Net forces versus net moments -- solving a beam problem

[yecko]

Homework Statement

Homework Equations

in photos

The Attempt at a Solution

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First approach:

Second approach:

The two approaches used net force and net moment respectively in order to find reaction force of O, yet they got different answers. Why?

Thank you very much!

 

[haruspex]

In your first approach, first equation (F_yx3...) you have tricked yourself by writing it in the form "sum of moments from linear forces equals moment from torque". Try writing it as "sum of all moments equals zero".

I have not checked your second approach. It is hard to read. Please take the trouble to type in your working. Images are really for textbook extracts and diagrams.

 

[yecko]

o thanks
wait let me correct it
i have got 3582N as final answer

 

[haruspex]

o thanks
wait let me correct it
i have got 3582N as final answer

You seem to be assuming there is no torque reaction at O.

 

[yecko]

Corrected

 

[haruspex]

Corrected
View attachment 212346

You are still assuming there is no torque reaction at O.

 

[yecko]

how can I have two variables in the equation? F(O,y) and M(O)?
Thanks

 

[yecko]

Oh... you mean I should consider the net force first in finding F(O,y) then find M(O) by torque equation?

 

[yecko]

Correction

I am not sure it is right to put two moments together in this way...
Thanks~

 

[haruspex]

Correction
View attachment 212354

I am not sure it is right to put two moments together in this way...
Thanks~

Your moments equation has a few problems. You have included the x component of the force at O, and the 15000 has become 1500.

 

[yecko]

Ok i corrected it...
But i am not sure how two moments about different points being put together into one equation...
Thanks

 

[yecko]

 

[haruspex]

Ok i corrected it...
But i am not sure how two moments about different points being put together into one equation...
Thanks

I am not suggesting taking moments about two different points.
The problem is that you find the net force at O by applying Pythagoras to the x and y components, but then use that as though it were a vertical force in your moments equation. Only the y component has a moment about points in the beam.

(Do you need to find the net force? Can't you just leave it as x component, y component and moment?
And it would have been a bit simpler to take moments about O.)

 

[yecko]

I am not sure what you are talking about for where using Pyth Thm...
But if i take about O, how can i find out moment about O?

The net force i mean is of y direction, in order to find y direction reaction force of O...

 

[helppls]

Try re-drawing your free body diagram.

 

[yecko]

 

[helppls]

oh sorry I didn't account for the weight of the beam in my help. I'll amend that.

 

[helppls]

Your moment isn't a resultant force in the way you've described, it is a moment acting at point O from moments and forces acting perpendicular (Fy) to the point along the beam.

For example at point 0,
The Sum of Moments at point 0 is = Mo + Ay*distancefrom0 + Mb(given) - Cy*distancefrom0 - W*distancefrom0
rearrange to solve for Mo.

Your Oy reaction force is correct. Have you solved for Ox (Fx)?

 

[helppls]

Moments are not effected by horizontal forces. think of a moment like using a spanner/wrench acting on a bolt: If you push directly on the end of the tool, the nut doesn't turn - only if you apply a force on the side of the tool does the nut turn( ...get rid of your Fresultant equation)

Also, if you take the moment about a point you ignore the forces acting at that point BUT you still include any moment(s) acting at that point in your equation. Hope this helps.

 

[haruspex]

where using Pyth Thm...

You found F_O,x and F_O,y, then combined them with √(F_O,x²+F_O,y²). I don't know why you bothered to do that, but more importantly it does not make sense to multiply that by 4.8m to take moments. That resultant is not vertical. Leave it as two separate component forces, multiply F_O,y by the 4.8 and the F_O,x by zero.

But if i take about O, how can i find out moment about O?

I don't understand your difficulty. You want to find three reactions at O, F_O,x, F_O,y and M_O.
If you take moments about O, M_O appears in the sum but not the other two.

 

[yecko]

But there is also external force at B creating moment about B... how can i take moment about O taking this moment force into account?
They are about different points... as they are rotation, its not simply upward or downward force for F(B) with reference to O...

 

[haruspex]

But there is also external force at B creating moment about B... how can i take moment about O taking this moment force into account?
They are about different points... as they are rotation, its not simply upward or downward force for F(B) with reference to O...

An applied moment, as shown at B, is the same about all axes. In practice, it generally arises from two equal and opposite forces with different lines of action. E.g. we could represent it by an upward force of 25/6 kN at C together with a downward force of 25/6 kN at A. That gives an anticlockwise moment of 15kNm about any point in the plane. Try it for axes O, A and C.

 

[yecko]

Correction

 

[haruspex]

View attachment 212445
Correction

Looks right, except that you should not quote so many significant figures.

 

[yecko]

Thank you very much for your help!