A uniform ladder of mass m and length L leans against a smooth wall, making an angle q with respect to the ground. The dirt exerts a normal force and a frictional force on the ladder, producing a force vector with magnitude F1 at an angle φ with respect to the ground. Since the wall is smooth, it exerts only a normal force on the ladder; let its magnitude be F2.
(a) Explain why φ must be greater than θ. No math is needed.
(b) Choose any numerical values you like for m and L, and show that the ladder can be in equilibrium (zero torque and zero total force vector) for θ=45.00° and φ=63.43°

Question

A uniform ladder of mass m and length L leans against a smooth wall, making an angle q with respect to the ground. The dirt exerts a normal force and a frictional force on the ladder, producing a force vector with magnitude F1 at an angle φ with respect to the ground. Since the wall is smooth, it exerts only a normal force on the ladder; let its magnitude be F2.
(a) Explain why φ must be greater than θ. No math is needed.
(b) Choose any numerical values you like for m and L, and show that the ladder can be in equilibrium (zero torque and zero total force vector) for θ=45.00° and φ=63.43°

anonymous · Answer

http://www.lightandmatter.com/html_books/lm/ch15/ch15.html #8

jamesj · Answer

The ladder makes an angle q or $ \theta $ to the ground?  In other words, is this angle the angle $ \theta $ in part a of the question?

anonymous · Answer

yes

anonymous · Answer

attachment

jamesj · Answer

Right.  So the answer to the first part is fairly straight-forward.  Consider the torque calculation about the center of mass of the ladder: i.e., the mid-point.

anonymous · Answer

i see

jamesj · Answer

If the ladder is stable, i.e., not rotating at all, then the net torque about that point must be zero.  If F1 were at an angle less than theta, then that would not be the case.

jamesj · Answer

As it is, the torque due to F1 must cancel the torque due to F2.

anonymous · Answer

i reached a point where : Fwall*sinθ=W*cosθ*1/2

jamesj · Answer

Step back -- what are you solving now.  Part b?  And if so, about which point are you calculating the torque?

anonymous · Answer

the floor

jamesj · Answer

Right.  That's one equation.  What you'll need to do is find another equation so you can calculate the magnitude of $ F_{wall} = F_2 $

anonymous · Answer

+W?

jamesj · Answer

About the center of mass?  Yes, that's the logical choice.

anonymous · Answer

nvmd i thought Fwall=F1:P

anonymous · Answer

Nwall = Ffriction
N=W

jamesj · Answer

No.  What is true when we calculate torque around the center is that 
$$ F_2 \sin(\theta) = F_1 \sin(\phi - \theta) $$

anonymous · Answer

soo simple!:D

anonymous · Answer

i did not take into acount φ so i calculated at the floor , center and top with no luck ,