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Forces on a supported beam

  1. Dec 8, 2011 #1
    1. The problem statement, all variables and given/known data

    A 120 kg beam is 12.0 m long and has 2 supports, each of which are placed 2.0 m from each end (leaving 8.0 m of the beam between the 2 supports). A 50 kg student stands at the extreme left end of the beam, 2.0 m left of the left support. A second student stands 1.0 m from the right end which is 1.0 m to the right of the right support.

    a. How much upward force does the right support exert on the beam?

    b. How much upward force does the left support exert on the beam?


    2. Relevant equations

    F= ma
    Torque?

    3. The attempt at a solution

    I found that each support holds 588 N due to the weighted rod but im not sure how to account for the people and their distances from the supports.
     
  2. jcsd
  3. Dec 8, 2011 #2
    You need your definition of torque and you need to decide if the system is moving. If it is or it isn't, what information does this give you?

    In other words what is the acceleration of the beam. You also need Newton's second law of rotational motion and you need to decide if the beam is rotating or not and what information that gives you.
     
  4. Dec 9, 2011 #3
    T=dF, system is not moving, im not sure what information this gives me, acceleration of the beam is 9.8 due to gravity?
     
  5. Dec 9, 2011 #4
    Wikipedia has a definition of torque. Look it up. And yes it's a formula
    The beam is resting on supports. Acceleration requires a change in velocity. If the beam is resting, is it moving? No. At rest means v = zero, change in v = zero since it stays at rest or motionless so it has no rotational motion either.

    You need newton's second law of rotational motion as well as linear motion which you have already given which is F = ma so you are on the right track.
    You just need to draw your forces and distances and use the two formulas for newton's second law - one with forces, the other with torques.
     
  6. Dec 9, 2011 #5
    Newtons second law says that the force on the pivot is the distance from the pivot * F. Therefore does that mean on the left support it would be 2 * 50 * 9.8 = 980N on left support? but there is no weight given for the second student?
     
  7. Dec 9, 2011 #6
    I think this is a question about a beam in equilibrium and the weight of the second student is not known.
    If this is the case then there are 2 conditions that must be satisfied for equilibrium
    1) There is no resultant vertical or horizontal force. This means that the sum of the forces (F1 and F2) at the supports must equal the total weight of the 2 students plus the beam itself.

    2) There is no resultant turning effect/moment/torque (I believe that moment is the preferred term in equilibrium examples.
    Moment = force x perpendicular distance from pivot.
    Take moments (clockwise and anti-clockwise) about one of the supports.
    Clockwise moment = anticlockwise moment
    These equations should enable you to solve for the mass (weight) of the second student and find the rest of the information requested.
     
  8. Dec 9, 2011 #7
    In other words torque is the force applied times its perpendicular distance from your chosen axis of rotation. Since it's not really rotating you can choose the rotaional point anywhere convenient, like at the point where the person of unknown mass is located. If you sum all your torques about that point, since the acceleration is zero, that sum will be zero. do the same for just the forces and you have two equations.
     
  9. Dec 9, 2011 #8
    Moment= 490 N * 2m = 980N on first support Moment on second support = Mass * 9.8 * 1m ?? how do i solve for mass of second student?
     
  10. Dec 9, 2011 #9
    So from the point of the person of unknown mass, torque on closest support is M*9.8*1m, and the torque on the furthest support is M*9.8*9m?
     
  11. Dec 9, 2011 #10
    It is absolutely true.... you can take any point you want and (I say 'take moments' !!!!)
    get expressions for the torques. There is zero resultant torque.
     
  12. Dec 9, 2011 #11
    88.2M + 9.8M + 9.8*120kg = Support1 + Support2?
     
  13. Dec 9, 2011 #12
    There are 3 unknowns...F1, F2 and the weight W of the second student.
    So you will need 3 equations to get a solution. The usual thing is to take moments about each of the unknowns to eliminate it. Take moments about F1 will give you one equation, moments about F2 will give you a second equation and moments about W (the second student) will give you the third equation.... Messy but should be able to solve these.
     
  14. Dec 9, 2011 #13
    wtf is a moment??? im so confused
     
  15. Dec 9, 2011 #14

    PhanthomJay

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    The problem cannot be solved without knowing the mass or weight of the 2nd student! Check the problem to see if it is given.
     
  16. Dec 9, 2011 #15
    jjd:
    I am not sure how much you know about 'moments' (torque is more or less the same thing)
    I started this question by drawing a diagram like this :
    Draw a horizontal line to represent the 12m plank. At the mid-point draw a vertical down arrow attached to the plank and label it 1200N ( I am going to take g = 10 so that I can do calculations quickly in my head, you will need to use 9.81 to get a more exact answer).
    At a point that looks like 2m on your plank line draw 2 supports.... 1 at each end....2m from the end.
    At each of these supports draw a vertical upwards arrow and label them F1 and F2 (I have F1 on the left) These are the forces at the supports on the plank.
    On the left hand end of the plank draw a vertical down arrow labelled 500N... this is the first student
    On the right hand end of the plank draw a vertical downward arrow what looks like 1m in from the end (1/2 way between end of plank and F2) label this W (this is the weight of the second student.
    You should have a diagram with 3 down arrows (500N, 1200N and W Newtons)
    and 2 up arrows F1 and F2.
    Now you need to do that magic thing .... take moments.
    Hope this helps.
     
  17. Dec 9, 2011 #16
    i already had that diagram drawn out, what i am having trouble with is taking these "moments" not sure how to do that
     
  18. Dec 9, 2011 #17
    I checked the problem, it is definately not given
     
  19. Dec 9, 2011 #18
    I will show you how I started BUT I have a feeling that Phantomjay is correct although I do not know why?????
    Here goes on how to take moments....(bear in mind I am using g = 10)
    Moments about the F1 support
    Clockwise (1200x4) + (Wx9) = 4800 +9W
    Anticlockwise (500x2) +(F2x8) = 1000 + 8F2
    Clockwise = anticlockwise so....
    4800 +9W = 1000 + 8F2
    3800 = 8F2 - 9W.........equ1
    Moments about F2 gave me
    9800 = 8F1 + W ..........equ2
    Moments about W gave me
    11500 = 9F1 + F2 ..........equ3
    There is no resultant force so:
    F1 + F2 = 1200 + 500 + W
    F1 + F2 = 1700 + W ........equ4

    I thought that with 4 equations relating 3 unknowns I would be able to solve this.... BUT I have not been able to and I dont know why !!!!!!
    Your problem is not solved but at least I hope you can see a technique for solving this sort of problem
     
  20. Dec 9, 2011 #19
    moment i.e.torque
    you can actually run 4 equations by putting your axis of rotation at each of the three unknowns and summing the torques and then using newton's second law for linear motion also. If there is a missing piece of info it will show up in the solution of the four equations.

    example. Put the axis of rotation at upper support 1 on the left

    You then have -50kg (g)(2m) + 120kg(4m) - Fn'(8m) + m(person)(g)(9m) = 0

    Repeat summing the torques with axis of rotation at Fn and again at m(person)(g) on the right)

    Then sum the total forces on the system and try to solve from there.

    Watch your signs. Designate counter clockwise torques negative and clockwise as positive.

    I did get a solution but could have made a calculation error. I used the torques around the two upper supports to solve for each of them in terms of the the unknown person's weight. From there I put it back into the linear equation to get the unknown person's weight. Then just substituted to get the upper supports.

    do you have the correct answer for it??
     
    Last edited: Dec 9, 2011
  21. Dec 9, 2011 #20
    Phanthomjay.... any input please
     
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