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Chapter 8: Problem 10

A meterstick is found to balance at the \(49.7-\mathrm{cm}\) mark when placed on a fulcrum. When a \(50.0\) -gram mass is attached at the \(10.0-\mathrm{cm}\) mark, the fulcrum must be moved to the \(39.2-\mathrm{cm}\) mark for balance. What is the mass of the meterstick?

Short Answer

Expert verified
The mass of the meterstick is \(50.0g \cdot \frac{10.0cm}{49.7cm - 39.2cm}\).

Step by step solution

01

Interpret the First Balancing Condition

Interpreting the first balancing condition, the meter stick is balanced at the \(49.7 cm\) mark without any additional mass. This gives us an equation: \[m_{meterstick} \cdot x_{meterstick} = m_{meterstick} \cdot 49.7cm\] since the center of mass equals the product of mass and distance.
02

Interpret the Second Balancing Condition

The second condition tells us that with the addition of a \(50.0 g\) mass at the \(10.0 cm\) mark, the fulcrum must be moved to the \(39.2 cm\) mark to maintain the balance. We can translate this into another equation as follows: \[(m_{meterstick} \cdot x_{meterstick} + m_{mass} \cdot x_{mass})= m_{meterstick} \cdot 39.2cm + 50.0g \cdot 10.0cm \]
03

Substitute and Solve

We can solve these two equations to find the mass of the meter stick, \(m_{meterstick}\). Substituting equation from step 1 into equation from step 2, we get: \[ 49.7cm \cdot m_{meterstick} + 50.0g \cdot 10.0 cm = 39.2cm \cdot m_{meterstick} + 50.0g \cdot 10.0cm \] Simplify it to find \(m_{meterstick}\), resulting in \[m_{meterstick} = \frac{50.0g \cdot 10.0cm}{49.7cm - 39.2cm} \]
04

Obtaining The Final Answer

Calculate the given equation to get the mass of the meterstick.

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Key Concepts

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

Center of Mass
The center of mass is a crucial element in understanding how objects balance. It is the specific point in an object where all of its mass is considered to be concentrated. For symmetrical objects with uniform density, such as a meterstick, the center of mass is typically at its geometric center. However, this might shift if the distribution of mass changes.

In the exercise, the meterstick initially balances at the 49.7 cm mark, indicating that its center of mass is located there. This information is key because it allows us to set up the initial equation needed to solve for the meterstick's mass. When an additional mass is added or the object is altered, the center of mass calculation will help determine how the fulcrum must be adjusted to maintain balance.
Lever Arm
The lever arm, or moment arm, is the perpendicular distance from the fulcrum to the line of action of the force. It is a fundamental concept in calculating torque, which is the rotational equivalent of linear force. Torque can be calculated using the formula \( \tau = F \cdot d \), where \( F \) is the force applied, and \( d \) is the lever arm.

In the context of the exercise, the lever arm changes when a new mass is added to the meterstick. Specifically, the position of the 50.0 g mass at the 10.0 cm mark creates a new lever arm that affects how the system must be balanced. By calculating the torques created by the meterstick and the added weight, it is possible to understand why the fulcrum shifts to a new position.
Rotational Balance
Rotational balance occurs when an object's sum of torques around a fulcrum is zero. This means that the clockwise and counterclockwise torques are equal, keeping the object balanced and stable

In the exercise, achieving rotational balance involves calculating the torques caused by both the meterstick and the added 50.0 g mass. With the fulcrum initially balancing the meterstick at the 49.7 cm mark (its center of mass), the addition of a mass requires the fulcrum's position to be adjusted to restore balance. This results in a new equation based on the lever arms and forces, allowing us to solve for the unknown mass of the meterstick.
Fulcrum Position
Understanding the fulcrum position is essential for achieving balance in a lever system. The fulcrum is the pivot point around which the object rotates. Adjusting its position can change the lever arms, thus altering the torques affecting the system.

In the given exercise, the original fulcrum position at 49.7 cm aligns with the center of mass, ensuring initial balance. When the 50.0 g mass is introduced at the 10.0 cm mark, the fulcrum must be moved to 39.2 cm to maintain equilibrium. This shift compensates for the added torque from the new mass, showing how adjusting the fulcrum can achieve balance by equalizing the system's torques.

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Most popular questions from this chapter

A \(40.0\) -kg child stands at one end of a \(70.0\) -kg boat that is \(4.00 \mathrm{~m}\) long (Fig. \(\mathrm{P} 8.69\) ). The boat is initially \(3.00 \mathrm{~m}\) from the pier. The child notices a turtle on a rock beyond the far end of the boat and proceeds to walk to that end to catch the turtle. (a) Neglecting friction between the boat and water, describe the motion of the system (child plus boat). (b) Where will the child be relative to the pier when he reaches the far end of the boat? (c) Will he catch the turtle? (Assume that he can reach out \(1.00 \mathrm{~m}\) from the end of the boat.)

A space station shaped like a giant wheel has a radius of \(100 \mathrm{~m}\) and a moment of inertia of \(5.00 \times 10^{8} \mathrm{~kg} \cdot \mathrm{m}^{2}, \mathrm{~A}\) crew of 150 lives on the rim, and the station is rotating so that the crew experiences an apparent acceleration of \(1 g\) (Fig. P8.64). When 100 people move to the center of the station for a union meeting, the angular speed changes. What apparent acceleration is experienced by the managers remaining at the rim? Assume the average mass of a crew member is \(65.0 \mathrm{~kg}\).

A bicycle wheel has a diamcter of \(64.0 \mathrm{~cm}\) and a mass of \(1.80 \mathrm{~kg}\). Assume that the wheel is a hoop with all the mass concentrated on the outside radius. The bicycle is placed on a stationary stand, and a resistive force of \(120 \mathrm{~N}\) is applied tangent to the rim of the tire. (a) What force must be applied by a chain passing over a \(9.00-\mathrm{cm}-\) diameter sprocket in order to give the wheel an acceleration of \(4.50 \mathrm{rad} / \mathrm{s}^{2} ?\) (b) What force is required if you shift to a \(5.60-\mathrm{cm}\) -diameter sprocket?

A playground merry-go-round of radius \(2.00 \mathrm{~m}\) has a moment of inertia \(I=275 \mathrm{~kg} \cdot \mathrm{m}^{2}\) and is rotating about a frictionless vertical axle. As a child of mass \(25.0 \mathrm{~kg}\) stands at a distance of \(1.00 \mathrm{~m}\) from the axle, the system (merrygo-round and child) rotates at the rate of \(14.0 \mathrm{rev} / \mathrm{min}\). The child then proceeds to walk toward the edge of the merry-go-round. What is the angular speed of the system when the child reaches the edge?

A \(12.0\) -kg object is attached to a cord that is wrapped around a wheel of radius \(r=10.0 \mathrm{~cm}\) (Fig. P8.70). The acceleration of the object down the frictionless incline is measured to be \(2.00 \mathrm{~m} / \mathrm{s}^{2}\). Assuming the axle of the wheel to be frictionless, determine (a) the tension in the rope, (b) the moment of inertia of the wheel, and (c) the angular speed of the wheel \(2.00 \mathrm{~s}\) after it begins rotating, starting from rest.
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