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Statics Problem, Simple 2d rectangle with forces acting on it.

  1. Sep 26, 2009 #1
    1. The problem statement, all variables and given/known data
    A rectangular plate is acted upon by the force and couple shown. This system is to be replaced with a single equivalent force.
    a) for a(alpha)=40 degrees, specify the magnitude and the line of action of the equivalent force.
    b)Specify the value of a(alpha) if the line of action of the equivalent force is to intersect line CD 300mm to the right of D.

    2. Relevant equations

    http://img188.imageshack.us/img188/3503/0926091619.th.jpg [Broken]

    3. The attempt at a solution
    Ok I've found the equivalent force-couple system at C which is -48N j, and +6.997 N.m. From there I'm lost, I don't know what to do. I've tried to use the cross product(CBxF) to find Moment at C then add up the moment... Am I going in the right direction?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 27, 2009 #2

    nvn

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    Hint 1: What should be the value of the vertical force component for the equivalent force system? Hint 2: What should be the value of the horizontal force component for the equivalent force system? Hint 3: Now where will you place this equivalent force?
     
  4. Sep 30, 2009 #3
    But, this is exactly what I don't get. Do I add force components from the force and the couple force together? Wouldn't the couple forces just cancel out, so you'd be left with 48 n?
    Any other hints would help.
     
  5. Oct 1, 2009 #4

    nvn

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    Yes, that's correct; -48 N. You got the vertical component of the equivalent force system. Keep going. See hints 2 and 3.
     
  6. Oct 1, 2009 #5
    ok, so vertial force would be -48 N and horizontal would be 0 N. Alrighty, Thus the line of action would be on BC. Ok so what about part b). If line of action was 300mm to the right of point d? Wouldn't alpha be the same as it is now, just because it's part of a couple force and it produces no resultant force?? Is this just one big trick question?
    I really do appreciate your help!!
     
  7. Oct 1, 2009 #6

    nvn

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    LP20: The line of action for part (a) is not BC. Try again. Hint 4: The equivalent force system must cause the same moment on the plate as the original force system. See hint 3. For part (b), alpha will not be the same.
     
  8. Oct 1, 2009 #7
    WEll, Tomorrow this assignment is due, but no worries. I would still LOVE to figure this one out, its seems so simple yet its killing me on the inside. So The Magnitude of the equivalent force is -48N. And I thought the line of action is where the force follows the direction its pointing (which would be line BC?). *update* So I found the moment of couple M=fd (did each x and y component separately)... it is 2.313 Nm i + 4.596Nm j. Right? I set it equal to the rectangle with 300mm length (part b) Then I substituted my given length and moments I found into the equation M=fd, Than found out my alpha must be 32.15 degrees... Is this correct?!?! the answer just hit my like a ton of bricks... I hope I'm right!!
     
  9. Oct 2, 2009 #8

    nvn

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    For part (b), after you get your answer of alpha = 32.15 deg, now compute the moment on the plate using the original force system with your alpha. Next, compute the moment on the plate using the equivalent force system. I think you will see the original and equivalent moments do not match, which tells you your answer for alpha is incorrect.

    Whenever you get an answer, you can then recalculate using that answer, to check the answer.
     
  10. Oct 2, 2009 #9
    Alright nvn, I think I know how to do it, I already turned in my assignment today. We actually had a test on moments too. It went very smoothly. But I'll probably be back asking more questions in the future.
    Thanks!
     
  11. Mar 27, 2011 #10
    I know that this topic hasn't been addressed in a while, but I was also looking for the answer, and thought I would share with anyone else who needs help.

    Part A
    We must first find the sum of the forces in each direction, as LP20 noted.

    The couple force of 15N has x-components that can be calculated using the α given (40°).

    ΣFx: -15cos(40°) + 15cos(40°) = 0

    Using the angle given again, we can calculate the y-components of the couple forces, in addition to the 48N force applied at B:

    ΣFy: -15sin(40°) + 15sin(40°) - 48 = -48N --> 48N downward force

    With the given angle of (40°):

    15cos(40°)*(.4) + 15sin(40°)*(.24) - 48(.4) = -48*d --> eqt. 1

    Thus, the 48N downward force is applied at .256 m to the right of point D.

    Part B

    We can use eqt. 1 above to solve for the new angle desired:

    15cos(α)*(.4) + 15sin(α)*(.24) - 48(.4) = -48*(.3)

    To solve this equation, we can use trig identities and the quadratic formula, as we get:

    6cos(α) = 4.8 - 3.6sin(α)

    Square both sides, use trig identities (cos2x = 1 - sin2x), combine like terms, and apply quadratic formula to find α = 77.63°
     
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