Magnitude of force to keep stick in equilibrium.

[zeralda21]

1. Homework Statement [/b
A straight and homogenous stick with mass m is pressed against a wall with the force F. The stick is horizontal perpendicular against the wall. Given that the friction between the wall and the stick is μ, determine the horizontal component of F in order for the stick to not fall down.

http://i.imgur.com/QFJq7wT.png

The Attempt at a Solution

Forces involved:

We have:

The gravitational force mg in the negative y-direction.
The normal force from the wall, N.
The friction force in the positive y-direction which is f=μN
The force F which acts in the positive x-direction
??From the figure, the force F should have a component in the y-direction but if I follow the description I don't necessarily think it has...

Anyway;

ƩF_x; F-N=0 <---> F=N
ƩF_y; mg-μN=0 <---> N=F=mg/μ

Hence F=mg/μ. Correct answer is however F=mg/(2μ)

 

[gneill]

Assume that the force does have some vertical component. Can you tell from the setup what value that component would have? Hint: Through what point does the weight of the stick act? How much of that weight has to be supported at each end of the stick?

 

[zeralda21]

Assume that the force does have some vertical component. Can you tell from the setup what value that component would have? Hint: Through what point does the weight of the stick act? How much of that weight has to be supported at each end of the stick?

That's precisely what I have done, leading to F_x: F-N=0 , F=N. Through what point does the weight of the stick act? Hmm, the midpoint I guess. Do you mean that the "torque-equation" should be used? Because I believe all "lines of action" for forces(F, N, f) go through this point.

 

[gneill]

The stick is horizontal and of uniform composition and density so you should be able to draw an informed conclusion based upon symmetry (without even resorting to calculation!) about how much of the stick's weight needs to be supported at its ends... Its full weight is mg and acts vertically through the midpoint of the stick. How much of the full weight needs to be supported at either end? Your initial calculation seemed to imply that the stick is pressing down by its full weight mg at both ends.

 

[zeralda21]

The stick is horizontal and of uniform composition and density so you should be able to draw an informed conclusion based upon symmetry (without even resorting to calculation!) about how much of the stick's weight needs to be supported at its ends... Its full weight is mg and acts vertically through the midpoint of the stick. How much of the full weight needs to be supported at either end? Your initial calculation seemed to imply that the stick is pressing down by its full weight mg at both ends.

I am very sorry gneill about my errors. Of course you are right, the vertical component of force F holds up half the stick, so the friction force only needs to hold up the other half of the stick's weight, not the whole weight. Of course...again, my apology.

My mistake was that I assumed the force to be horizontal with no vertical component. In that case, my calculations is correct I think.

 

[gneill]

I am very sorry gneill about my errors. Of course you are right, the vertical component of force F holds up half the stick, so the friction force only needs to hold up the other half of the stick's weight, not the whole weight. Of course...again, my apology.

Hey, no need to apologize! The important thing is getting to the right answer and hopefully gaining experience along the way!

My mistake was that I assumed the force to be horizontal with no vertical component. In that case, my calculations is correct I think.

If F had no vertical component, how could the stick remain horizontal?