Recent content by sprouser

OK. Sum Torque = 0 = Tr(6.0m) - 800(6.0m)(cos(37)) - 400(3.0m)(cos(37))

yes, There is a vertical wall. A diagnol beam is attached at one end to the bottom of the vertical wall and attached at the other end with a ropw that is connected to the top of the vertical wall. It is a confusing diagram becuase of the periods.

I don't understand how the rope has a vertical component

Here is my FBD ...T---./ .../.| AY.../...M |.../ W |/____AX Tr represents tension in the rope

Alright Sum of forces in y = 0 = Ay - 800N - 400N Sum of forces in x = 0 = Ax - Tr Sum Torque = 0 = Tr(6.0m) - 800N(6.0m) - 400N(cos(38))(3.0m) Tr = 960N Ax = 960N Ay = 1200N I think that is right?

Alright, could someone give me an answer to see if I applied the method correctly?

Alright I don't know how to solve the problem. All I know is that sum of forces in xa nd y are 0 and sum of torque is 0. I don't know how to draw the FBD.

HELP! Static Equilibrium |_______/ |.../| |.../ 0 400N |.../ |/_37? | Uniform pole of 6.0 m w/ weight 800N is attached to a vertical wall. A load of 400N hangs from other end of pole. The pole it held by a wire with an angle of 37?. WHAT is the TENSION of the wire and the HORIZONTAL and...