Recent content by bnoone

Ohhh okay that makes sense. I forgot to include the sine of the angle between them. Thank you!

Isn't it just the length of the ladder?

Should I use 6 (the horizontal distance from the base of the ladder to the wall) instead?

Okay so after fixing that I have ∑\tau = 0 = F_{A}sin(53) +5(180)sin(37) - 10(F_{A} - (.38)(180)) Solving for F_{A}, I got 118.23 N which is still wrong. Did I make a mistake in summing the torques?

Homework Statement A uniform ladder is 10 m long and weighs 180 N. In the figure below, the ladder leans against a vertical, frictionless wall at height h = 8.0 m above the ground. A horizontal force is applied to the ladder at a distance 1.0 m from its base (measured along the ladder). (a)...