need help calculating angle with coefficient of static frictionby Fireant Tags: angle, coefficient, friction, static 

#1
Feb2712, 09:12 AM

P: 22

There's a box sitting on a plank. The coefficient of static friction is 0.33. I want to figure out at which angle the box will begin to slide if the plank is tilted.
I know the factors that come into play are FN, Fg or Fgx, and Fs. So far I have Fnet= FNFsFg =FN μssin∅ =cos∅0.33sin∅ This doesn't look right... can anyone help me out? 



#2
Feb2712, 09:20 AM

PF Gold
P: 1,153

What are the conditions for impending motion using Newton's second law?
I would suggest looking at the horizontal and vertical scalar equations of motion of the block and solving for [itex]\theta[/itex]. I'm having a bit of difficulty following your work. Could you clarify your last expression? (It looks like [itex]F_{net} = \cos\theta0.33\sin\theta[/itex] to me.) 



#3
Feb2712, 09:29 AM

P: 22

Im assuming there's no acceleration, so Fnet= 0N.
This is the symbol for theta: θ I used this symbol for "angle":∅ """"I'm having a bit of difficulty following your work. Could you clarify your last expression? (It looks like F_{net} = \cos\theta0.33\sin\theta to me.) """" That's pretty much it! Except using angle to substitute for theta. I don't know how to get to the angle... I'm missing mass from the equation too right? 



#4
Feb2712, 09:35 AM

PF Gold
P: 1,153

need help calculating angle with coefficient of static frictionThere's one more condition dealing with static friction. What would that be? (Remember how static friction works; is there a maximum value? What is it?) Next, sum the forces in the [itex]x[/itex] and [itex]y[/itex] directions (I would suggest using a coordinate system aligned with the ramp.). From these two equations (plus an expression relating the frictional force and the normal force) you should be able to find [itex]\theta[/itex]. 



#5
Feb2712, 09:48 AM

P: 22

x Fnet= Fgx =mg(sin∅)Fs =mg(sin∅)μs and y Fnet=FN 0= mgcos∅ So far so good? One more thing: The weight of the box is 10kg 



#6
Feb2712, 09:52 AM

PF Gold
P: 1,153

[itex]\mu_s[/itex] is the coefficient of static friction, not the force of friction. So, what is the force of friction in this case?
You're almost there! 



#7
Feb2712, 10:05 AM

P: 22

So.... since
μs= Ff/FN 0.33=Ff/mg Ff=0.33(10)(9.8) Ff=32.34N ? 



#8
Feb2712, 10:11 AM

PF Gold
P: 1,153

Static friction is given by [itex]F_s \leqslant \mu_s N[/itex], so [itex]F_{s,max}=\mu_s N[/itex]. So, from this and the FBD we can get the friction force as a function of mass, gravitational acceleration, and the angle [itex]\theta[/itex].
What is this force, and how can we substitute it to find [itex]\theta[/itex]? Hint: the final answer will be independent of mass and gravitational acceleration. 



#9
Feb2712, 10:52 AM

P: 22

You're saying
Fsmax=m(μs)FN and i'm not sure what to subsitute the force with... 



#10
Feb2712, 11:05 AM

P: 22

so technically....
if i know Fg and FN, I can figure out the angle? So tan∅= 32.34/9.8 ∅=73° ? 



#11
Feb2712, 11:23 AM

PF Gold
P: 1,153

Close, but not quite.
[itex]F_{s,max} = \mu_sN = \mu_s mg\cos\theta[/itex]. Then if [itex]ma = 0 = mg\sin\theta  \mu_s mg\cos\theta[/itex], what is [itex]\theta[/itex]? Sorry for the latency in response, just got out of class. 



#12
Feb2712, 11:29 AM

P: 22

No I really appreciate your help... getting my noggin working...
Fsmax=musN =mus(mg)(costheta) 0=mgsinθmμs(mg)cosθ that's the part where i'm confused... how do i isolate theta? 


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