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Mechanics-principle of transmissibility,resolution of force into couple and force

  1. Jan 22, 2010 #1
    I have been doing mechanics in college and in the present chapter,two of the concepts that I found rather confusing are principle of transmissibility and resolution of force into couple and force.

    1. In resolution of a force,acting,say at a point A on a body,suppose we have to resolve the force into a force and couple acting at another point,say O on the body,the translational force is simply transferred to the point O---how is this possible?

    I thought that a force acting at a point is not equivalent to the same force acting at another arbitary point---for one,the torque will be different.

    Besides,if a force is applied at a particular point on a body,it can only be transferred to the centre of mass...we can't simply change the location of the force to any other point.

    Besides,please could someone tell me intuitively what we are actually doing in resolution of a force into an equivalent force and couple?

    2. In principle of transmissibility,it again says that applying a force at any point on the line of action is equivalent to the original force applied at the original point.

    Perhaps this is all confusing me,since till now,I thought that any force applied on a body can be considered to be acting only at the centre of mass.

    Thanks in advance for any help.
     
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  3. Jan 22, 2010 #2

    tiny-tim

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    Hi Urmi Roy! :smile:
    Intuitively, you're adding an equal-and-opposite pair of forces at the same point.

    So a single force FA at point A is the same as:
    FA at A plus FO at O plus -FO at O,​
    and if we choose FA and FO to be parallel and equal in magnitude, then FA and -FO is a torque, giving
    FA = FO + torque :wink:
    Yes, if the line AO is the line of FA, then FA and -FO cancel, leaving FA = FO.
    No, if the line of a force doesn't go through the centre of mass of a rigid body, then the body will start turning.
     
  4. Jan 23, 2010 #3
    Wow! Thanks tiny-tim, those were really good explanations to my doubts!! :)
     
  5. Jan 24, 2010 #4
    Hi,
    I was just doing some sums,and while doing them I found some points that I'm still hazy about...

    1. Suppose there is a system of forces and moments acting in a particular region...in the problem,we have to resolve all the forces into a single force and a moment.....now,as far as the forces are concerned,I know that I have to find the resultant of all the forces and then follow what tiny-tim said...but in this particular sum, there is a moment (represesnted as a vector arrow) acting at a point A......how do I find the moment of that moment about an arbitrary point O ( I don't know what force is causing the moment-all I know is its magnitude,direction and the point which it is acting)?

    2.In my book it says that the main difference between the moment of a force and a couple is that the moment is a sliding vector and the couple is a free vector...what does that mean? (I tried to find out by googling but didn't really succeed.)

    3. What is the basic point that differentiates stable,unstable and neutral equilibrium from eachother..(I found lots of different explanations in different books,but there must be one particular criteria)?
     
  6. Jan 24, 2010 #5

    tiny-tim

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    Hi Urmi Roy! :smile:
    What book did you get this from?

    I'm guessing that they mean that a couple is the same about any point, and in that sense a couple is "free".

    to see this, suppose you have a couple about a point A, and you choose another point B … we'll keep to two dimensions …

    draw two equal circles round A and B, and the common tangents meeting them in LA LB RA and RB (L and R for left and right :wink:) …

    then we can represent the couple as two parallel equal and opposite forces FL and FR through LA and RA

    now draw at point LB a two equal and opposite forces GL and -GL, with GL equal to FL (so they both point along the same line).

    Then FL and -GL are in line, and equal and opposite, so they completely cancel: FL + -GL = 0.

    And also GL and -GL are in line, and equal and opposite, so they completely cancel: GL + -GL = 0.

    Similarly, draw forces GR and -GR at RB.

    Then also FR + -GR = 0, and GL + -GR = 0.

    So FL + FR

    = FL + GL + -GL

    + FR + GR + -GR

    = GL + GR.

    ie the couple made of the parallel equal and opposite forces FL and FR about A

    is the same as the couple made of the parallel equal and opposite forces GL and GR about B. :smile:
    I don't understand this … do you mean a couple? :confused:

    If not, can you explain what the situation is?

    "equilibrium" means that if you do nothing, it will stay there.

    "stable equilibrium" means that if you do nothing, it will stay there, and if you do something small, it will stay near there.

    For example, a ball inside a sphere and at the bottom will go backward and forward near the bottom if you push it slightly.

    "unstable equilibrium" means that if you do nothing, it will stay there, but if you do something small, it will move away.

    For example, a ball outside a sphere and at the top will slide down the sphere if you push it slightly.

    I'm not sure what "neutral equilibrium" means :redface:, but I'll guess it means like a ball on a flat horizontal surface.
     
