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Simple Torque Problem With Picture, Help!

  1. Oct 4, 2008 #1
    1. The problem statement, all variables and given/known data

    [​IMG]

    2. Relevant equations

    Torque(n/m)=Force(N)*Length(L)
    Basic Trigonometry

    3. The attempt at a solution

    The concept of torque is new to me, so I just need a push in the right direction here. lbs and ft are used so we are using non-metric? Should I convert them to metric? How do lbs convert to newtons? One pound-force is 4.448222N right? Am I going in the right direction?

    T=LF

    Length=10ft
    Force=250lbs

    We can't just do LF without some trig, and this where I don't know where to start.
     
  2. jcsd
  3. Oct 4, 2008 #2

    tiny-tim

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    Hi raven2783! :smile:

    Why are you worrying about the units? If the question is in lbs, then you answer in lbs! :rolleyes:

    Torque (moment of force) = force times perpendicular distance.

    For that, you will need to find the angle between the force and PA.

    Just work out the angle of that 3,4,5 triangle, and combine it appropriately with the given angle. :smile:
     
  4. Oct 4, 2008 #3
    I know how to do all the math, it is the conceptual part I don't have... can someone explain to me the picture in terms of the question??
     
  5. Oct 4, 2008 #4
    What is PA? Point angle, I'm so confused.
     
  6. Oct 5, 2008 #5

    tiny-tim

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    PA is the line from P to A on the diagram.

    You need to multiply the force (250 lbs) by the distance PB, where B is the foot of the perpendicular from P to the line of force.

    So you need to use trig to find the angle PAB, and then use that to find the distance PB. :smile:
     
  7. Oct 5, 2008 #6
    So AB would be straight down from initial point A? And what is the 3,4,5 triangle telling me? The proportions of PAB? If so then AB would be the 3 and PA would be the 5?

    What if I just did cos(30)*10, wouldn't that give me PB? Then I could do 250*PB. But I need the perpendicular don't I...so

    sin(30)10 will give me the perpendicular to the x-axis AB then I can do

    arccos(sin(30)10/10) to get PAB. Then Law of sines

    sin(30)10/30deg=PB/PAB then to get the answer...

    PB*250.. right?
     
  8. Oct 5, 2008 #7

    tiny-tim

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    Hi raven2783! :smile:
    Nooo … that B is the foot of the perpendicular from P to the y-axis.

    I said the foot of the perpendicular from the line of the force to P to the line of the force.

    (I couldn't give it letters because there isn't a letter on the diagram for the endpoint).

    The reason is that the torque of the force (about P) is defined as the force times the perpendicular distance from P … that is, times PB. :wink:
    No … it tells you the angles between the line of force and the axes …

    which you need to find the angle PAB, which you need to find the distance PB.

    Try again! :smile:
     
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