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Need help calculating angle with coefficient of static friction

by Fireant
Tags: angle, coefficient, friction, static
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Fireant
#1
Feb27-12, 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= FN-Fs-Fg
=FN- μs-sin∅
=cos∅-0.33-sin∅

This doesn't look right... can anyone help me out?
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jhae2.718
#2
Feb27-12, 09:20 AM
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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\theta-0.33-\sin\theta[/itex] to me.)
Fireant
#3
Feb27-12, 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\theta-0.33-\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?

jhae2.718
#4
Feb27-12, 09:35 AM
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Need help calculating angle with coefficient of static friction

Quote Quote by Fireant View Post
Im assuming there's no acceleration, so Fnet= 0N.
That's right. Impending motion, if you are unfamiliar with the term, is the point right before motion starts.

There'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?)

Quote Quote by Fireant View Post
That's pretty much it! It's wrong... I don't know how to get to the angle...

I'm missing mass from the equation too right?
The first place to start is by drawing a free body diagram of the situation. We know the block is tilted at some angle [itex]\theta[/itex] at which the box will just start to move, so what forces act on the block?

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].
Fireant
#5
Feb27-12, 09:48 AM
P: 22
Quote Quote by jhae2.718 View Post

There'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?)
0.33 ?

The first place to start is by drawing a free body diagram of the situation. We know the block is tilted at some angle [itex]\theta[/itex] at which the box will just start to move, so what forces act on the block?
Force of gravity in the x compenent, the normal force, and the force of friction.

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].
So you mean :

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
jhae2.718
#6
Feb27-12, 09:52 AM
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[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!
Fireant
#7
Feb27-12, 10:05 AM
P: 22
So.... since

μs= Ff/FN
0.33=Ff/mg
Ff=0.33(10)(9.8)
Ff=32.34N

?
jhae2.718
#8
Feb27-12, 10:11 AM
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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.
Fireant
#9
Feb27-12, 10:52 AM
P: 22
You're saying

Fsmax=m(μs)FN

and i'm not sure what to subsitute the force with...
Fireant
#10
Feb27-12, 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

?
jhae2.718
#11
Feb27-12, 11:23 AM
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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.
Fireant
#12
Feb27-12, 11:29 AM
P: 22
No I really appreciate your help... getting my noggin working...

Fsmax=musN
=mus(mg)(costheta)

0=mgsinθ-mμs(mg)cosθ

that's the part where i'm confused... how do i isolate theta?
jhae2.718
#13
Feb27-12, 11:46 AM
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Quote Quote by Fireant View Post
No I really appreciate your help... getting my noggin working...

Fsmax=musN
=mus(mg)(costheta)

0=mgsinθ-mμs(mg)cosθ

that's the part where i'm confused... how do i isolate theta?
Maybe you could try moving [itex]\mu_s mg \cos\theta[/itex] to the other side and using trigonometry? Also, what can you do with the [itex]mg[/itex] term?


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