A sign is suspended from two chains. Determine moment and force.?

Question

A sign is suspended from two chains AE and BF. Knowing that the tension in BF is 200 N, determine (a) the moment about A of the force exterted by chain B, (b) the smallest force applied at C which creates the same moment about A.

Diagram:
http://i236.photobucket.com/albums/ff35/bugmenot_07/6.jpg

Answer 1

First step is to find the distance from A to B:
A-B=sqrt(2^2+(1.35-.95)^2)

Next, compute the angle between A-B and B-F:

Atan(2/.4)+radians(30)

the angle, th, that is the angle subtended by B-F and a perpendicular to d is
th=Atan(2/.4)-radians(60)

now compute the moment
=A-B*200*cos(th)

=386 N m

b) I will assume the force at C will be straight up
The distance A-C=sqrt(1.35^2+4)

The angle between A-C and vertical is
Atan(2/1.35)

the angle subtended by vertical and perpendicular to A-C is
th = radians(90)-Atan(2/1.35)

therefore
386=A-C*cos(th)*f
where f is the force
f= 193 N

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