  7. Jan 25, 2010 #6
    In the sum, there is a digram,with different forces,represented by vectors acting at different points within space. One vector in the diagram does not represent force,but it represents moment about the point that it is drawn at. Now, I have to find out the resultant force and couple (as we did in resolution of a force into a force and moment) at a given point O (which has certain coordinates) of all these forces and this moment.....but how can I find the component of the given moment (the one that is represented by the vector arrow in the diagram) about O? Does it make sense to find this component at all? The thing is,we don't even know what forces are causing this moment!

    Thanks for your explanation of the first question,though I'll have to go through it a couple of times before I completely understand...it's a little tricky.
     
  8. Jan 26, 2010 #7

    tiny-tim

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    Yes, that'll be a couple. The couple, of course, turns about the line of that vector … it's not a force that you have to take a moment of about some point.

    (A couple is the same about any point, so it doesn't actually matter where they put the arrow.)

    And you don't need to know what forces it's made of … all the information is combined into the one vector. :smile:
     
  9. Jan 27, 2010 #8
    I didn't understand that....suppose we have a line,and we take it as axis.....we have two opposite forces,causing a couple about that line......how can the couple have any effect,let alone have any component about a point which isn't even on the axis(that's what the situation is in the sum,isn't it?)?

    I seem to be in a mess!! Help!
     
  10. Jan 27, 2010 #9
    I was just doing more sums...and I think I found something about what tiny-tim was saying in his last post...it says,'couple at any arbitraty point = moment about the arbitrary point'....but I still don't understand it!
     
  11. Jan 27, 2010 #10

    tiny-tim

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    Think of a couple as being a rotating circular brush

    you push it against something, and the friction makes that thing turn.

    For a slightly simpler example, suppose you have a horizontal disc free to rotate on a fixed vertical axle.

    Place two vertical wheels, with rubber rims, on top of the disc, a fixed distance d apart, and rotate these wheels at the same speed but in opposite directions.

    Obviously, the rubber will cause equal and opposite friction forces, F, from each wheel, and will be a couple of magnitude Fd about the midpoint.

    But that will make the disc turn, because, wherever you place the midpoint, and whichever direction the wheels face, so long as they are at that fixed distance d from each other, the moment about the axle will be the same, Fd.

    Why? :rolleyes: Because if the perpendicular distance from the line of one wheel to the centre is x, then the perpendicular distance from the line of the other wheel will be x + d, and so the total moment about the axle is F(x+d) - Fx = Fd. :smile:

    In other words: a couple Fd about any point has the same effect as a couple Fd about the axle (and therefore any other point). :wink:

    The effect of a particular couple is the same about any point.​
     
  12. Jan 30, 2010 #11
    Wow tiny-tim,you're really good at explaining things..I think I got it now. Thanks for your infinite patience in answering my questions!
     
  13. Feb 8, 2010 #12
    Hi,I'm back with some more difficulties...(just small ones this time.)

    1. In a sum that I was doing,there were 2 forces acting on an object...one was Fa at an angle 20 degree to the horizontal,and we have to find the direction and magnitude of the least possible value of the other force Fb ,given that their resutant is 20kN along the positive direction of the horizontal axis.
    Obviously,for least value of Fb,it should be the altitude of the right angled force triangle formed by Fa,Fb and R.
    In this sum,is it necessary to assume that the resultant R is the hypotenuse of the right angled force triangle?(If Fb is the altitude,it may be perpendicular to either Fa or R,so how do we know which case to take)

    2.If a structural member does has negligible mass,isn't finding the forces and moments acting on it meaningless....force on the member =ma...but m=0,so it doesn't make sense!

    3. I thought that there was nothing such as 'moment about a point',since even when you're claculating moments about a point,you can always consider an axis through the point,so it's always 'moment about an axis' that matters...is this right?

    4.In another sum that I was doing,there is a rod,of given dimensions,standing vertically on the ground (touching the ground at point A)..there's another rod which is fixed on this vertical rod in such a way that it is horizontal w.r.t the ground...like a half T.....there is a weight at the free end of the horizontal rod....this weight obviously has a certain moment about point A on the ground.
    In the sum,we are asked to calculate the reactive force and moment at A.

    I understand reactive force at A,but when we are considering all the forces w.r.t A,we can't have a reactive moment at A w.r.t itself!
    Am I getting it wrong somewhere?
     
  14. Feb 8, 2010 #13

    tiny-tim

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    Hi! :smile:
    "the altitude of the right angled force triangle formed by Fa,Fb and R" is a very bad way of putting it, because,as you say, it's ambiguous.

