- #1
eurekameh
- 210
- 0
A uniform ladder of mass m = 30.0 kg and length l = 2.50 m leans against a frictionless vertical wall. The ladder forms an angle theta = 75.0 degrees with the rough floor. The coefficient of static friction μ between the floor and the ladder is equal to 0.288. Determine the force Fw exerted on the ladder by the wall, and the frictional force f that the floor exerts on the ladder.
So here's my force diagram: http://imageshack.us/photo/my-images/90/gha.png/
Here are my equilibrium equations:
Fnet,x = Fw - f = 0
Fnet,y = Fv - mg = 0
Tnet,z = -(Fw)(2.4148) + (mg)(0.3235238) = 0
From the sum of the torques, I found Fw:
Fw = [(mg)(0.3235236)] / (2.4148) = 39.4 N.
Fw = f = 39.4 N.
What did I do wrong?
So here's my force diagram: http://imageshack.us/photo/my-images/90/gha.png/
Here are my equilibrium equations:
Fnet,x = Fw - f = 0
Fnet,y = Fv - mg = 0
Tnet,z = -(Fw)(2.4148) + (mg)(0.3235238) = 0
From the sum of the torques, I found Fw:
Fw = [(mg)(0.3235236)] / (2.4148) = 39.4 N.
Fw = f = 39.4 N.
What did I do wrong?