Find Tension of cable and momentum about point

[Pete_01]

Homework Statement

A door is held open by a cable, which is vertical, and the door is 200kg (its center of gravity is midway along the door). Calculate the tension of the rope and the momentum at the hinge A for angle theta = 30 degrees. Picture here: http://i54.tinypic.com/160wqxk.jpg"

Homework Equations

sum of forces = (sum of forces in x)+(sum of forces in y) = 0
sum of momentum(about A) = rxF

The Attempt at a Solution

First I attempted to find the sums of forces in the x and y directions:
(sum forces x dir): Ax - Wsin30
(sum forces y dir): Ay+T-Wcos30
where the hinge has two forces, and Ax and Ay acting on it.

(sum momentum about A): This is where I get stuck.

Am I approaching this right? I feel like I need to break up the Tension (T) into x and y?

Thanks.

 

[RTW69]

Find the component of T and force of door due to gravity that is perpendicular to the door. T wants to rotate the door clockwise. The force due to gravity wants to rotate the door ccw around A. You can solve for T directly.

 

[Pete_01]

Would the angle I use for T be 30 degrees?

 

[PhanthomJay]

You are using the wrong terminology...it's moment (or torque), not momentum.

If you draw a free body diagram of the door, the weight force of the door produces a moment about A, and the tension force in the vertical rope produces a counterbalancing moment about A. There is no horizontal force at the hinge.

 

[Pete_01]

You are using the wrong terminology...it's moment (or torque), not momentum.

If you draw a free body diagram of the door, the weight force of the door produces a moment about A, and the tension force in the vertical rope produces a counterbalancing moment about A. There is no horizontal force at the hinge.

I guess I'm still a bit confused. My book shows a pin support (a hinge) with two force components, Ax and Ay.

 

[RTW69]

Yes, what is the component of T perpendicular to the door and creating a torque cw around A?

 

[RTW69]

Both T and F_g due to the mass of the door are in the vertical, (Y) direction. There is no R_x unless there is a force in the horizontal, (x-direction). What are the sum of the forces in the x-direction?

 

[Pete_01]

Both T and F_g due to the mass of the door are in the vertical, (Y) direction. There is no R_x unless there is a force in the horizontal, (x-direction). What are the sum of the forces in the x-direction?

Wait, what do you mean by R_x? The sum of the forces in the x-dir would be A_x+W_x+T_x correct?

 

[PhanthomJay]

Wait, what do you mean by R_x? The sum of the forces in the x-dir would be A_x+W_x+T_x correct?

Yes. Note we are considering the x direction as the horizontal direction. Since in the free body diagram of the beam the weight and tension have no horizontal components, then A_x must be ____?