/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 53 A \(1.40-\mathrm{kg}\) bottle of... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A \(1.40-\mathrm{kg}\) bottle of vintage wine is lying horizontally in the rack shown in the drawing. The two surfaces on which the bottle rests are \(90.0^{\circ}\) apart, and the right surface makes an angle of \(45.0^{\circ}\) with respect to the ground. Each surface exerts a force on the bottle that is perpendicular to the surface. What is the magnitude of each of these forces?

Short Answer

Expert verified
The force on the left surface is approximately 6.47 N, and the force on the right is 9.15 N.

Step by step solution

01

Understand the System

The bottle is in equilibrium, resting on two surfaces that exert perpendicular forces. The angle between these surfaces is 90°. We need to find the forces exerted by each surface.
02

Identify the Forces

The weight of the wine bottle, \[ W = mg = 1.40 \text{ kg} \times 9.8 \text{ m/s}^2 = 13.72 \text{ N}, \]acts downward. Let \( F_1 \) and \( F_2 \) be the forces exerted by the surfaces. \( F_1 \) acts perpendicular to the left surface, and \( F_2 \) acts perpendicular to the right surface.
03

Set Up Equilibrium Equations

Since the bottle is in equilibrium, the sum of the forces in both horizontal and vertical directions is zero. Horizontally, \( F_1 = F_2 \cos(45°) \). Vertically, \( F_1 \sin(45°) + F_2 = W \).
04

Solve for Forces

Substitute \( F_1 \) from the horizontal equation into the vertical equation:\[F_2 \cos(45°) \sin(45°) + F_2 = 13.72 \text{ N}.\]Simplify and solve for \( F_2 \):\[F_2 \left( \frac{1}{2} + 1 \right) = 13.72,\]\[F_2 \left( \frac{3}{2} \right) = 13.72,\]\[F_2 = \frac{13.72}{1.5} \approx 9.15 \text{ N}.\]Now substitute \( F_2 \) back to find \( F_1 \):\[F_1 = F_2 \cos(45°) \approx 9.15 \times \frac{\sqrt{2}}{2} \approx 6.47 \text{ N}.\]
05

Verify the Solution

Check equilibrium equations with values:- Horizontal: \( 6.47 \approx 9.15 \times \frac{\sqrt{2}}{2} \).- Vertical: \( 6.47 \frac{\sqrt{2}}{2} + 9.15 \approx 13.72 \text{ N}\).The values satisfy both equilibrium conditions.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Forces in Equilibrium
When an object is in equilibrium, it means that all the forces acting on it are balanced. This can be visualized as the object being perfectly still or moving at a constant speed without tumbles or swerves. In the case of our wine bottle, it remains still on the rack because the forces interacting with it come into perfect balance.
  • There is no net force pushing the bottle in any direction.
  • This balance of forces results from equal and opposite forces that cancel each other out.
The forces involved here are the gravitational force pulling the bottle downward, and the two contact forces exerted by the surfaces of the rack. These contact forces are perpendicular to the surfaces, meaning they push directly away from the surface at a 90-degree angle. The position and orientation of these forces are crucial as they must counteract the downward pull of gravity to keep the bottle still.
Understanding equilibrium is vital because it helps ascertain that the forces are truly balanced, which can be calculated and observed through the resulting zero net force.
Newton's Laws of Motion
Newton's Laws of Motion provide the fundamental principles that explain how forces operate. The first law, often called the law of inertia, tells us that an object at rest will stay at rest unless acted upon by a force. In our scenario, the static wine bottle is a perfect demonstration of this law. It stays put because the forces are balanced, which means the net force is zero.
The second law formalizes the relationship between force, mass, and acceleration through the famous equation: \[ F = ma \]This tells us that the force on an object is equal to the mass of the object times its acceleration. In equilibrium, since there's no motion or change of motion, the acceleration is zero, reinforcing the net force as zero.
The third law states that for every action, there is an equal and opposite reaction. This means when the surfaces of the rack push against the wine bottle, the bottle pushes back with an equal force. These interactive forces are crucial for equilibrium conditions.
By applying these laws, we can better understand how the forces are configured to maintain the bottle in a state where it does not move.
Problem Solving in Physics
Problem-solving in physics involves understanding the scenario, breaking it down into manageable parts, and applying the relevant principles. Here's how you can tackle problems about forces in equilibrium using a systematic approach:
  • **Identify the Given Information:** Start by restating what you know about the problem. In our example, we have the weight of the wine bottle and the angles given.
  • **Identify All Forces:** List all the forces acting on the object. For the wine bottle, these were the gravitational force and the forces from the surfaces.
  • **Apply Equilibrium Conditions:** Use the fact that the sum of forces in both directions (vertical and horizontal) is zero to set up equations.
  • **Solve Systematically:** Substitute known values into your equations and solve for the unknowns. Check your solutions against the conditions provided to ensure they make sense.
Breaking down a physics problem into smaller pieces aids comprehension and provides a clear pathway toward the solution. By mastering these steps, the process becomes less intimidating and more accessible.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Review Interactive LearningWare 4.3 at in preparation for this problem. The helicopter in the drawing is moving horizontally to the right at a constant velocity. The weight of the helicopter is \(W=53800 \mathrm{~N}\). The lift force \(\overrightarrow{\mathrm{L}}\) generated by the rotating blade makes an angle of \(21.0^{\circ}\) with respect to the vertical. (a) What is the magnitude of the lift force? (b) Determine the magnitude of the air resistance \(\overrightarrow{\mathrm{R}}\) that opposes the motion.

