so T=0? Since -209.132 + -114.717 = -323.849.. the positive torque is 323.849 so they cancel and equal 0.
so. 0=F(1.29)(sin27.2172).. does that mean the force of the wall is 0? Since 1.29F(sin27.2172)=.59. 0=.59F. And 0/.59=0.
T of man on ladder=(603N)(.390m)(sin62.7828)=-209.132
T of ladder=(200N)(.645m)(sin62.7828)=-114.717
T of friction on floor=0
T of force exerted by floor=0
So. T=F(1.29)(sin27.2172)? Would T be the -209.132 and -114.717 added?
So would the angle be 62.8 instead of 27.2? I'm not exactly sure on what you mean by the radial line.
T=F(1.29)(sinθ) is what you mean to find the force caused by the wall, right? Will i end up with an answer that contains both T and F? Or will it be a number?
I've decided that i'll try to pick the point where the ladder is on the floor, since it elimates the friction of floor and force of floor, since r=0. I guess I'm confused about the forces 603-N and 200-N acting on the ladder... should i add those together? Or should the equation look like this...
Ok, well, i think i'll pick where the man is on the ladder as the pivot point, since I'm trying to set both sides equal to each other, correct?(equilibrium?). The torque equation is t=Frsinθ... I'm not exactly sure how to go about finding the torques on a point.
Alright, i redid the image. I was sure about the frictional force that the floor exerts, so i have not added that in. Which way is the force for the friction the floor exerts?
http://img526.imageshack.us/img526/2717/equilibrium3im.gif
A 1.29-m long ladder weighing 200N rests against a vertical wall so that the top of the ladder is at a height of .590m. A 603-N man stands on the ladder at a distance of .390m along the ladder. Calculate the force exerted by the wall on the ladder. Assume that the wall is perfectly smooth...