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Multiple forces acting on a rigid body

  1. Mar 17, 2008 #1
    Hi,

    I'm attempting to do a simulation of rigid bodies dynamics, and have ran into a problem.

    1. The problem statement, all variables and given/known data
    It can and at some point will occur that multiple forces will be acting on a rigid body, at different points with different strength and under different angles.
    Obviously, before translating these forces into translational and rotational acceleration, they should be somehow added for a correct result.
    I do know how to add forces acting on 1 point, but I think this is a bit more tricky. I need to find out what the resulting force will be (my guess is the same as when they act upon 1 point), and secondly, what the new point will be this added force acts upon.

    2. Relevant equations
    not really an equation, but for multiple forces at one point you can use the 'head to tail' adding method. Or for computation, make an weighed avarage of both the x and y components of all forces.

    3. The attempt at a solution
    As mentioned before, I guess the resulting force will be the same as when all forces would be acting on one point. Furthermore, I have the idea that the point where the resulting force will be acting upon is an avarage of all points, somehow weighed by the strength of each of those forces. I can't seem to find nor figure out the exact equations.

    Some help would be highly appreciated :smile:
     
  2. jcsd
  3. Mar 17, 2008 #2

    PhanthomJay

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    You are on the right track, and you are correct in that the magnitude and direction of the resultant force will be the same as if all forces were applied at the same point. But now you need to know at what point the resultant force acts. This is determined by summing torques of the original forces about any point. Then the resultant force is located by dividing this summed torque value by the resultant force value. The result gives you the perpendicular distance from the point you chose for summing moments. It is often easier to split the individual forces into its x and y components. Then sum moments of the x components of the forces about a point, and divide the sum by the x component of the resultant force, to determine the y value of the resultant from that point; similarly, sum the moments of the y components of the forces about that same point, and divide its sum by the y component of the resultant force, to get the x value of the resultant force from that point.
     
  4. Mar 17, 2008 #3
    Thank you very much for your reaction,

    think I get the idea, just have a few questions to be sure.
    By 'the summed torque value', I take it you mean splitting each torque into it's x and y components, and adding those components up. In my view it would make little sense summing torques up directly
    Then you tell that I should devide the summed torque value by the resultant force value. As you did mention at the end, I take it that also here you mean deviding the summed x/y components of the torques by the summed x/y components of the resultant force. I could ofcourse devide the total forces, but that would not take angles into account and wouldn't provide me with a vector.
    This takes me to my last question, you say this result gives me the perpendicular distance from the point I chose to calculate torques from (let's say the center of mass), but there is no edge to a point so also nothing perpendicular. I guess here you mean that the point the resultant force acts upon, is the the oposite point the vector I got by the devision leads to, when it starts at my chosen point (COM). I hope I made my question clear.

    Thanks in advance
     
    Last edited: Mar 18, 2008
  5. Mar 17, 2008 #4

    PhanthomJay

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    I think I get your question, but i am not sure. When you sum moments (torques) abot any point, the chosen point does not have to be the COM, it can be any point on or off the object. It is sometimes best to choose a point thru which one of the forces act, to simplify calculations, and your calculated distances will be with respect to that point. As far as defining perpendicular distance , the moment of a force about a point is the force times the perpendicular distance from the 'line of action' of the force to the point. When you break up each force into its x and y components, you automatically get these perpendicular distances when doing your calcs for the x and y distances from the chosen point, where x = sum of torques of y components of forces divided by y component of resultant force, and where y = sum of torques of x components of forces divided by x component of resultant force.
     
  6. Mar 18, 2008 #5
    Ok thanks again. I've tried to use your method with an example. Just made up a random body with random forces and numbers to test it.
    Now, I think by what you told me I did it correctly up to point 4. I end up with 2 numbers, 1 for the x devision and one for the y devision, but what to do now still don't really get. I've used the point of force A to calculate the torques about, which is a good point since now I have one torque less to calculate. I don't know for sure if I calculated the resultant torque the right way, please explain to me if I didn't.
    Now for the last step, please tell me how I can derive the location based on my results and the location of point A.
    =============================================
    [​IMG]
    =============================================

    Thanks for the effort
     
  7. Mar 18, 2008 #6

    PhanthomJay

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    Step 1 looks good, but you lost me in 2. I don't know what that green line is, and when you say for example "FA = 10N under 160 degrees at 30,40", is that '30,40' the x and y coordinates? Well, to calculate the torques of each X component of the forces about A, you multiply the x component of the Force at B times its y distance to A, and get a number. Then you multiply the x componnet of the force at C times its y distance to A, and get a number. Then algebraically add those 2 numbers together, and divide the result by it by the x componnet of the resultant, to get a resultant y value. Do the same for the y components to get a resultant x value. Then the resultant force , or its force components, passes through that calculated (x,y) point.
     
  8. Mar 18, 2008 #7
    Ok, I think I was basically doing what you said, but in a way too complicated way. If I hadn't rounded down to 4 significant numbers I think results would've been the same.
    Indeed, by 'at 30,40' I meant x,y coordinates. And the green lines are the torque arms so to say, at which the torques are perpendicular. But because of your explaination I realized that in this case I do not need to calculate the actual torques at all for this, just the x and y components. I did calculate them, and translated them back in x and y again, that's where all the sines and cosines come from.

    By your method I get the folowing result:

    B(x): -11,82*30 = -354,6
    C(x): 1,743*5 = 8,715
    resultant y = (-354,6 + 8,715) / -19,48 = 17.76

    B(y): -2,084*5 = -10,42
    C(y): -19,92*50 = -997,5
    resultant x = (-10,42 + -997,5) / 18,58 = 54,25

    These are quite close to my original results. Close enough to make me believe I did the right thing the wrong way.

