Moments & Pivots: Find Force T for Level Board

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Homework Help Overview

The problem involves a uniform board that is pivoted at one end and held level by an upward force at the opposite end. The weight of the board is given, and the task is to find the force T using the concept of moments about the hinge.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of moments and the balance of forces. There is a mention of using the moment equation and a specific attempt at calculating T, which raises questions about the correctness of the approach. Some participants express confidence in the initial calculation while others inquire about angular acceleration and its relation to the problem.

Discussion Status

The discussion is ongoing with various interpretations being explored. Some participants affirm the correctness of the initial calculation, while others seek clarification on related concepts such as angular acceleration and Newton's laws. There is no explicit consensus on the final answer, but productive dialogue is occurring.

Contextual Notes

Participants note that angular acceleration has not been covered in their studies, which leads to questions about its calculation and relevance to the problem at hand. The discussion also touches on the relationship between linear and angular motion.

Sarahborg
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Homework Statement


A uniform board of length 90cm is pivoted at a hinge at one end. It is kept level by an upward vertical force T applied at the opposite end. The weight of the board is 6N. Take moments about the hinge to find T when the board is level.

Homework Equations


Moment = Force x Perpendicular distance

The Attempt at a Solution


I tried doing 45 x 6 = 270, and then 270/90 which is 3N. I think it's wrong though (I got the 45 by dividing 90 by 2)
 
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Hello Sarah, :welcome:

Much better !
What makes you doubt the result ?
Your moment balance looks like ( -45 * 6 + 90 * 3) cm * N and that is zero, so no angular acceleration.
 
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+1

Welcome to the forum. It looks right to me as well.
 
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CWatters said:
+1

Welcome to the forum. It looks right to me as well.
Thank you very much
 
How do you calculate angular acceleration, because we haven't covered it yet, and got curious.

Thank you very much for your help :)
 
Pardon the link: :smile: there is a parallel between linear motion and angular motion.
 
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Although in this case there is no angular acceleration because you were asked to arrange for the net torque to sum to zero. See also Newton's laws.
 
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BvU said:
Pardon the link: :smile: there is a parallel between linear motion and angular motion.
Thanks
 
How are Newton's Laws related to angular acceleration?

Thanks
 
  • #10
Have you come to the conclusion the answer you found in post #1 is correct ? Oh well, I hinted as much in post #2.

For Newton we have $$ \vec F = m\vec a $$ and the (almost carbon copy)https://www.boundless.com/physics/textbooks/boundless-physics-textbook/uniform-circular-motion-and-gravitation-5/angular-vs-linear-quantities-59/angular-vs-linear-quantities-272-6253/in angular motion is $$ \vec \tau = I\vec \alpha $$
(see table 1 https://www.boundless.com/physics/textbooks/boundless-physics-textbook/rotational-kinematics-angular-momentum-and-energy-9/problem-solving-88/problem-solving-techniques-332-6291/)
## \vec \tau ## is the torque
## I ## is the moment of inertia
##\vec \alpha## is the angular acceleration
 
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