How can I find the critical mu of friction?

[student34]

Homework Statement

A lady pushes a box with mass (m) with force (F) at a horizontal angle of θ. There is a static friction (μ)

Homework Equations

ƩFy = Fn - m*g - F*sinθ = 0

ƩFx = F*cosθ - μFn = 0, where μFn is the force of friction.

The Attempt at a Solution

F*cosθ - μ*Fn = 0

μ*Fn = F*cosθ; and Fn = m*g + F*sinθ

So, μ(m*g + F*sinθ) = F*cosθ

μ = (F*cosθ)/(m*g + F*sinθ)

But the textbook's answer is μ = 1/tanθ. Is it wrong? I almost get their answer except for the m*g. And when I use dummy values I don't get the same answer as the book, so I am pretty sure that my problem is from not simplifying the equation enough.

 

[rude man]

The answer must have F in it somewhere! So yeah, the book answer is wrong. I believe.
My answer has tan(theta) in it, but it also has mg and F.
(Ooh, can't wait to be corrected on this one by some smart ME! )

 

[student34]

The answer must have F in it somewhere! So yeah, the book answer is wrong. I believe.
My answer has tan(theta) in it, but it also has mg and F.
(Ooh, can't wait to be corrected on this one by some smart ME! )

Is your answer equivalent to mine?

 

[rude man]

Is your answer equivalent to mine?

'Fraid not.

Don't even bother to write separate equations in x and y.

The lady has to overcome two forces as she's pushing the mass uphill. What are those two forces (along the ramp)?

 

[haruspex]

A lady pushes a box with mass (m) with force (F) at a horizontal angle of θ. There is a static friction (μ)

Is she pushing down at angle theta or up at that angle? I'll guess it's up.
You haven't said what the question asks. I'll guess again: it asks for the angle which minimises F.

 

[student34]

Is she pushing down at angle theta or up at that angle? I'll guess it's up.
You haven't said what the question asks. I'll guess again: it asks for the angle which minimises F.

She pushes down at an angle θ below the horizontal.

 

[haruspex]

You're still not saying what the book actually asks. Pls post the whole question verbatim.

 

[student34]

You're still not saying what the book actually asks. Pls post the whole question verbatim.

I am not sure if I have to do this, but I am going to reference the question just in case I am not committing some kind of plagiarism. So just ignore the reference.

"A large crate with mass m rests on a horizontal floor. The coefficients of friction between the crate and the floor are μ(static) and μ(kinetic). A woman pushes downward at an angle θ below the horizontal on the crate with force F. If μ(static) is greater than some critical value, the woman cannot start the crate moving no matter how hard she pushes. Calculate this critical value of μ(static)" (P167).

Referenced from Sears and Zemansky's University Physics (Young and Freedman 13th Edition)

 

[SammyS]

I am not sure if I have to do this, but I am going to reference the question just in case I am not committing some kind of plagiarism. So just ignore the reference.

"A large crate with mass m rests on a horizontal floor. The coefficients of friction between the crate and the floor are μ(static) and μ(kinetic). A woman pushes downward at an angle θ below the horizontal on the crate with force F. If μ(static) is greater than some critical value, the woman cannot start the crate moving no matter how hard she pushes. Calculate this critical value of μ(static)" (P167).

Referenced from Sears and Zemansky's University Physics (Young and Freedman 13th Edition)

Showing the complete problem does help very much.

Start by drawing a free body diagram for the crate.

 

[haruspex]

no matter how hard she pushes.

That's the piece of information that was missing. You obtained this equation correctly (rude man seems to have misread the set-up as being a ramp):

μ = (F*cosθ)/(m*g + F*sinθ)​

With the understanding that F is not a specific value and is unlimited in magnitude, can you now derive the desired result from your equation?

 

[student34]

That's the piece of information that was missing. You obtained this equation correctly (rude man seems to have misread the set-up as being a ramp):

μ = (F*cosθ)/(m*g + F*sinθ)​

With the understanding that F is not a specific value and is unlimited in magnitude, can you now derive the desired result from your equation?

The only way that I can get rid of F is by using ƩFy = Fn - m*g - F*sinθ, where Fn = (F*cosθ)/μ. Then I isolate F which equals (μ*m*g)/(cosθ - μ*sinθ), and substitute this equation in for F from our answer μ = (F*cosθ)/(m*g + F*sinθ). But it turns into a massive mess that I can't simplify to 1/tanθ, and these questions never expect us to do that kind of math. So I know that I must be missing a way that is much more simple.

 

[haruspex]

The only way that I can get rid of F is by using ƩFy = Fn - m*g - F*sinθ, where Fn = (F*cosθ)/μ.

You're not grasping the implication of "no matter how hard she pushes".
Consider your equation μ = (F*cosθ)/(m*g + F*sinθ) as defining μ as a function of F. What happens to μ as F increases?

 

[SammyS]

Start by drawing a free body diagram for the crate.

A free body diagram should help you get an expression for the normal force -- which does depend upon F .

