# Unsolvable Fricton Problem

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• Nov 21st 2006, 09:45 AM
ThePerfectHacker
Unsolvable Fricton Problem
How do I solve this,

"Given that the static fricton is $\displaystyle \mu_s =.25$ between the object weighing 60 lbs and the inclined plane. Find the force $\displaystyle P$ and angle $\displaystyle x$ such that motion is impending (about to happen).

(Note the Force Diagram I drew did not take shown rotation of axes. The coordinate system I am using is the same as in first diagram)

I resolve everything into component form.
$\displaystyle \vec{W}=<0,-60>$
$\displaystyle \vec{F}=<-.87F,-.5F>$
$\displaystyle \vec{N}=<-.5N,.87N>$
$\displaystyle \vec{P}=<P\cos(30^o+x),P\sin(30^o+x)>$
We also know that since motion is about to happen,
$\displaystyle F=\mu_s N=.25N$
But that is not sufficent to solve this problem.
• Nov 21st 2006, 10:27 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
How do I solve this,

"Given that the static fricton is $\displaystyle \mu_s =.25$ between the object weighing 60 lbs and the inclined plane. Find the force $\displaystyle P$ and angle $\displaystyle x$ such that motion is impending (about to happen).

(Note the Force Diagram I drew did not take shown rotation of axes. The coordinate system I am using is the same as in first diagram)

I resolve everything into component form.
$\displaystyle \vec{W}=<0,-60>$
$\displaystyle \vec{F}=<-.87F,-.5F>$
$\displaystyle \vec{N}=<-.5N,.87N>$
$\displaystyle \vec{P}=<P\cos(30^o+x),P\sin(30^o+x)>$
We also know that since motion is about to happen,
$\displaystyle F=\mu_s N=.25N$
But that is not sufficent to solve this problem.

Resolve everything into components normal to the plane and parallel to
the plane (then ignore the components normal to the plane)

To me this looks like it leaves you with the equation if the block is about
to move up the plane (frictional force act down the plane):

P*cos(x)-mu*sqrt(3)*30g-30g=0

and if about to move down the plane (friction acts up the plane):

P*cos(x)+mu*sqrt(3)*30g-30g=0

(you will have to check that these are the equations that you get
also check that you dont lift the block off of the plane).

RonL
• Nov 21st 2006, 12:20 PM
topsquark
Quote:

Originally Posted by ThePerfectHacker
How do I solve this,

"Given that the static fricton is $\displaystyle \mu_s =.25$ between the object weighing 60 lbs and the inclined plane. Find the force $\displaystyle P$ and angle $\displaystyle x$ such that motion is impending (about to happen).

(Note the Force Diagram I drew did not take shown rotation of axes. The coordinate system I am using is the same as in first diagram)

I resolve everything into component form.
$\displaystyle \vec{W}=<0,-60>$
$\displaystyle \vec{F}=<-.87F,-.5F>$
$\displaystyle \vec{N}=<-.5N,.87N>$
$\displaystyle \vec{P}=<P\cos(30^o+x),P\sin(30^o+x)>$
We also know that since motion is about to happen,
$\displaystyle F=\mu_s N=.25N$
But that is not sufficent to solve this problem.

Your second diagram troubles me. Either you rotated by 30 degrees and have the direction of the weight incorrect, or you didn't rotate anything and have the directions of the normal force and friction incorrect.

Just to be clear: the weight vector is always directed toward the center of the Earth. The normal force is always perpendicular to the plane formed by the two surfaces that meet. The friction force is always parallel to the plane formed by the two surfaces that meet.

I would highly recommend CaptainBlack's coordinate system, but you can use any you like. Using what you appear to be using, for example:
$\displaystyle \sum F_{horizontal} = Fcos(30) + Nsin(30) - Pcos(x + 30) = 0$
$\displaystyle \sum F_{vertical} = -W - Fsin(30) + Ncos(30) + Psin(x + 30) = 0$

Given that we are looking for the maximum static friction force we have (as you correctly stated)
$\displaystyle F = \mu N$

We now have three equations in 4 unknowns (P, x, F, and N) so I agree, you can't solve this system.