    If your book says this, then your book is wrong to do so. :frown:

    The correct description is "the distance (or perpendicular distance) from the point R to the line Fa". :smile:
    ah, but "negligible" isn't the same as zero … so long as it's non-zero, F = ma still makes sense! :wink:
    No, you're wrong … there's no such thing such as moment of inertia about a point, but moment about a point is correct.

    Moment of inertia has to be about an axis (though in 2D cases, we often say 'about a point', meaning about the axis through that point perpendicularly out of the page :wink:).

    But moment (of a force F, for example) is defined as r x F, where r is the position vector from (obviously) a point.

    What you're thinking of is the component of the moment in a particular direction …

    in the vector equation torque (= moment of force) = Iα, α (the angular acceleration) is usually about a fixed axis (with unit vector k say), so we just use the components along that direction: τk = Iαk.

    There, τk = τ.k = (rxF).k = (r - µk)xF).k for any number µ, is the component of moment about any point µk on the axis, and so can be unambiguously and correctly called the moment about that axis.

    But always remember that the definition of moment starts "r x", and so is fundamentally about a point. :smile:
    The reactive force is equal and opposite to the resultant (total) force on the system.

    Similarly, there is a resultant moment (the vector sum of all the individual moments) on the system, and since the system is in equilibrium, that must be balanced by a reactive moment somewhere …

    there is a twisting force at A …

    for example, if you use a spanner (wrench) to try to turn a nut that won't move :yuck:, you exert a torque on the nut, but the nut also exerts a reactive torque on you, doesn't it? :smile:
     
    Last edited: Feb 8, 2010
  15. Feb 9, 2010 #14
    ...okay, then what should I do about this sum? Will there be two cases to consider...considering the rwo different possible orientations of Fb?


    I'm sorry,this point is still not clear.

    We take the reactive torque of the nut only when we are considering the moments about the point of application of the external force (at the end of the spanner)...but in this sum,we are taking all the forces and torques with respect to the point on the ground (A).

    Thanks tiny-tim.
     
  16. Mar 25, 2011 #15
    Hi Tiny Tim,

    Now I'm doing FIXED BEAMS in college and the explanation in our text seems to be related to your post on how the moment about any point is the same around another point....but I'm not sure, so I was wondering if you could help me agian....

    1. I've attached a diagram, and you'll notice that the fixed beam is condidered equivalent to a Simply supported beam (SSB) with a point load at its mid point(and thus having reactions Ra* and Rb* at its ends) and another beam with no external load (as the end reactions are both R in magnitude but opposite in direction), but the forces R on this beam are what cause the end moments Ma and Mb......Now, I know that the function of the end moments Ma and Mb is to neutralise the deflection produced on the beam and thus their directions are shown as in the diagram...

    ...the problem is that from the direction of the forces 'R' at either end of the beam (which are supposed to cause the end moments), the direction of the moments Ma and Mb do not neutralise the deflection of the SSB......Any idea what might be wrong?(I've picked this up from my book exactly as its given)
     

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  17. Mar 25, 2011 #16

    tiny-tim

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    Hi Urmi Roy! :smile:

    Structural engineering like this isn't really my field :redface:,

    but I think they're saying that if you have a beam between two walls, then generally the load won't be symmetric, so there'll be a couple as well as a vertical load …

    in that case you can regard the couple and the vertical load as completely separate. :wink:
    I think equilibrium is preserved by torsion in the beam created by and balancing the couple.
     
  18. Mar 25, 2011 #17
    But, if you see the diagram, the load is just at the mid point, so it is symmetrical....

    The main point of query is as to how a couple is formed by forces acting at the ends of the beam so that it neutralises the deflection at the ends.....perhaps you could look at the diagram and come up with something based on the principles of Physics....?
    Please try once more....:(
     
  19. Mar 25, 2011 #18
    I've been thinking about this...and I was wondering, if we could resolve the moment due to the forces 'R' acting at the ends of the beam into two equal moments (as shown in the diagram attached) by introducing two forces +R and -R at the centre of the beam, then the couple in purple would serve to increase the deflection of the beam already present (due to the load at the centre)...whereas at A(on the left), the couple in pink would serve to decrease the deflection of the beam....(imagine that there's a point load P at the centre of the beam that causes some deflection, causing a U shape in the beam).....but that still doesn't solve the problem :-(
     

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  20. Mar 25, 2011 #19

    tiny-tim

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    Sorry, but as I said earlier, structural mechanics really isn't my field (and this wasn't the original topic of the thread) …

    you'll have to ask someone else (maybe start a new thread). :redface:
     
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