Net forces versus net moments -- solving a beam problem

Click For Summary
SUMMARY

This discussion focuses on solving a beam problem using net forces and net moments to determine the reaction force at point O. The participants highlight discrepancies in their approaches, particularly in the application of torque and the treatment of forces. Key insights include the necessity of separating vertical and horizontal components when calculating moments and the importance of accurately accounting for all forces acting on the beam. The final correct reaction force at O is determined to be 3582N after addressing these issues.

PREREQUISITES
  • Understanding of static equilibrium principles in mechanics
  • Familiarity with free body diagrams and their construction
  • Knowledge of torque and moment calculations
  • Proficiency in vector decomposition of forces
NEXT STEPS
  • Study the method of resolving forces into components using vector analysis
  • Learn about calculating moments about various points in a beam
  • Explore the application of Pythagorean theorem in force analysis
  • Review examples of static equilibrium problems in structural engineering
USEFUL FOR

Students and professionals in mechanical engineering, civil engineering, and physics who are working on static equilibrium problems and beam analysis.

yecko
Gold Member
Messages
275
Reaction score
15

Homework Statement


螢幕快照 2017-10-05 下午9.10.05.png


Homework Equations


in photos

The Attempt at a Solution


[/B]
First approach:
05763487-2c0a-4b8b-9dc1-816c03665cde.jpeg


Second approach:
aa.jpeg


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!
 
Last edited by a moderator:
Physics news on Phys.org
In your first approach, first equation (Fyx3...) 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.
 
o thanks
wait let me correct it
i have got 3582N as final answer
 
yecko said:
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.
 
Corrected
image.jpg
 
how can I have two variables in the equation? F(O,y) and M(O)?
Thanks
 
Oh... you mean I should consider the net force first in finding F(O,y) then find M(O) by torque equation?
 
Correction
image.jpg


I am not sure it is right to put two moments together in this way...
Thanks~
 
  • #10
yecko said:
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.
 
  • #11
Ok i corrected it...
But i am not sure how two moments about different points being put together into one equation...
Thanks
 
  • #12
image.jpg
 
  • #13
yecko said:
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.)
 
  • #14
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...
 
  • #15
Try re-drawing your free body diagram.
 

Attachments

  • IMAG1252.jpg
    IMAG1252.jpg
    27 KB · Views: 527
  • #16
661E5F1A-747D-47F7-9122-936F73E7B329.jpeg
 
  • #17
oh sorry I didn't account for the weight of the beam in my help. I'll amend that.
 
  • #18
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)?
 
  • #19
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.
 
  • #20
yecko said:
where using Pyth Thm...
You found FO,x and FO,y, then combined them with √(FO,x2+FO,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 FO,y by the 4.8 and the FO,x by zero.
yecko said:
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, FO,x, FO,y and MO.
If you take moments about O, MO appears in the sum but not the other two.
 
  • #21
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...
 
  • #22
yecko said:
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.
 
  • #23
image.jpg

Correction
 
  • #24
yecko said:
View attachment 212445
Correction
Looks right, except that you should not quote so many significant figures.
 
  • #25
Thank you very much for your help!
 

Similar threads

Find the mass M so that the net torque is zero
  • · Replies 6 ·
Replies
6
Views
2K
Understanding Net Force and Net Moment in Statics
  • · Replies 4 ·
Replies
4
Views
4K
Net force and moment on a wooden bar fixed at one end
  • · Replies 16 ·
Replies
16
Views
3K
Tension and net torque in a cable
  • · Replies 10 ·
Replies
10
Views
5K
Net Force acting on block on incline
  • · Replies 1 ·
Replies
1
Views
2K
Lifting a 3m Beam: Calculating Force, Friction, & Net Force
  • · Replies 2 ·
Replies
2
Views
1K
Solve Force & Moment Homework: Why Don't Consider P at C?
  • · Replies 1 ·
Replies
1
Views
1K
Book plots position & force on the same graph, still gets right answer
  • · Replies 8 ·
Replies
8
Views
998
Torque: find the force necessary for a body to be in equilibrium
  • · Replies 6 ·
Replies
6
Views
2K
Net force acting on a charged particle ##+Q##
  • · Replies 9 ·
Replies
9
Views
3K