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Chapter 12: Problem 9

A meter stick balances horizontally on a knife-edge at the 50.0 cm mark. With two 5.00 g coins stacked over the 12.0 cm mark, the stick is found to balance at the 45.5 cm mark. What is the mass of the meter stick?

Short Answer

Expert verified
The mass of the meter stick is 74.5 g.

Step by step solution

01

Understand the Problem

We are given a balanced meter stick with known weights placed on it, and we need to determine the mass of the stick itself. The balance point changes when coins are added, indicating a shift in equilibrium.
02

Set Up Torque Equations

The torque about the new balance point (45.5 cm) must be zero. The left side has two coins, each with mass 5.00 g placed at 12.0 cm. The torque from the coins is \(Torque_{coins} = (0.005 \, \text{kg}) \times 2 \times 9.8 \, \text{m/s}^2 \times (0.455 - 0.120) \, \text{m} \).
03

Calculate Torque Due to Coins

Convert distance to meters and solve: \[ Torque_{coins} = 9.8 \times 0.01 \times 0.335 = 0.03283 \, \text{Nm}. \]
04

Set Up Torque for Meter Stick's Mass

Assuming the meter stick's mass, \(M\), acts uniformly about its center of mass at 50.0 cm, the torque it exerts about the 45.5 cm mark is \[ Torque_{stick} = Mg(0.500 - 0.455). \]
05

Balance Equations and Solve for Mass

Since the sum of torques must be zero: \[ M \times 9.8 \times 0.045 = 0.03283. \] Solving for \(M\): \[ M = \frac{0.03283}{9.8 \times 0.045} \approx 0.0745 \, \text{kg}. \] So the mass of the meter stick is 74.5 g.

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Key Concepts

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

Equilibrium
Equilibrium is a state where everything is perfectly balanced. Imagine a playground seesaw. If two people of equal weight sit on either end, the seesaw is balanced at equilibrium. In physics, specifically mechanical equilibrium, this means all forces acting on an object are balanced and there is no net movement.

In the meter stick problem, it's crucial to understand equilibrium to solve for the unknown mass. Before adding coins, the meter stick balanced at its center, the 50 cm mark. Adding coins changed this, illustrating how new forces affected the stick's balance point, shifting it to where all torques (a turning or twisting force) are again at equilibrium.
  • Equilibrium = No net force or movement.
  • Key in solving problems involving balancing forces.
  • Involves understanding forces and torques working together.
Understanding equilibrium helps unpack how additional weights shift balance, critical in this type of physics problem.
Balance Point
The balance point is where an object can be perfectly balanced and is crucial in solving physics problems related to equilibria. For our meter stick example, the balance point began at the 50.0 cm mark.

However, when two coins were added, it shifted. This was because the additional weight created more torque on one side of the balance point.

The balance point moved to 45.5 cm, and this shift is central to calculating the mass of the stick.
  • Balance point = Equal torques on all sides.
  • Determining balance point aids in solving for weights and masses.
  • Shifted balance point indicated new forces acting on the stick.
Getting the balance point right allows us to set up torque equations that show how added forces affect the overall balance, leading to solving for the unknowns.
Center of Mass
Understanding the center of mass simplifies the analysis of motion for rigid bodies, like a meter stick. For objects where mass is distributed uniformly, the center of mass is the midpoint.

In the problem, it was the 50.0 cm mark before any external weight was added.

The center of mass tells us where the total mass of an object concentrates and how it will react to forces.
  • Center of mass = "average" location of mass.
  • Important for understanding balance and equilibrium.
  • Helps predict how objects will move or balance when forces are applied.
Once the concept of center of mass is clear, it's easier to see how adding weights to one side allows us to calculate the mass of the entire object, using the shift in balance to set up equations. This concept is key to many physics problems involving motion and balance.

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Most popular questions from this chapter

A pan balance is made up of a rigid, massless rod with a hanging pan attached at each end. The rod is supported at and free to rotate about a point not at its center. It is balanced by unequal masses placed in the two pans. When an unknown mass \(m\) is placed in the left pan, it is balanced by a mass \(m_{1}\) placed in the right pan: when the mass \(m\) is placed in the right pan, it is balanced by a mass \(m_{2}\) in the left pan. Show that \(m=\sqrt{m_{1} m_{2}}\)

A makeshift swing is constructed by making a loop in one end of a rope and tying the other end to a tree limb. A child is sitting in the loop with the rope hanging vertically when the child's father pulls on the child with a horizontal force and displaces the child to one side. Just before the child is released from rest, the rope makes an angle of \(15^{\circ}\) with the vertical and the tension in the rope is \(280 \mathrm{~N}\). (a) How much does the child weigh? (b) What is the magnitude of the (horizontal) force of the father on the child just before the child is released? (c) If the maximum horizontal force the father can exert on the child is \(93 \mathrm{~N},\) what is the maximum angle with the vertical the rope can make while the father is pulling horizontally?

Figure 12-50 shows a 70 kg climber hanging by only the crimp hold of one hand on the edge of a shallow horizontal ledge in a rock wall. (The fingers are pressed down to gain purchase.) Her feet touch the rock wall at distance \(H=2.0 \mathrm{~m}\) directly below her crimped fingers but do not provide any support. Her center of mass is distance \(a=0.20 \mathrm{~m}\) from the wall. Assume that the force from the ledge supporting her fingers is equally shared by the four fingers. What are the values of the (a) horizontal component \(F_{h}\) and (b) vertical component \(F_{v}\) of the force on each fingertip?

A \(75 \mathrm{~kg}\) window cleaner uses a \(10 \mathrm{~kg}\) ladder that is \(5.0 \mathrm{~m}\) long. He places one end on the ground \(2.5 \mathrm{~m}\) from a wall, rests the upper end against a cracked window, and climbs the ladder. He is \(3.0 \mathrm{~m}\) up along the ladder when the window breaks. Neglect friction between the ladder and window and assume that the base of the ladder does not slip. When the window is on the verge of breaking, what are (a) the magnitude of the force on the window from the ladder, (b) the magnitude of the force on the ladder from the ground, and (c) the angle (relative to the horizontal) of that force on the ladder?

The leaning Tower of Pisa is \(59.1 \mathrm{~m}\) high and \(7.44 \mathrm{~m}\) in diameter. The top of the tower is displaced \(4.01 \mathrm{~m}\) from the vertical. Treat the tower as a uniform, circular cylinder. (a) What additional displacement, measured at the top, would bring the tower to the verge of toppling? (b) What angle would the tower then make with the vertical?
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