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Really simple Moment

  1. Dec 28, 2007 #1
    [SOLVED] Really simple Moment

    Edit: So I have solved this, but still need some help, skip to post #4 or#5 where I have questions concerning cross-product. I would love to talk about some more questions about cross product too if I get any responses. Thanks!






    I just need a check. My answer seems logical to me, but it does not match the solution. I am supposed to find the moment of the force at A about P.
    [​IMG]

    Now I am using the component r that is perpendicular to a component of F.

    I tried (4sin30+6)(12/13)(520)=3840 but the solution is 3150 N*m.

    Am I correct or is my technique flawed?
     
    Last edited: Dec 29, 2007
  2. jcsd
  3. Dec 28, 2007 #2

    D H

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    Your technique is flawed. The correct answer is 3147 Nm, or 3150 Nm to three places. You are using the component of r that is perpendicular to a component of F, not the component of r that is perpendicular to F.
     
  4. Dec 28, 2007 #3
    Gotcha. I just thought this problem would be way easier than it is. I had to use a lot of trig to find r and the component of F perp to r.

    I got it....but it just feels like there may have been an easier way. I'll scan my work in when I get home. Maybe someone will see a more efficient way.
     
  5. Dec 28, 2007 #4

    D H

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    Trig? Yech. Use the vector cross product:

    [tex]\vec P = -4(\sin30^{\circ}\hat x + \cos30^{\circ}\hat y)
    = -2 \hat x + 2 \surd3 \hat y[/tex]

    [tex] \vec r = 6\hat x - \vec P
    = 8 \hat x - 2 \surd3 \hat y[/tex]

    [tex]\vec F = \frac {520}{13} (-5\hat x + 12 \hat y)[/tex]

    [tex]\vec\tau = \vec r \times \vec F
    = \frac {520}{13}(8*12-(-2\surd3)*(-5))\hat z
    = \frac {520}{13}(96-10\surd 3) \hat z\approx 3147\hat z[/tex]
     
  6. Dec 28, 2007 #5
    I have only used cross product a few times. What is [itex]\vec P[/itex] ? I am assuming it is the position vector of point P.

    Now [itex]\vec r[/itex] is from P to A. I think I see now. so [itex]\vec r=\vec P+\vec A=-4(\sin30\hat x+\cos30\hat y)+6\hat x[/itex]

    Then you wrote F as a Cartesian vector by similar triangles and used [itex]\vec r \times \vec F[/itex]
     
    Last edited: Dec 28, 2007
  7. Dec 28, 2007 #6
    So if I want to use the determinant, I think that's what it is called, how would I apply it?

    I know:
    [tex]\vec r \times \vec F=\left[\begin{array}{ccc}i & j & k \\ r_x & r_y & r_z \\ F_x & F_y & F_z \end{array}\right][/tex]

    Now since their are no z components of either original vectors. Do treat that column as 0? or like they weren't there?

    That is, how do I expand this determinant?
     
    Last edited: Dec 28, 2007
  8. Dec 28, 2007 #7
    Or is this just now a 2 X 2 determinant where [itex]\vec r \times \vec F=(r_XF_y-r_YF_x)[/itex] ?

    Also, why did D H put a [itex] \hat z[/itex] in the solution? What does that imply?
     
    Last edited: Dec 28, 2007
  9. Dec 29, 2007 #8
    Is the [itex]\hat z[/itex] in D H's solution because the resultant vector is in the direction perpendicular to the x-y plane in which [itex]\vec r[/itex] and [itex]\vec F[/itex] lie?
     
  10. Dec 29, 2007 #9

    PhanthomJay

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    If you don't like cross products, you might want to use the alternate (equivalent) definition of moment: Moment = force times the perpendicular distance from the line of action of the force to the point you're taking the moment about, and clockwise is plus, counterclockwise is minus, or vice versa. It's simpler to break up the force into its x and y components, then sum the torques of each component torque. In your first attempt, you corectly got the torque of the y component; now just get the torque of the x component, and sum them algebraically.
     
  11. Dec 29, 2007 #10

    I have solved this problem using trig to find the component of F perp to r. It was a pain in the arse.

    I did not mean to imply that I dislike cross-product, rather that I want to learn how to expand the 2X2 determinant properly i.e., is this correct [itex]\vec r \times \vec F=(r_XF_y-r_YF_x)[/itex] ?

    Also P-Jay, are you implying Varignon's theorem here?
     
  12. Dec 29, 2007 #11

    PhanthomJay

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    Oh, sorry, I thought you were looking for another 'simpler' way, but what's simple to some is complex to others.' I'm not much into names, determinants, and cross products, so I can't help much more here.
     
  13. Dec 29, 2007 #12
    Well let me ask you this. Varignon's theorem just says that the resultant moment of a bunch of forces or to make it simple....just one force you can take the sum of the moments of the components. That is to say given some force F and some distance r I can say that [itex]M=r_xF_y+r_yF_x[/itex]

    Now I know to figure the signs out I use the right-hand rule for the torques. But here is a question: If the torque spins counterclockwise thus being positive,but r lies on the negative x axis.....what is the sign? To I take -rF or +rF?
     
  14. Dec 30, 2007 #13

    PhanthomJay

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    If the torque spins counterclockwise and you call it positive, that's fine, as long as if it spins clockwise, you call it negative. It doesn't matter that r may lie on the negative axis...the sign is determined from the clockwise or counterclockwise sense. You can also call counterclockwise negative if you call clockwise as positive, it doesn't matter at all as long as you are consistent in your convention.
     
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