    But you were saying these coordinates form the actual point where the resultant force acts upon? That would mean this point actually lies outside the region within the original points, even outside the body. Ofcourse that is possible, and if this is right, I'm happy and I thank you. If this is not right, I feel stupid and hope you will point it out, and be happy anyway.

    EDIT: I did some more tests en figured I wrongly swapped the x and y coordinates in the end. x=17,76 and y=54,25 lies on a point one would expect. I think I got the idea.
     
    Last edited: Mar 18, 2008
  9. Mar 18, 2008 #8

    PhanthomJay

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    Looking good, but you're not quite there. Your 'resultant y' value of 17.76 looks correct (you didn't swap them as you have indicated in your edit. The resultant force will pass thru a point located 17,76 units below point A). But you have an error in the calculation of the 'resultant x' value. The y component of the force at B produces a clockwise torque about point A. The y component of force C produces a counterclockwise moment about point A. That implies that you don't add the torques, you must subtract them: 'resultant x ' = (-10,42 + 997,5)/18,58 = 53,12 units to the left of point A. Thus the resultant force acts through a coordinate point ( -23,12 , 57,76), which lies off the body, but that's OK. (I'd be happier if you chose easier numbers to work with, and if you used decimal points instead of commas! My apologies for not explaining the clockwise/counrterclockwise stuff earlier). You've done quite well thus far, this can get confusing. Happy?
     
  10. Mar 19, 2008 #9
    Thanks, that clears something up.
    As for the clockwise/counter clockwise thing, the same would go for the x components then. The x component of force B causes a clockwise torque, while the x component of force C a counter clockwise one. I see that I accidentally calculated that the right way, but I should not work with positive or negative torque components, rather with cw or ccw ones to decide weather to add or substract them (in practice this would still be positive or negative, but in a different context than ordinary forces).
    Sorry for the hard numbers to work with, I just wanted a realistic example, so easier numbers that could turn out to be the same for different components couldn't be confused.
    Well, I think this is it then. I'll do some more tests with it and I'm happy.
    I'll post again if I've got questions, but for now, thanks a lot for your help :biggrin:

    EDIT: after some tests I figured out that all this is not working when the resultant force = 0 for both x and y. So in that case you have to calculate the total torque about the axis of rotation, pick a point yourself not being that axis and apply a force perpendicular to the line from the axis to your point to apply this torque with one combined force. That would be for example 10 units to the right of the axis, and torque/10 N upwards.
    I got this all working nicely now in Java.
     
    Last edited: Mar 19, 2008
  11. Mar 19, 2008 #10

    PhanthomJay

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    Nice observation. When the resultant force (or its components) is 0, you end up having to divide by 0 in the equations, which is forbidden. When the resultant force is 0, there is no resultant force, but there may be a resultant torque (called a couple). For example, consider a horizontal bar 10 units long, with a vertical force of 100 applied downward at the right end, and with a vertical force of 100 applied upward at the left end. The resultant force is 0. The resultant torque, however, about the left end, is (100)(10) = 1000 . So for a rigid body, you can pick any point where this resultant "couple" of 1000 may be applied, but there is no resultant force acting through this arbitrarily chosen point.
     
  12. Mar 21, 2008 #11
    Yes, or I could make it a bit easier so I can use my existing algorythm. If I detect the resultant force is 0, I just take one force, and calculate the components acting on rotational and translational acceleration on the rigid body. Then I move the point the force acts on to twice the original distance from the axis of rotation. Then I calculate the new force by using the same translational force, and half the rotational force, which would make for the exact same result. Now the resultant force will not be 0 anymore, since the rotational component of the force has been halved, and I can use the original method again.
     
  13. Mar 21, 2008 #12

    PhanthomJay

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    I don't follow. If the resultant force is zero, it is what it is....zero. How can the resultant force not be zero if it is zero? For the simple example I gave, using units of Newtons and meters, there is one resultant solution and one only....the two forces of 100N up and 100N down are equivalent to a couple of 1000 N.m applied anywhere on the rod. There is no resultant force.....hence there is no translation, only rotation.
     
  14. Mar 21, 2008 #13
    That is not correct. Any force will ultimately be split in the part that acts along the radius (axis of rotation to point of force), the translation, and perpendicular to this radius, which when multiplied by the radius creates the torque. If the resultant force is 0, there will neither be torque nor translation. Still, if the situation is as you scetched it, there obviously will be torque. 100*10 about the left point indeed.

    What I wanted, was to create a situation where the effect of both forces could still be captured in one resultant force. And what happens if I double the distance from point A to point B, and halve the force of B. There would be a torque about point A of 20*50 which just like 100*10 equals 1000, and still results in no translational force. This means the exact same result, but now with a resultant force of 50 N upwards, by which the point of effect can also be calculated. That is what I meant.

    I made working code of this principle and it works fine. I encountered another exeption though, which is when 2 forces are equal but point straight away from or towards eachother. This also results in 0 N resultant force, but also 0 torque. Since that's handeled, I think I got an air tight algorythm for this.
     
    Last edited: Mar 21, 2008
  15. Mar 21, 2008 #14

    PhanthomJay

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    That is wrong. See my example. The resultant force is zero. But there is a resultant torque.
    Correct. About any point, actually.
    You should try out your alogarithm using this simple example. If it yields a resultant ficticious force, it is wrong.
     
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