Uniform Beam (equivalent systems)

[ste234]

Homework Statement

A uniform beam |AE| at 2m from A there's a 500N force at 60degrees to the negative x-axis direction. At 3.5m from A (1.5m from the 600N) There's a 200j N force and at 4.5m from A there's a -100i N force acting 0.5m above the beam.i)Determine the single force which is passing through B (2m from A) and the resuting moment which is equivalent

ii)Reduce the system in i to a singular force (no couple) and determine where its line of action lies on AE using A as a point of reference for distance

(THis is B in the attached PDF)

Homework Equations

For equilibrium:
sum of forces=0
sum of moments=0

The Attempt at a Solution

i)
The 600N force can be broken down to 500Cos60 i - 500sin j = 250i-433j N
Sum of forces in Y= -433+200=-233
Sum of forces in X= 250+100=350
Sum of moments about B= 200(1.5)+100(0.5)=350Nm
Therefore, the equivalent system is 350i-233j N and a 350Nm moment at B.
ii)
I'm not very sure about this. Can I move the moment a certain distance so that it become a force of 350/(some distance from B)
Edited:
I found an equation which says d=M/F (moment divided by resultant force) so in this case it would be 350/(350i-233j)
so would that be 350/420=5/6 m but to which side of B? is 5/6m to the left since its a positive moment. (Anti-clockwise being positive)

 

[jbsteele]

Therefore, the line of action of the equivalent force is 5/6m from A in the negative x-axis direction