Mechanics & Free Body Diagrams

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Homework Statement

Please see the attachment, Q2

Homework Equations

$$\sum$$F_x = 0
$$\sum$$M_A = 0

The Attempt at a Solution

for part A, i get 14123N as the tension, moment at C about B, 9.8x1000x5 = 49050
49050/7 = 7007N
tan^-1 (4/7) = 29.74^o
TSin(29.74) - 7007 = 0
TSin(29.74) = 7007
T=7007/sin(29.74) = 14123N

Is this correct?

Also I had a go at part b, however, went round in circles, so can anyone help me?

Cheers

Andrew

 

[pongo38]

There are 3 forces that matter to the beam. First the tension in the cable; second the 1T weight; third the reaction at C. The lines of action of these three must meet at one point if the object is to be in equilibrium from a moment point of view. If you can identify that point of intersection, that will give you the direction of the reaction at C. If you now move those three lines apart to form a triangle, the sides will be in proportion to the forces. You should be able to get it from there. That will incidentally check your first answer for yourself. I accept that there are more algebraic ways of doing this, but with triangles etc, you can see what is going on.