Coefficient of friction and forces proof

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Homework Help Overview

The problem involves determining the minimum force required to move a box of weight W on a horizontal surface, considering the angle theta at which the force is applied and the coefficient of static friction mu. The original poster attempts to derive a specific formula relating these variables.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the applied force, frictional force, and normal force, with various attempts to set up equations based on horizontal and vertical force balances. Questions arise regarding the correct interpretation of forces and the notation used for the normal force.

Discussion Status

There is ongoing exploration of the problem with multiple attempts shared. Some participants provide guidance on setting up the equations correctly, while others question the assumptions made about the forces involved. The discussion reflects a mix of interpretations and approaches without a clear consensus on the final solution.

Contextual Notes

Participants note that the angle theta is above the horizontal and express confusion regarding the notation for the normal force, which is referred to as both R and n. There is also mention of differing interpretations of the forces acting in the vertical direction.

physfan
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Homework Statement



A box of weight W is pushed by a force F on a horizontal floor (the diagram below shows the force on the box is at an angle theta to the horizontal)
If the coefficient of static friction is mu, show that the minimum value of F that will move the crate is given by:

F=(mu*W*sec theta)/(1-mu*tan theta)

Homework Equations



The only maths needed i think is trig functions such as sec=1/cos, 1=sin^2+cos^2 etc.
For physics, I have called the frictional force acting against the box fr and the reaction against the horizontal plane R. fr=mu*R is the only equation I could find with the exception of the equations that are results of Newton's thid law (which are listed in my attempted solution)


The Attempt at a Solution



Horizontally I worked out F*cos theta=fr=mu*R
Vertically, F*sin theta=W=n (not sure which values to use so I plugged them all in there)

1st attempt: R is rective force and W=mg so R=W.
fr=mu*W
F*cos theta=mu*W
F=(mu*W)/cos theta

But obviously the result is more complicated than that so I then tried:

fr=R*tan theta (taken from previous class notes but I don't know if this value is suitable for this problem...)

F*sin theta=W=n

F*sin theta*tan theta=mu*W=(F*sin^2 theta)/cos theta

F=(mu*W*cos theta)/sin^2 theta (from rearranging the above line)

F=(mu*W*cos theta)/(1-cos^2 theta) then I just trailed off...
I have made many more attempts but have rubbed them out as they were yet more unsuccessful...

Please help!
 
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physfan said:

Homework Statement



A box of weight W is pushed by a force F on a horizontal floor (the diagram below shows the force on the box is at an angle theta to the horizontal)
Is it above the horizontal or below?
If the coefficient of static friction is mu, show that the minimum value of F that will move the crate is given by:

F=(mu*W*sec theta)/(1-mu*tan theta)

Homework Equations



The only maths needed i think is trig functions such as sec=1/cos, 1=sin^2+cos^2 etc.
For physics, I have called the frictional force acting against the box fr and the reaction against the horizontal plane R. fr=mu*R is the only equation I could find with the exception of the equations that are results of Newton's thid law (which are listed in my attempted solution)


The Attempt at a Solution



Horizontally I worked out F*cos theta=fr=mu*R
:confused: Iguess that by R you mean the normal force. This is not standard notation.
Vertically, F*sin theta=W=n (not sure which values to use so I plugged them all in there)

No. You must set the net force along y equal to zero. You can't set the three forces eqaul to one another!
1st attempt: R is rective force and W=mg so R=W.
the normal force is certainly not equal to the weight here since there is another force with a y component.
 
nrqed said:
Is it above the horizontal or below?

:confused: Iguess that by R you mean the normal force. This is not standard notation.


No. You must set the net force along y equal to zero. You can't set the three forces eqaul to one another!

the normal force is certainly not equal to the weight here since there is another force with a y component.

Yeah for R I do mean the normal force, I've heard it as both R and n depending on the teacher... I prefer n to be honest but my teacher today is an R person.
Looking at y now I'm thinking F*sin theta - n = 0 (hence F*sin theta = n) but that doesn't seem to get me any further than my previous attempts...
 
