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Apologies if I have done something wrong, first time posting on a stack exchange website.

Four forces act perpendicular to the surface of a circular slab (assume mass of slab is 0 and middle of slab is at 0,0,0)(all forces in -y direction)
Force 1 is located at 5,0,0 and has a magnitude of 77.4kN
Force 2 is located at 0,0,5 and has a magnitude of 48.6kN
Force 3 is located at 0,0,-5 and has a magnitude of 24.4kN
Force 4 is located at -5,0,0 and has a magnitude of 97.6kN
What is the coordinates of the resulting force (x,y,z) to two decimal places?

Visual Representation of Question



My working currently and my initial answer Working and answer



The given answer is (0.49,0,-0.41) and I don't know where I went wrong.

After talking to my lecturer, she said the question was supposed to be determining the z coordinate only, which is why it would say the answer was correct when 0.49 was put first, meaning that the answers I calculated were correct.

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    $\begingroup$ If you want help, show us what you've done so far. This isn't a "do my homework for me" site. $\endgroup$ – Fred Mar 29 at 9:37
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    $\begingroup$ Sorry about that, thanks for letting me know, fixed. $\endgroup$ – Max Mar 29 at 10:16
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    $\begingroup$ The book answer looks wrong IMO. Without doing any calculations, Force D is bigger than force C. Imagine the slab is balanced like a seesaw at x = 0. It would tip towards D, therefore the x position of the resultant will be negative. Also force A is bigger than force C so the z position of the resultant will be positive. $\endgroup$ – alephzero Mar 29 at 18:47
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We start by the forces along the x-axis about D and call the resultant distance from D, X.

$\Sigma M_{about\ D}=0 \rightarrow X*(97.6+24.4+77.4+48.6 ) \\= 73*5+77.4*10 \quad \rightarrow X=\frac{365+774}{248}=4.59$

$ 5-4.59=0.41$ X is 0.41 meters to the left of the origin $ \ X=-0.41m \quad $ as per the book.

The same steps apply to the Y-axis. Please check my numbers. also, I assumed the radius 5m not 6m as shown, because you asked 5m.

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  • $\begingroup$ The reason the distance was 5 instead of 6 is because the question is worded weirdly, the radius of the slab is 6m but the forces are placed 5m away from origin. Thanks for the response. $\endgroup$ – Max Mar 30 at 4:11
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I haven't looked at your specific numbers but on quick review it appears that your approach is correct by summing the moments (force x distance) in each of the two directions, and that both directions are independent from each other.

Two things

  1. To be an engineer, you have to learn to trust YOURSELF. The option that you are right is always an option. The book could be wrong. Find another way to do the math to double check your answer or confirm the books answer
  2. An engineer solves problems, yes? Or how about this: engineers create solutions. The former is passive, the later is assertive. You are doing this problem passively 'by the book.' However, you can be assertive and find your own approach and use it if it is valid. I'll use your question as an example. Note that the distances are all the same... figured out where I'm going yet? The distance is not a variable in this problem, it's a fixed number. Therefore, the problem is a simple algebra problem. All you need is the ratios of the loads and then apply that to the distance. If the two loads equal themselves, the ratio is 1:1 and the coordinate is directly between them, since the distance is the same for both from the origin, then the resultant is at 0. What happens when the ratio is not 1:1, on what side is the resultant and how far over? If nothing else, with this, you can come up with a very good approximate answer in seconds, in your head!
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    $\begingroup$ Thanks for your answer, I guess I automatically assumed the given answer was correct because it came from my lecturer (I realise that they can make mistakes too but that was why I made the assumption). I sort of did what you said in the second point a couple hours after I posted the question and looked at the opposing forces along the x and z axis and sort of realised that my answer must be right. But even then, I probably don't trust myself maybe as much as I should. Sorry for the long response, thanks again! $\endgroup$ – Max Mar 30 at 4:07

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