At a time when mining asteroids has become feasible, astronauts have connected a line between their 3500 -kg space tug and a 6200 -kg asteroid. Using their ship's engine, they pull on the asteroid with a force of \(490 \mathrm{~N}\). Initially the tug and the asteroid are at rest, \(450 \mathrm{~m}\) apart. How much time does it take for the ship and the asteroid to meet?

A block is pressed against a vertical wall by a force \(\vec{P}\), as the drawing shows. This force can either push the block upward at a constant velocity or allow it to slide downward at a constant velocity, the magnitude of the force being different in the two cases, while the directional angle \(\theta\) is the same. Kinetic friction exists between the block and the wall. (a) Is the block in equilibrium in each case? Explain. (b) In each case what is the direction of the kinetic frictional force that acts on the block? Why? (c) In each case is the magnitude of the frictional force the same or different? Justify your answer. (d) In which case is the magnitude of the force \(\overrightarrow{\mathrm{P}}\) greater? Provide a reason for your answer. The weight of the block is \(39.0 \mathrm{~N}\), and the coefficient of kinetic friction between the block and the wall is \(0.250\). The direction of the force \(\overrightarrow{\mathrm{P}}\) is \(\theta=30.0^{\circ}\). Determine the magnitude of \(\overrightarrow{\mathrm{P}}\) when the block slides up the wall and when it slides down the wall. Check to see that your answers are consistent with your answers to the Concept Questions.

In a supermarket parking lot, an employee is pushing ten empty shopping carts, lined up in a straight line. The acceleration of the carts is \(0.050 \mathrm{~m} / \mathrm{s}^{2}\). The ground is level, and each cart has a mass of \(26 \mathrm{~kg}\). (a) What is the net force acting on any one of the carts? (b) Assuming friction is negligible, what is the force exerted by the fifth cart on the sixth cart?

A supertanker (mass \(=1.70 \times 10^{8} \mathrm{~kg}\) ) is moving with a constant velocity. Its engines generate a forward thrust of \(7.40 \times 10^{5} \mathrm{~N}\). Determine (a) the magnitude of the resistive force exerted on the tanker by the water and (b) the magnitude of the upward buoyant force exerted on the tanker by the water.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.