College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
expand_more
expand_more
format_list_bulleted
Related questions
Concept explainers
Question
thumb_up100%
pivot Position = 1.0 cm
center of mass position = 50.0 cm
weight of meter stick = 1.517 N
Transcribed Image Text:### Data Table: Torque Calculation
This data table presents measurements related to torque calculations where force is applied at different string positions. Below is a detailed transcription of the table:
| **String Position (cm)** | **F (N)** | **r (m)** | **τ (Nm)** |
|--------------------------|-----------|-----------|------------|
| 90.0 | 0.842 | 0.89 | 0.74938 |
| 80.0 | 0.945 | 0.79 | 0.74655 |
| 70.0 | 1.082 | 0.69 | 0.74658 |
| 60.0 | 1.272 | 0.59 | 0.75048 |
| 50.0 | 1.525 | 0.49 | 0.74725 |
| 40.0 | 1.917 | 0.39 | 0.74763 |
| 30.0 | 2.581 | 0.29 | 0.74849 |
| 20.0 | 3.882 | 0.19 | 0.73758 |
#### Table Explanation:
- **String Position (cm):** Indicates the position along the string in centimeters where the force is applied.
- **F (N):** Represents the force applied in Newtons.
- **r (m):** Denotes the lever arm distance in meters.
- **τ (Nm):** Indicates the torque produced in Newton-meters.
The table shows the relationship between the applied force, the distance from the pivot point, and the resulting torque. The data suggests that as the string position decreases, the force increases while the distance decreases, resulting in slight variations in torque values.
![**Educational Content:**
### Torque and Equilibrium Analysis
**10. Free Body Diagram:**
- **Diagram Explanation:**
- The diagram shows a horizontal meter stick with a pivot point on the left.
- Vertical (V) and horizontal (H) components of the force are shown at the pivot point.
- The stick is labeled with a length marked in centimeters (× CM).
**11. Calculating Torque Due to Gravity:**
- Use the formula for torque: \(\tau = r \cdot F \cdot \sin \theta\).
- Determine the torque due to the weight of the meter stick and calculate the result.
\[\tau_{\text{gravity}} = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \text{Nm}\]
**12. Torque Equilibrium Analysis:**
- The torques due to gravity (\(\tau_{\text{gravity}}\)) and the tension in the string (\(\tau_{\text{tension}}\)) should be close since they balance each other for equilibrium (no rotation).
- To find the percent difference between two values (\(v_1\) and \(v_2\)), use:
\[
\text{\% difference} = \frac{v_1 - v_2}{v_{\text{avg}}} \times 100
\]
\[\text{\% difference} = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_]](https://content.bartleby.com/qna-images/question/56747038-88d2-45a2-87b0-6373049be49b/58aba01a-9a22-402e-b677-7b0ea25bde6a/vxo1ycb_processed.png)
Transcribed Image Text:**Educational Content:**
### Torque and Equilibrium Analysis
**10. Free Body Diagram:**
- **Diagram Explanation:**
- The diagram shows a horizontal meter stick with a pivot point on the left.
- Vertical (V) and horizontal (H) components of the force are shown at the pivot point.
- The stick is labeled with a length marked in centimeters (× CM).
**11. Calculating Torque Due to Gravity:**
- Use the formula for torque: \(\tau = r \cdot F \cdot \sin \theta\).
- Determine the torque due to the weight of the meter stick and calculate the result.
\[\tau_{\text{gravity}} = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \text{Nm}\]
**12. Torque Equilibrium Analysis:**
- The torques due to gravity (\(\tau_{\text{gravity}}\)) and the tension in the string (\(\tau_{\text{tension}}\)) should be close since they balance each other for equilibrium (no rotation).
- To find the percent difference between two values (\(v_1\) and \(v_2\)), use:
\[
\text{\% difference} = \frac{v_1 - v_2}{v_{\text{avg}}} \times 100
\]
\[\text{\% difference} = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_]
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 3 steps with 3 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
- On the extreme left side of a (15 m) long beam sits a (50 kg) weight. On the extreme right side of the beam sits a (20 kg) weight. At what distance from each weight must the fulcrum be placed such that the system is in equilibrium?arrow_forwardDetermine the net torque (about point P at one end) of the 4.0-m-long uniform beam shown in the Figure below. All forces are shown. 60 N 32 50 N 2 m 45° 2 m 40 Narrow_forward370 The beam (of length 2.9 m) shown in the figure has a mass of 2.1 kg. You hang plants on it as shown. The string attaches to the middle of the beam and passes over a pulley. The plant hanging from the string coming down from the pulley has a mass of 7.1 kg. 17.1 кдarrow_forward
- If a mass of 23 kg acts downward 0.3 m from the axis of rotation on one end of a board and another force of 82.5 N also acts downward. What is the distance of point where the force is applied to balance the force?arrow_forwardFind the net torque on the wheel in Figure about the axle through O, taking a=10.0 cm and b=25.0 cm.arrow_forwardA minimum force of P = 56 lb is required to hold the cylinder from slipping against the belt and the wall. The coefficient of friction between the belt and cylinder is He = 0.3 and slipping does not occur at the wall. (Eigure 1) %3D Figure < 1 of 1 30 0.1 ftarrow_forward
- Find the magnitude of the torque exerted by F on the bolt at P if PQ| = 8 in. and |F| = 30 lb. 450 Parrow_forwardProblem 1 0.21 and r2 Two solid spherical masses mị end of a massive rigid rod of mass m2 = cylinder of mass m4 = axle perpendicular to the rod through point Pa distance l1 = 0.83 m and l = 0.52 m. The bar is released from rest in the horizontal position at t = 0. 0.14, are attached to the 1.00 kg, as shown in the below figure. The rod is attached to a massive 0.14 m, and is free to rotate in a vertical plane about a frictionless 2.51 kg and mз — 1.48 kg with radii ri = 3.2 kg and radius r4 = a) Find the moment of inertia of the rod with all three masses. b) Find the torque acting on the system at t = 0. c) Find the angular acceleration of of the system at t = 0. d) Find the maximum kinetic energy of the system during the swing. e) Find the maximum angular velocity of the system. m3 m2 l1arrow_forwardA little league player balances a bat 67.0 cm from one end. She then hangs her 0.540-kg batting helmet on that end of the bat and the balance point moves 24.0 cm toward the helmet. Determine the mass of the bat.arrow_forward
- A uniform plank is supported by two vertical wires attached to its ends. A small mass rests on the plank. The plank is balanced horizontally. It has a mass of 250 kg and is 5.50 m long. A small mass of 11.0 kg is 2.00 m from the left end of the plank. What is the tensions in the left wire? Assume the units are Newtons, and type in the numerical answer to three significant figures.arrow_forwardSam and Joe carry a beam, walking to the right. Calculate the force that Sam exerts, in pounds. The beam's weight is 163 lb. The beam's length is 22 ft. Sam is 5 ft from the back. Joe is 5 ft from the front.arrow_forward17. A 0.25 kg meterstick has a 2kg mass suspended from the 80 cm mark and a 0.5 kg mass suspended from the 5 cm mark. At what mark should a pivot be placed to perfectly balance the meterstick? Answer 17:arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- College PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage Learning
University Physics (14th Edition)PhysicsISBN:9780133969290Author:Hugh D. Young, Roger A. FreedmanPublisher:PEARSON
Introduction 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 Learning
Lecture- Tutorials for Introductory AstronomyPhysicsISBN:9780321820464Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina BrissendenPublisher:Addison-Wesley
College 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