Thursday, December 15, 2011

Statics: Still Hanging On... ...



We’ve been studying statics in our physics class. We have examined how to calculate the forces in ropes and cables that support hanging objects. Up to this point, we’ve only considered the cases where the cables are symmetric (make the same angle with the horizontal).
Consider (the somewhat unorthodox) case where a sign is hung by two cables that *are  not* symmetric—here, the cables each make different angles with the horizontal. Let’s do some work to calculate the force in each cable:
  1. Suppose the sign’s mass is 5 kg. What is the weight of the sign?
  2. Call the tension in the left cable FL and the tension in the right cable FR. Write an expression for FLx, FLy, FRx, and FRy.
  3. Write an equation relating the forces in the vertical direction. (HINT: Think balanced.)
  4. Write an equation relating the forces in the horizontal direction. (HINT: Think balanced.)
  5. Notice you should now have two equations with two unknowns FL and FR. Solve the resulting system to find the tension in each cable.
Posted by Brian Bearss at 7:52 AM

10 comments:

  1. NessDecember 15, 2011 at 11:48 AM

    1. The weight of the sign is 49 N.
    2. Left cable : Fx = F cos 35, Fy = F sin 35
    Right cable : Fx = F cos 55, Fy = F sin 55

    ReplyDelete
  2. ilovesportsDecember 17, 2011 at 2:22 PM

    If the sign is not moving, the vertical forces must be balanced and the horizontal forces must also be balanced.
    The equation for balanced vertical forces is:
    (F*sin 35) + (F*sin 55) = 49N
    The equation for balanced horizontal forces is:
    F*cos 35 = F*cos 55

    ReplyDelete
  3. AnonymousDecember 18, 2011 at 9:53 PM

    1. the weight of the sign is 49N.
    2. left cable: Fx= F*cos(35). Fy= F*sin(35)
    right cable: Fx= F*cos(55). Fy= F*sin(55)
    the vertical and horizontal forces should both be balanced.

    Mason Smeznik

    ReplyDelete
  4. TFDecember 19, 2011 at 7:33 AM

    1. 49N 2. Flx=Fl*cos35 Fly=Fl*sin35 Frx=Fr*cos55 Fry=Fr*sin55 3. Fup=Fdown Fl*sin35+Fr*sin55=49 4. Fleft=Fright Flcos35=Frcos55

    ReplyDelete
  5. Useful_idiotDecember 19, 2011 at 7:55 AM

    First off, the weight would be 49N. To find the tension in each wire fist find how much the vertical weight would be if the wires were going straight up,(which would be 24.5N in each wire)and make a triangle with the angle and the tension wire. If you divide 24.5 by the sin of each angle,(in the left side 24.5/sin 35, and the right side 24.5/sin 55) this will give u the tensoin force in each wire.

    ReplyDelete
  6. UnknownDecember 19, 2011 at 7:55 AM

    The weight is 49N
    Right Cable Fx=F*cos55, Fy=sin55
    Left Cable Fx=cos35, Fy=Sin35
    fsin35+fsin55=49 (Left) fcos35=fcos55+49 (Right)

    ReplyDelete
  7. iloveweekendsDecember 19, 2011 at 8:00 AM

    1. 49N 2. Flx=Fl*cos35 Fly=Fl*sin35 Frx=Fr*cos55 Fry=Fr*sin55 3. Fup=Fdown Fl*sin35+Fr*sin55=49 4. Fleft=Fright Flcos35=Frcos55

    ReplyDelete
  8. Brian BearssDecember 19, 2011 at 8:05 AM

    Thanks for the comments this week. It looks like most of you have (incorrectly) assumed that the tension must be equal in both wires. You will need to have both FL and FR in your equations--you can't just use F because the left and right forces are different.

    ReplyDelete
  9. Brian BearssDecember 19, 2011 at 8:37 AM

    1. Fg = mg = 5 * 9.8 = 49

    2. FLx = FL * cos(35)
    FLy = FL * sin(35)
    FRx = FR * cos(55)
    FRy = FR * sin(55)

    3. Fup = Fdown
    FLy + FRy = Fg
    ( FL * sin(35) ) + ( FR * sin(55) ) = 49

    4. Fleft = Fright
    ( FL * cos(35) ) = ( FR * cos(55) )

    5. from equation 4, notice that:
    FL = (FR*cos(55))/(cos(35))
    substitute this in to equation 3:
    ( (FR*cos(55))/(cos(35)) * sin(35) ) + ( FR*sin(55) ) = 49
    1.22077*FR = 49
    FR = 40.1

    Plugging back in to find FL:
    FL = 28.1

    So, the tension in the right cable is about 40 N, and the left cable is about 28 N.

    ReplyDelete
  10. Brian BearssDecember 19, 2011 at 8:38 AM

    Notice how the tensions in the cables are *not* equal, but that the net force on the sign is still zero.

    ReplyDelete

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