Finding resultant force and moment about point O

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LMW - Dk
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Homework Statement



I have to determine the resultant R and the moment about O.

Homework Equations



see the image.

The Attempt at a Solution



See the image.
 
Attachments
  • Isometric diagram of a steel I-beam cantilevered from a wall, with axes x, y, z and origin O; a 36 kN force acts horizontally on the web, a 20 kN force at 30° acts at the lower flange, and dimensions 200 mm, 150 mm, 100 mm and 100 mm are marked.
    statics.PNG
    39.4 KB · Views: 474
  • Handwritten-style worked solution: position vector r = (0.1, -0.35, 0), moment formula M_O = sum(r x F) = r x R, and cross product giving (12.6, -3.6, 5.062), shown beside a 3D sketch of a wide-flange beam fixed to a wall with forces 36 kN, 20 kN, a 30 degree angle and dimensions 200 mm, 150 mm, 100 mm.
    Udklip 1.PNG
    21.5 KB · Views: 513
  • Handwritten-style Danish worked solution: components of 36 kN and 20 kN forces from F cos θ relations, giving Fx36=0, Fy36=0, Fz36=36 and Fx20=17.32, Fy20=-10, Fz20=0, then resultant R = (17.32, -10, -36) kN and magnitude |R| = 1695.98 kN.
    Udklip.PNG
    18.5 KB · Views: 465
Hi LMW - Dk! Welcome to PF! :wink:

Show us what you've tried and where you're stuck, and then we'll know how to help! :smile:
 
See the pictures. I have found the moment and the resultant. However I think it's the wrong method ^^
 
sorry, i didn't notice your third picture :redface:

your Fz36 should be minus 36, shouldn't it? :smile:

(btw, i think most people find it easier to use cos and sin, of angles < 90°, and then use common-sense to decide whether its + or -)

i don't understand what you did to find the moment :confused: …
can you please type it out in the post?​
 
Well can't you tell me how you would do this problem? how would you find the resultant and the moment?
 
If you click on the second picture then you can see that i have drawn a direction vector. Then i have found the cfross product of the direction vector and the resultant force.
 
LMW - Dk said:
If you click on the second picture then you can see that i have drawn a direction vector. Then i have found the cfross product of the direction vector and the resultant force.

ah, now i understand …

no, you've crossed the resultant with the position vector for only one of the vectors: you need to cross each vector with its own position vector, and then add (ie, there's no short-cut)

(and did the minus 36 help?)​