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Moment Of A Force

  • Thread starter Struggling
  • Start date
52
0
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 [Broken]


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 im going wrong i have obviously misunderstood something

thanks
 
Last edited by a moderator:

Answers and Replies

Pyrrhus
Homework Helper
2,160
1
Your thinking is right, check the numbers and maybe the signs.

[tex] \vec{M}_{o} = (6 \vec{i}) \times (-520 \frac{5}{13} \vec{i} + 520 \frac{12}{13} \vec{j}) [/tex]
 
52
0
thanks Cyclovenom
 

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