^{2}\) about an axis passing through the bottom end of the ladder. The ladder's center of gravity lics halfway between the top and bottom ends. (a) What is the net torque acting on the ladder? (b) What is the ladder's moment of inertia?

Just after a motorcycle rides off the end of a ramp and launches into the air, its engine is turning counterclockwise at 7700 rev/min. The motorcycle rider forgets to throttle back, so the engine"s angular speed increases to 12.500 rev/min. As a result, the rest of the motorcycle (including the rider) begins to rotate clockwise about the engine at 3.8 rev/min. Calculate the ratio \(I_{\mathrm{L}} / I_{\mathrm{M}}\) of the moment of inertia of the engine to the moment of inertia of the rest of the motorcycle (and the rider). Ignore torques due to gravity and air resistance.

Two spheres are each rotating at an angular speed of \(24 \mathrm{rad} / \mathrm{s}\) about axes that pass through their centers. Each has a radius of \(0.20 \mathrm{m}\) and a mass of \(1.5 \mathrm{kg}\). However, as the figure shows, one is solid and the other is a thin-walled spherical shell. Suddenly, a net external torque due to friction (magnitude \(=0.12 \mathrm{N} \cdot \mathrm{m}\) ) begins to act on each sphere and slows the motion down. Concepts: (i) Which sphere has the greater moment of inertia and why? (ii) Which sphere has the angular acceleration (a deceleration) with the smaller magnitude? (iii) Which sphere takes a longer time to come to a halt? Calculations: How long does it take each sphere to come to a halt?
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