- #1
DrBanana
- 47
- 3
- Homework Statement
- A uniform rod/ladder rests against a smooth wall, with its foot resting on a smooth plane that makes an angle ##\theta## with the wall. The wall is perpendicular to the horizontal x axis, and the ladder is at an angle ##\alpha## with this x axis. Show that ##tan\theta=2tan\alpha##
- Relevant Equations
- ##\sum F = 0##, ##\sum M = 0##
In the picture you can see what I'm talking about. This is a simple problem to do if you can use moments, but alas my education system seems to have something against those.
##AB## is the uniform rod/ladder, G is its centre of mass, BE is the horizontal x axis, AD is the wall, BD is the smooth surface. S is the normal force on the foot of the ladder, so it's perpendicular to BD. R is the normal force from the wall. BC and GO are just geometric constructions, and they are both parallel to the wall.
My book claims that since the net force on the ladder is zero, the line of action of S will have to go through O. I want to know why that should be the case, where does it come from, is it actually a valid way to solve problems, if so then does it actually have a physical meaning, and if it does have a physical meaning, then how. And if it doesn't have a physical meaning, then why does this 'method' still work.
The proof continues by showing ##\angleBOC=\theta##, and then that O is the midpoint of AC (because G is the midpoinnt of AB and OG and BC are parallel). So ##tan\theta=\frac{BC}{\frac{1}{2}AC}=2tan\alpha##.
So in the same picture we represented forces, positions and also arbitrary (as in, with no physical existence) lines, and it all sort of worked out. For context, all of our statics proofs are like this, and it's a predicament that's shared by the bloke who made this post six years ago: https://physics.stackexchange.com/q...which-the-resultant-of-two-unlike-unequa?rq=1