- #36
2013
- 92
- 0
thank you
Is this the complete solution?
Is this the complete solution?
I asked2013 said:T=Fn*(L^2-R^2)
Fn=mgr/(L^2-R^2) ?
Neither of what you've written above appear to be attempted answers to those questions.- what equation can you write relating θ, L and R?
- what is the horizontal component of T?
[In response to post #40, which I'm assuming is a reply to post #39]2013 said:Is this right?
If those (T, Fn, Fg) were the only forces then your first equation would be right, but the second is wrong. But please, please, stop trying to jump ahead. Take one step at a time, as I keep asking you. It'll be quicker in the long run.2013 said:And yet?
I'm unable to comprehend what you've posted above. Suddenly adopting a new set of symbols for the forces doesn't help. Please stick to:2013 said:When drawn from the outside in addition to the weight force acts, the tensile force.
Same size and in the same direction, the frictional force down (due to the deflection).
So this results in an effective force to bottom:
F = F (g) + 2 F (r), this causes the pushing force F on the wall (E) = F tan (alpha),
from which the friction force F (r) = mu * F (D) results.
so:
mu * (F (g) + 2F (r)) (alpha) = F (r) * tan, switch to F (r), done.
Is it right?
Excuse me but I have only two days time to solve the problem, otherwise I fall through the semesters and must repeat the year.
Can you please help me?
Hooray!2013 said:1. sin(θ) = R/L
I asked for the horizontal components of 5 forces, so there should be 5 items before the equation that combines them. Pls don't skip questions I ask, I ask them for a reason.2.+3.
Fn = sin(θ) * L
You're still not doing what I ask. Don't write any equations yet - you're not ready.2013 said:Fg=Fn/tan(θ)
Fz=Fr
Try to write the corresponding statements for the other four forces: Fz, Fr, Fg, and T. The last one, T, will involve theta.The force Fn acts horizontally. Therefore it has a vertical component 0 and a horizontal component Fn. The horizontal component acts away from the wall.
All the above are correct2013 said:------------------vertical component -------------------- horizontal component
Fn ---------------------------- 0 ---------------------------- Fn
Fz ---------------------------- Fz ---------------------------- 0
Fr ---------------------------- Fr ---------------------------- 0
Fg ---------------------------- Fg ---------------------------- 0
No. The answer should only involve T and θ.T ---------------------------- Fg/cos(θ) --------------------- Fn/sin(θ)
2013 said:T ---------------------------- T*cos(θ) --------------------- T*sin(θ)
And then?
Yes.2013 said:horizontal components:
Fn - T*sin(θ)=0
No, it makes no sense to put the same force on both sides.vertical components:
Fz+Fr+Fg-T*cos(θ)=Fz+Fr
It could be, depending on which option you are adopting for the signs on the forces. Please choose one.2013 said:M1=Fr*R
M2=Fz*R
M1=M2
Fz=Fr
Is this right?
2013 said:I would like to choose option 3.
I don´t know which case we discuss already, but I thought we speak about where the paper is hanging over.
But I need both cases, because I should say which is from the physical side better.
Consider which way Fr acts when the paper hangs away from the wall.2013 said:It applies to when the paper hang next to the wall.
hang away from wall:
?
Yes, so how does that change the equation for vertical forces?2013 said:Fr acts the same way like Fz.
Both acts down.
2013 said:Fz+Fg+Fr = T*cos(θ)
?