-Dan
• Nov 21st 2006, 12:28 PM
topsquark
I have posted what I laughingly call a diagram. I hope it comes out clear!

-Dan

(Sigh) I blew the image size up by 300% and it looks small as ever. Anyway, the force pointing down is W, the force down the slope is F, the force up and to the right is P and the force up and to the left is N. (I gotta get a new scanner.)
• Nov 22nd 2006, 10:36 AM
ThePerfectHacker
So maybe I copied the diagram incorrectly from my text. See what my Professor says tomorrow. If he says there is a solution I will tell him a physics professor told me it cannot be solved.
-------------------------------------------
I had class today and my professor had to apologize for giving an unsolvable problem. His edition stated (the one I used) how it is worded. It should have said "What is the minimum value of the force...". He wanted us to solve an optimization problem.
• Nov 22nd 2006, 05:02 PM
TriKri
Quote:

Originally Posted by ThePerfectHacker
How do I solve this,

"Given that the static fricton is $\displaystyle \mu_s =.25$ between the object weighing 60 lbs and the inclined plane. Find the force $\displaystyle P$ and angle $\displaystyle x$ such that motion is impending (about to happen).

(Note the Force Diagram I drew did not take shown rotation of axes. The coordinate system I am using is the same as in first diagram)

I resolve everything into component form.
$\displaystyle \vec{W}=<0,-60>$
$\displaystyle \vec{F}=<-.87F,-.5F>$
$\displaystyle \vec{N}=<-.5N,.87N>$
$\displaystyle \vec{P}=<P\cos(30^o+x),P\sin(30^o+x)>$
We also know that since motion is about to happen,
$\displaystyle F=\mu_s N=.25N$
But that is not sufficent to solve this problem.

If the angle of the plane is $\displaystyle \alpha$,
$\displaystyle mg = 60\ lbs \cdot g$
$\displaystyle F = sin(\alpha) \cdot mg$
$\displaystyle W = cos(\alpha) \cdot mg$
$\displaystyle N$ is the normal force
$\displaystyle f$ is the friction
$\displaystyle f = \mu_s \cdot N\ \Rightarrow\ N = f/\mu_s$

Then $\displaystyle P_x = P \cdot cos(x)$
$\displaystyle P_y = P \cdot sin(x)$
Then $\displaystyle F + f = P_x\ \Rightarrow\ f = P_x - F$
$\displaystyle N = W - P_y \ \Rightarrow \ P_y = W - N$
$\displaystyle P_y = W - f/\mu_s$
$\displaystyle P_y = W + (F - P_x)/\mu_s$
$\displaystyle P_y = W + F/ \mu _s - \frac{1}{\mu_s} \cdot P_x$

I didn't really understand the question, but isn't this an approach to anything?
• Nov 22nd 2006, 05:08 PM
topsquark
Quote:

Originally Posted by ThePerfectHacker
So maybe I copied the diagram incorrectly from my text. See what my Professor says tomorrow. If he says there is a solution I will tell him a physics professor told me it cannot be solved.
-------------------------------------------
I had class today and my professor had to apologize for giving an unsolvable problem. His edition stated (the one I used) how it is worded. It should have said "What is the minimum value of the force...". He wanted us to solve an optimization problem.

Now that makes more sense. :)

I was thinking about this again. There is a calculable value for the component of P along the plane, but there is a range of possible values for the perpendicular component. In the terms of the problem that you should have been given, then the angle x = 0 to give the smallest P.

-Dan
• Nov 22nd 2006, 05:13 PM
topsquark
Quote:

Originally Posted by TriKri
If the angle of the plane is $\displaystyle \alpha$,
$\displaystyle mg = 60\ lbs \cdot g$
$\displaystyle F = sin(\alpha) \cdot mg$
$\displaystyle W = cos(\alpha) \cdot mg$
$\displaystyle N$ is the normal force
$\displaystyle f$ is the friction
$\displaystyle f = \mu_s \cdot N\ \Rightarrow\ N = f/\mu_s$

Then $\displaystyle P_x = P \cdot cos(x)$
$\displaystyle P_y = P \cdot sin(x)$
Then $\displaystyle F + f = P_x\ \Rightarrow\ f = P_x - F$
$\displaystyle N = W - P_y \ \Rightarrow \ P_y = W - N$
$\displaystyle P_y = W - f/\mu_s$
$\displaystyle P_y = W + (F - P_x)/\mu_s$
$\displaystyle P_y = W + F/\mu_s - \frac{1}{\mu_s} \cdot P_x$

I didn't really understand the question, but isn't this an approach to anything?

