
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
expand_more
expand_more
format_list_bulleted
Concept explainers
Question

Transcribed Image Text:### Physics Problem: Shelf with Supports and Weight Distribution
---
#### Problem Statement:
You build a shelf by putting a 0.90 m board across two supports. One support is at the left end of the board and the other is 0.10 m from the right end. The board has a mass of 1.25 kg and you place a potted plant with a mass of 3.40 kg, 0.65 m from the left end of the board.
#### Diagram Explanation:
The diagram shows a horizontal board positioned across two vertical supports. Specific measurements and masses are marked on the diagram as follows:
- **Board Length:** 0.90 m
- **Left Support Position:** At the left end of the board (0 m mark)
- **Right Support Position:** 0.10 m from the right end of the board
- **Mass of the Board (m):** 1.25 kg centered along its length
- **Mass of the Potted Plant (m):** 3.40 kg located 0.65 m from the left end of the board
The board and potted plant are shown with arrows indicating the positions and distances from the left end of the board.
#### Tasks:
a. Draw a free-body diagram labeling all the forces and distances. Also, choose and mark a pivot point for your shelf on the drawing.
b. Find the force exerted by both supports.
---
### Physics Concepts Involved:
- **Torque and Equilibrium:** Utilize principles of torque and static equilibrium to solve for the forces at the supports.
- **Free-Body Diagram (FBD):** A graphical illustration used to visualize the applied forces, moments, and reactions on a body.
To solve the problem:
1. **Determine the Pivot Point:**
- Typically, one support or the center of mass of the system can be used as a pivot.
2. **Calculate Forces:**
- Apply the equilibrium condition for both forces and torques:
- ΣF = 0 (sum of vertical forces must be zero)
- Στ = 0 (sum of torques around any point must be zero)
3. **Equations:**
- For instance, if taking the left support as the pivot,
- Consider the torque due to the board’s weight and the plant’s weight about the left support.
- Calculate individual contributions to the net torque to find the unknown forces.
Use these principles
Expert Solution

This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution
Trending nowThis is a popular solution!
Step by stepSolved in 2 steps with 2 images

Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Similar questions
- A uniform tabletop with a mass of 415kg has a length of 2.50m. A 775kg box is placed 0.800m from one side of the table and a 265kg box is placed 1.80m from the same side of the table. What are the forces from the table legs on the tabletop?arrow_forwardThe diagram above shows a 3.00 m long uniform beam. The left end of the beam is attached to a wall by a frictionless pivot. The beam is supported by a string that keeps the beam horizontal. The string makes an angle of 35.0° relative to vertical. The beam has a mass of 7.00 kg. A 2.00 kg mass is located at the end of the beam. A 6.00 kg mass is located 1.00 m from the 2.00 kg mass, as shown in the diagram. a. What is the tension in the string? b. What is the y-component of force applied to the beam by the pivot?arrow_forwardTwo children are trying to balance on a seesaw, which includes a fulcrum and a board that is 10m long. Boy A has a mass of 50kg, and boy B has a mass of 35kg. How far from the pivot point should boy B be to maintain stable equilibrium?arrow_forward
- B4arrow_forwardTo get up on the roof, a person (mass 87.0 kg) places a 5.60 m aluminum ladder (mass 14.0 kg) against the house on a concrete pad with the base of the ladder 2.00 m from the house. The ladder rests against a plastic rain gutter, which we can assume to be frictionless. The center of mass of the ladder is 2 m from the bottom. The person is standing 3 m from the bottom. What are the magnitudes (in N) of the forces on the ladder at the top and bottom? top N bottom Narrow_forwardA horizontal 8 kg plank is 3.0 m in length and has its center of mass at the geometrical center. A 2 kg mass is placed 1.0 m from one end and the plank is supported from the ends. Calculate the forces of support. 1.0 m 2 kg 3.0 marrow_forward
- Bob and Anna sit on opposite ends of a uniform see-saw which is 12.0 m in length. The see-saw is balanced by a fulcrum which is located 4.00 m from Bob. Bob has a mass of 85.0 kg, and the see-saw itself has a mass of 100 kg. Anna’s mass is unknown. NOTE THAT THE FULCRUM IS NOT PLACED AT THE CENTER OF MASS OF THE SEE- SAW!!! a) What is the net torque on the see-saw? Explain clearly and briefly how you know. b) Draw a properly labeled extended force diagram of the see-saw c) Solve for Anna’s mass. Show your work clearly!arrow_forwardA 100-N board 12-m long, rests on two supports, one support is 1m from the end and other is 3 m from the end of the board. A 400-N block is placed on the board 3 m from one end. Find the force exerted by each support on the board.arrow_forwardBob and Anna sit on opposite ends of a uniform see-saw which is 12.0 m in length. The see-saw is balanced by a fulcrum which is located 4.00 m from Bob. Bob has a mass of 85.0 kg, and the see-saw itself has a mass of 100 kg. Anna’s mass is unknown. NOTE THAT THE FULCRUM IS NOT PLACED AT THE CENTER OF MASS OF THE SEESAW!!! a) What is the net torque on the see-saw? Explain clearly and briefly how you know. b) Draw a properly labeled extended force diagram of the see-saw. c) Solve for Anna’s mass.arrow_forward
arrow_back_ios
arrow_forward_ios
Recommended textbooks for you
- College PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningUniversity Physics (14th Edition)PhysicsISBN:9780133969290Author:Hugh D. Young, Roger A. FreedmanPublisher:PEARSONIntroduction To Quantum MechanicsPhysicsISBN:9781107189638Author:Griffiths, David J., Schroeter, Darrell F.Publisher:Cambridge University Press
- Physics for Scientists and EngineersPhysicsISBN:9781337553278Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningLecture- Tutorials for Introductory AstronomyPhysicsISBN:9780321820464Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina BrissendenPublisher:Addison-WesleyCollege Physics: A Strategic Approach (4th Editio...PhysicsISBN:9780134609034Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart FieldPublisher:PEARSON

College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning

University Physics (14th Edition)
Physics
ISBN:9780133969290
Author:Hugh D. Young, Roger A. Freedman
Publisher:PEARSON

Introduction To Quantum Mechanics
Physics
ISBN:9781107189638
Author:Griffiths, David J., Schroeter, Darrell F.
Publisher:Cambridge University Press

Physics for Scientists and Engineers
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning

Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:9780321820464
Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:Addison-Wesley

College Physics: A Strategic Approach (4th Editio...
Physics
ISBN:9780134609034
Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:PEARSON