Calculate Moment of Force at Point A: 2.88kN.m

[Struggling]

hi, i have tried to work this question out it seems so easy but i must be missing something.

determine the magnitude and directional sense of the moment of the force at A about point O.
http://img229.imageshack.us/img229/9895/statics19hv.jpg


these are 3 of my calculations for the question

F = 520 sin 30(6) - 520 cos 30(0)
F = 1560N or 1.56kN
also

F = 520 sin 120(6) - 0
F = 2.7kN

i also tried using vectors

r = {6i- 0j}m
F = {520sin30i - 520cos30j}N
= {260i - 450.33j}
The moment is therefore

Mo = r x F = 0i - 0j + [6(-450.33)-(0)(200)]k
= {-2.7k} kN.m

the answer in the books telling me 2.88 kN.m can someone point me to where I am going wrong i have obviously misunderstood something

thanks

 

[Pyrrhus]

Your thinking is right, check the numbers and maybe the signs.

$$\vec{M}_{o} = (6 \vec{i}) \times (-520 \frac{5}{13} \vec{i} + 520 \frac{12}{13} \vec{j})$$

 

[Struggling]

thanks Cyclovenom