That depends.

First, the weight is 60 lbs. That is NOT m, but W. "lbs" is the unit of force in the British system of units, not mass which is usually taken to be in "slugs" if I remember correctly.

Second, W = mg by definition, not mg*cos(alpha). The weight component along the plane has this magnitude. Is this what you meant?

Finally, what is F? ThePerfectHacker was using this symbol for the static friction force and you are using f for that purpose.

I'm not saying your approach is wrong, I'm saying I don't understand what you are trying to do with it.

-Dan
• Nov 22nd 2006, 08:48 PM
CaptainBlack
[quote=topsquark]

That depends.

First, the weight is 60 lbs. That is NOT m, but W. "lbs" is the unit of force in the British system of units, not mass which is usually taken to be in "slugs" if I remember correctly.[quote]

Wonderfully system, you have two options pounds are mass and poundals are
force, or pounds are force and slugs are mass. My experience is that the
first of these is more commonly used (in a bizarre usage of the term common
- we never use imperial or customary units over here any more).

units in maths or science homework whenever I see it, telling the student to
go back to their professor/lecturer/teacher and telling them off on my behalf.

RonL
• Nov 23rd 2006, 02:05 AM
topsquark
Quote:

Originally Posted by CaptainBlack
Wonderfully system, you have two options pounds are mass and poundals are
force, or pounds are force and slugs are mass. My experience is that the
first of these is more commonly used (in a bizarre usage of the term common
- we never use imperial or customary units over here any more).

units in maths or science homework whenever I see it, telling the student to
go back to their professor/lecturer/teacher and telling them off on my behalf.

RonL

(Shakes his head in despair) Never heard of "lb - mass" before! :eek: I have heard of poundals, though, so I should have suspected it.

I suppose that also gives a reason for the (mis)usage of "kg" as a unit of weight.

-Dan
• Nov 23rd 2006, 03:38 AM
TriKri
> Wonderfully system
Thank you CaptainBlack!

Haha, ok maybe I missunderstood ThePerfectHacker's use of variables names, this is how I imagined it:
http://i60.photobucket.com/albums/h2...rictionbox.png
Hope that makes any sense ;)

And by the way, by setting out the forces in the system I meant their absolute values and nothing else. Maybe I should have stated that.
• Nov 23rd 2006, 04:54 AM
topsquark
Quote:

Originally Posted by TriKri
> Wonderfully system
Thank you CaptainBlack!

Haha, ok maybe I missunderstood ThePerfectHacker's use of variables names, this is how I imagined it:
http://i60.photobucket.com/albums/h2...rictionbox.png
Hope that makes any sense ;)

And by the way, by setting out the forces in the system I meant their absolute values and nothing else. Maybe I should have stated that.

Gotcha. Yes, that is valid. We still have the problem, though, that we have 4 unknowns (Px, Py, f, and N) and only 3 independant equations to work with.

-Dan
• Nov 23rd 2006, 05:04 AM
TriKri
Maybe |P| shall be minimized? I agree, the problem statement is a little bit fuzzy.
• Nov 23rd 2006, 06:07 AM
ThePerfectHacker
Quote:

Originally Posted by TriKri
Maybe |P| shall be minimized? I agree, the problem statement is a little bit fuzzy.

It turns out that the force is minimized when the angle is minimized. One of my classmates said he solved the problem. When I asked him how, he simply said, "I ignored the angle".
• Nov 23rd 2006, 06:13 AM
topsquark
Physics problem solving 101:

When you can't solve the problem, change it to something that can be solved. :) (Actually, I've noted Mathematicians using this one too. They simply redefine the problem into one they know how to solve. Physicists, on the other hand, tend to use "spherical horses," as the joke goes.)

-Dan
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