Forces Free Body Diagram & Finding μ

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Homework Help Overview

The discussion revolves around a physics problem involving forces, free body diagrams, and the coefficient of friction (μ) in the context of a plank balanced on a step. Participants are tasked with finding the components of forces acting on the plank and expressing one of the forces in terms of known quantities.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the calculation of torque and the implications of the plank's center of mass on their equations. Questions arise regarding the frictional nature of the contact point and whether certain assumptions about angles and perpendicularity can be made.

Discussion Status

Some participants have offered guidance on the assumptions regarding friction and the relationship between forces. There is an acknowledgment of the simplicity of certain expressions, but uncertainty remains about the implications of the assumptions made regarding angles and friction.

Contextual Notes

Participants note that the problem does not explicitly state whether the contact is frictionless, leading to discussions about the validity of their assumptions. The known quantities and their roles in the equations are also under examination.

jumbogala
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Homework Statement


See the diagram here: http://i47.photobucket.com/albums/f173/xallip/bar2-1.jpg

Known quantities: L, Ff.

a) Find the x and y components of A and B.
b) μ is a known quantity; give an expression for A in this case.

Homework Equations


The Attempt at a Solution


a) Rx = Ff
N + Ry - mg = 0
net torque about the blue point must also be zero.

I'm unsure how to calculate the torque, since the entire length of the plank isn't spanned by the step and floor. Does gravity still act at the center of mass, 0.5L? If so then net torque:
0 = 0.9LR - mg0.5Lcosθ

Then I can solve for R: R = (mg0.5Lcosθ)/0.9L = (5/9)mgcosθ

And knowing R and Rx I can get Ry (pythagoras' thm), and from Ry I can find N. Is that correct?

b) This is what's really confusing me! I thought Ff = μN so N = Ff/μ, but I don't think it can be that simple.
 
Last edited by a moderator:
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Is the contact between the corner of the step and the plank frictionless? What does the problem say about that?
 
It doesn't say that directly. It just shows Ff in the diagram acting on the blue spot (the floor).

So it doesn't explicitly say it's not frictionless, but I assumed it was. I'm not sure if that's okay, but if it's not I can't assume A is perpendicular to the bar. And if A isn't perpendicular to the bar my torque equation gets messed up...

and it also doesn't show a 90 angle symbol between A and the bar. =S
 
Last edited:
jumbogala said:
It doesn't say that directly. It says "the friction force that stops the bar from sliding has magnitude Ff", and then it shows Ff in the diagram acting on the blue spot (the floor).

So it doesn't explicitly say it's not frictionless, but I assumed it was. I'm not sure if that's okay, but if it's not I can't assume R is perpendicular to the bar. And if R isn't perpendicular to the bar my torque equation gets messed up...

OK. I think you can safely assume that the contact is frictionless - I was just checking. For part (b), yes you can write N = Ff/µ and it's that simple. Usually with this sort of problem you are given µ and you are asked to find the angle at which slipping is just about to start or you are given the critical angle and are asked to find µ. Here you are given both it seems, so it is that simple as long as what you have down as "known quantities" for part (a) do not change in part (b).
 
No, the only new known quantity seems to be µ. I guess you could also write B = Rx/µ, but that would probably be pointless.

And it is true that the center of mass is still at 0.5L, and gravity acts there, no matter where the bar is balanced?
 
Last edited:
jumbogala said:
And it is true that the center of mass is still at 0.5L, and gravity acts there, no matter where the bar is balanced?

Yup.
 
Okay, thank you so much!
 

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