Oh and the angle theta is above the horizontal.
 
Oh hang on, F*sin theta + W = n doesn't it... sorry, I don't know why I didn't write that before. Blonde moment... I'll have another go.
 
Right, I'm now getting:

For y: F*sin theta+W - n=0
so F*sin theta+W=n

For x: F*cos theta - mu*n=0
so F*cos theta = mu*n

Substituting in the first equation for n:
F*cos theta=mu*(F*sin theta+W)
F*cos theta=mu*F*sin theta+mu*W
F*(cos theta - mu*sin theta)=mu*W

so F=(mu*W*cos theta)/sin^2 theta
=(mu*W*cos theta)/(1-cos^2 theta)

and then the trail goes cold again...
 
physfan said:
Oh hang on, F*sin theta + W = n doesn't it... sorry, I don't know why I didn't write that before. Blonde moment... I'll have another go.

This does not look quite right (unless some of your quantities there are negative)

The way to do it is to start with \sum F_y = m a_y
In your case a_y is zero. Therefore the net force along y is zero. Therefore
F \sin \theta + n - mg = 0
I had never heard of "R" for a normal. In what country are you, just out of curiosity?
 
I'm from England, there's just one teacher I have who calls it the "reaction" to the plane.

So if we take the direction of n to be positive, why is Fsin theta also positive... isn't that acting in the opposite direction? i.e. pushing down, therefore adding to the weight?
 
Last edited:
nrqed said:
This does not look quite right (unless some of your quantities there are negative)

The way to do it is to start with \sum F_y = m a_y
In your case a_y is zero. Therefore the net force along y is zero. Therefore
F \sin \theta + n - mg = 0
I had never heard of "R" for a normal. In what country are you, just out of curiosity?

I obtain the equation you quoted except that I have a division by 1+ \mu \tan \theta instead of with a minus sign like you gave. Are you sure you typed the rigt formula?
 
  • #10
nrqed said:
I obtain the equation you quoted except that I have a division by 1+ \mu \tan \theta instead of with a minus sign like you gave. Are you sure you typed the rigt formula?

You mean the right result?
Yes, it's definitely supposed to be 1-mu*tan theta on the bottom...
 
Last edited:
  • #11
nrqed said:
In your case a_y is zero. Therefore the net force along y is zero. Therefore
F \sin \theta + n - mg = 0
The force must be acting downward, so that should be:
-F \sin \theta + n - mg = 0

n = F \sin \theta + mg
 
  • #12
Yup, thought so, but I'm still not getting the right result...
I used n=Fsintheta + mg earlier in post #6... which shows my latest (unsuccessful) attempt. I can't really see how to carry it on.
 
  • #13
physfan said:
Right, I'm now getting:

For y: F*sin theta+W - n=0
so F*sin theta+W=n
OK.

For x: F*cos theta - mu*n=0
so F*cos theta = mu*n
OK.

Substituting in the first equation for n:
F*cos theta=mu*(F*sin theta+W)
F*cos theta=mu*F*sin theta+mu*W
F*(cos theta - mu*sin theta)=mu*W
Good.

so F=(mu*W*cos theta)/sin^2 theta
=(mu*W*cos theta)/(1-cos^2 theta)
Huh?

Just divide by (cos theta - mu*sin theta) and you're done. (Almost.)
 
  • #14
Doc Al said:
OK.


Huh?

Just divide by (cos theta - mu*sin theta) and you're done. (Almost.)

Sorry! My notes are rather messy and I typed out the wrong bit.
I meant:

F(cos theta - mu*sin theta)=mu*W
so F=(mu*W)/(cos theta-mu*sin theta). which is where I got stuck but if that's right I'll have another look first.
Thanks!
 
  • #15
You are one small step away from the final answer. Hint: What do you have to do the denominator to make it the way you want?
 
  • #16
Oh yeah you just divide throughout by cos theta!

Right, got it, thanks again

(Edit: and for the record, I did that before I saw your post...)
 

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