Physics, help needed!!?

Question

A long uniform rod which weighs 5.00 X 10^2 N leans against a smooth vertical wall. It leans with an angle of 60.0 degrees with respect to the horizontal floor. The rod is 1.5 meters long.

a) Find the Normal Force exerted by the wall.
b) Find the horizontal and vertical components of the force exerted by the floor assuming the rod is in equilibrium.
c) If the rod begins sliding such that it has an angular acceleration of 3.1 rad/s^2, what will the net torque be? (The moment of inertia of the rod can be calculated using I= (1/3) ML^2.)

I can't figure out how you came up with 144.3 N...Thanks for the help

Answer 1

The best way to solve this problems is as follows.

A) The torque acting about the base of the rod due to the weight of the rod is 500 N X (1.5 cos 60)/2 = 187.5 Nm This torque is balanced by the reaction force of the wall times its moment arm or Fwall X 1.5 sin 60 = 187.5. Thus F wall = 144.3 N

B) The horizintal force at the base of the rod must be equal and opposite the reaction force of the wall so it is 144.3 N. Because the wall smooth (frictionless) the vertical force at the base of the rod must be equal and opposite to the weight of the rod or 500 N.

C) Torque equals the moment of inertia times the angular acceleration thus the net torque = 3.1 rad/s^2 X (1/3) (500/9.8) (1.5 ^2) = 1186.2 Nm

Hope this helps, Mike R

Answer 2

For a static object, the sum of the x and y forces from any point must be equal to zero. Since the object is uniform, you can assume its weight is at its center. So draw a diagram and do the summation of the forces from the floor. Can you do this or shall I help you with this?

Write comments I will check back later.