
#1
Feb1910, 02:18 PM

P: 400

1. The problem statement, all variables and given/known data
See the diagram here: http://i47.photobucket.com/albums/f1...lip/bar21.jpg Known quantities: L, F_{f}. 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. 2. Relevant equations 3. 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. 



#2
Feb1910, 03:07 PM

HW Helper
PF Gold
P: 3,444

Is the contact between the corner of the step and the plank frictionless? What does the problem say about that?




#3
Feb1910, 03:10 PM

P: 400

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 



#4
Feb1910, 03:26 PM

HW Helper
PF Gold
P: 3,444

Forces Free Body Diagram & Finding μ 



#5
Feb1910, 03:35 PM

P: 400

No, the only new known quantity seems to be µ. I guess you could also write B = R_{x}/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? 



#7
Feb1910, 04:47 PM

P: 400

Okay, thank you so much!



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