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Chapter 9: Problem 46

You are holding a shopping basket at the grocery store with two \(0.56-\mathrm{kg}\) cartons of cereal at the left end of the basket. The basket is \(0.71 \mathrm{m}\) long. Where should you place a \(1.8-\mathrm{kg}\) half gallon of milk, relative to the left end of the basket, so that the center of mass of your groceries is at the center of the basket?

Short Answer

Expert verified
Place the milk 0.576 m from the left end of the basket.

Step by step solution

01

Understand the Problem

We need to find the position of the milk in the basket such that the center of mass of all items is at the center of the basket. We have two 0.56 kg cereal cartons at the left end and a 1.8 kg milk carton whose position we need to determine. The basket is 0.71 m long.
02

Identify and Define Variables

The masses of the cereal cartons are each 0.56 kg. Thus, the total mass of both cereals is \(2 \times 0.56 \ kg = 1.12 \ kg\). The mass of the milk is 1.8 kg. Let \(x\) be the distance from the left end of the basket to where the milk is placed. The center of the basket is \(\frac{0.71}{2} = 0.355 \ m\).
03

Set Up the Center of Mass Equation

The center of mass of the items can be found using the equation: \(\frac{m_1x_1 + m_2x_2}{m_1 + m_2} = x_{cm}\), where \(m_1\) and \(x_1\) are the masses and positions of the cereal cartons, \(m_2\) and \(x_2\) are the masses and positions of the milk, and \(x_{cm}\) is the center of the basket (0.355 m). The cereals are at the left end, so their position is 0 m. Substitute these into the equation:\[\frac{1.12 \times 0 + 1.8 \times x}{1.12 + 1.8} = 0.355\]
04

Solve for Milk Location

Solve \(\frac{1.8x}{2.92} = 0.355\) for \(x\):\[1.8x = 2.92 \times 0.355\]\[1.8x = 1.0366\]\[x = \frac{1.0366}{1.8} = 0.576\ m\]
05

Interpret the Result

The milk should be placed 0.576 m from the left end of the basket, ensuring the center of mass of the groceries is in the center of the basket at 0.355 m.

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Key Concepts

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

Understanding Mass Distribution
Mass distribution refers to how mass is spread out in any given system. In our grocery basket problem, mass distribution helps us find out where to place the milk so that the center of mass is centered.

When we talk about the center of mass, it's like the balance point of all the masses involved.
  • The two cereal cartons are positioned at one end, with their combined mass being fixed at that point.
  • The milk needs to be positioned so that this balance point stays right in the middle of the basket.
By understanding how we can distribute the weight (mass) across the basket, we can use simple equations to solve for the right spot for the milk, ensuring the entire setup stays balanced as desired.
Finding Equilibrium
Equilibrium in physics often means balance. In this context, the equilibrium condition is met when the mass center is aligned with the mid-point of the basket, making the basket easier to carry.

This problem involves calculating such a balance. We ensure that the sum of the moments (mass times distance) about any point equals zero.
  • For our basket, this involves balancing the two cereal cartons at one end with the milk placed at the calculated distance.
  • This balance check means the center of mass remains at the basket's midpoint, creating equilibrium.
Taking these steps helps maintain stability, allowing the groceries to be carried without tipping to one side.
Approaching Physics Problem Solving
Physics problem solving often involves breaking down complex scenarios into simpler steps, just like the grocery basket situation. Here鈥檚 how such problems can typically be approached:
  • **Understand the Problem:** Identify what is being asked. Here, it's about finding the milk's position for a balanced load.
  • **Identify Variables and Set Up Equations:** Determine what values are given (like mass of cereals and milk, length of basket), and what needs to be calculated, followed by forming relevant equations.
  • **Solve and Interpret:** Utilize algebra to solve the equations and interpret the results. For instance, finding the distance where milk should be placed.
This methodical approach helps in systematically tackling problems and reaching logical solutions in physics, quite beneficial for students aiming to sharpen their analytical skills.

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Most popular questions from this chapter

Two air carts of mass \(m_{1}=0.84 \mathrm{kg}\) and \(m_{2}=0.42 \mathrm{kg}\) are placed on a frictionless track. Cart 1 is at rest initially, and has a spring bumper with a force constant of \(690 \mathrm{N} / \mathrm{m}\). Cart 2 has a flat metal surface for a bumper, and moves toward the bumper of the stationary cart with an initial speed \(v=0.68 \mathrm{m} / \mathrm{s}\). (a) What is the speed of the two carts at the moment when their speeds are equal? (b) How much energy is stored in the spring bumper when the carts have the same speed? (c) What is the final speed of the carts after the collision?

Force A has a magnitude \(F\) and acts for the time \(\Delta t\) force B has a magnitude \(2 F\) and acts for the time \(\Delta t / 3,\) force \(C\) has a magnitude \(5 F\) and acts for the time \(\Delta t / 10,\) and force \(D\) has a magnitude \(10 F\) and acts for the time \(\Delta t / 100 .\) Rank these forces in order of increasing impulse. Indicate ties where appropriate.

Suppose you throw a rubber ball at an elephant that is charging directly at you (not a good idea). When the ball bounces back toward you, is its speed greater than, less than, or equal to the speed with which you threw it? Explain.

A cart of mass \(m\) moves with a speed \(v\) on a frictionless air track and collides with an identical cart that is stationary. If the two carts stick together after the collision, what is the final kinetic energy of the system?

A 0.726-kg rope 2.00 meters long lies on a floor. You grasp one end of the rope and begin lifting it upward with a constant speed of \(0.710 \mathrm{m} / \mathrm{s}\). Find the position and velocity of the rope's center of mass from the time you begin lifting the rope to the time the last piece of rope lifts off the floor. Plot your results. (Assume the rope occupies negligible volume directly below the point where it is being lifted.)
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