 

[student34]

You're not grasping the implication of "no matter how hard she pushes".
Consider your equation μ = (F*cosθ)/(m*g + F*sinθ) as defining μ as a function of F. What happens to μ as F increases?

After trying some values of F, I noticed that μ increases less and less as F increases. So, now I do understand how F increases so does μ therefore so does friction pushing against her push. But I can't make the connection as to how they get μ = 1/tanθ.

It all makes good sense except for how they get μ = 1/tanθ.

 

[student34]

A free body diagram should help you get an expression for the normal force -- which does depend upon F .

I have made a FBD. In it Fn = m*g + F*cosθ, and Fn = (F*cosθ)/μ. I can't find anymore useful relationships to Fn.

 

[mukundpa]

the woman cannot start the crate moving no matter how hard she pushes.

for maximum force F applied toslide depends on angle θ. for critical value of coefficient of friction dF/dθ must be equal to zero for some angle θ.

 

[SammyS]

What equation do you get from the horizontal component of force?

 

[student34]

What equation do you get from the horizontal component of force?

ƩFx = F*cosθ - μFn = 0, where μFn is the force of friction, and Fn = m*g + F*cosθ, and Fn = (F*cosθ)/μ

 

[SammyS]

I have made a FBD. In it Fn = m*g + F*cosθ, and Fn = (F*cosθ)/μ. I can't find anymore useful relationships to Fn.

That should be F_n = mg + F sin(θ) .

ƩFx = F*cosθ - μFn = 0, where μFn is the force of friction, and Fn = m*g + F*cosθ, and Fn = (F*cosθ)/μ

Put those together.

 

[haruspex]

After trying some values of F, I noticed that μ increases less and less as F increases. So, now I do understand how F increases so does μ therefore so does friction pushing against her push. But I can't make the connection as to how they get μ = 1/tanθ.

It all makes good sense except for how they get μ = 1/tanθ.

What you might also notice is that μ does not go on increasing without limit. What value does it tend to as F tends to infinity?

 

[student34]

That should be F_n = mg + F sin(θ) .



Put those together.

Ok but when I do, I just get my original answer of μ = (F*cosθ)/(m*g + F*sinθ). The answer in the textbook is μ = 1/tanθ.

 

[ehild]

Isolate F.

ehild

 

[student34]

What you might also notice is that μ does not go on increasing without limit. What value does it tend to as F tends to infinity?

I will go out on a limb and make the derivative of μ equal zero, but I don't think that they expect us to do this yet. So μ' = 0 = (-F*sinθ)/(m*g + F*sinθ) + (F*cosθ)/-(m*g + F*sinθ)^2, and when F is isolated, its value should be a derivative of zero.

 

[ehild]

Isolate F instead of mu from μ = (F*cosθ)/(m*g + F*sinθ). You know that F must be positive, being the magnitude of the applied force.

ehild

 

[student34]

Isolate F.

ehild

Ok, F = (μ*m*g)/(cosθ - μ*sinθ), but I am not sure why I want to do this.

 

[ehild]

Because F can not be negative. What dos it mean for (cosθ - μ*sinθ)?

ehild

 

[student34]

Because F can not be negative. What dos it mean for (cosθ - μ*sinθ)?

ehild

I see; cosθ ≥ μ*sinθ. But that is so beyond any kind of critical thinking that the textbook has ever expected in any other questions. This is very humbling.

 

[haruspex]

I will go out on a limb and make the derivative of μ equal zero, but I don't think that they expect us to do this yet. So μ' = 0 = (-F*sinθ)/(m*g + F*sinθ) + (F*cosθ)/-(m*g + F*sinθ)^2,

You seem to have got confused between differentiating wrt F and wrt θ. Wrt F will give you F=∞, which is correct. I.e., μ will always increase as F increases. But that isn't what I asked - the question is whether there is a limit to the value of μ. For that, let F tend to infinity in your equation for μ. This will make the mg term so small in comparison that it can be ignored. What does that leave you with?

 

[mukundpa]

I see; cosθ ≥ μ*sinθ. But that is so beyond any kind of critical thinking that the textbook has ever expected in any other questions. This is very humbling.

cosθ ≥ μ*sinθ
or μ*sinθ ≤ cosθ
or μ ≤ (1/tanθ)

this means that 1/tanθ is the critical value for μ. if μ is greater than this value the force required to move the block is infinite i.e.can not be moved.

 

[SammyS]

Ok but when I do, I just get my original answer of μ = (F*cosθ)/(m*g + F*sinθ). The answer in the textbook is μ = 1/tanθ.

Write your expression for μ as

\displaystyle \ \mu=\frac{1}{\displaystyle\frac{mg}{F\cos(\theta)}+\frac{F\sin(\theta)}{F\cos(\theta)}}

\displaystyle=\frac{1}{\displaystyle\frac{mg}{F\cos(\theta)}+\tan( \theta)}\ .​

Then follow haruspex's advice in post #20.