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Homework Help: Static Equilibrium of a beam and hinge

  1. Jan 5, 2010 #1
    1. The problem statement, all variables and given/known data

    A 90 N sign hangs on the end of a 55 N beam supported by a wire as shown. The beam is attached to the wall by a hinge. What are the horizontal and vertical componetns of the force on the hinge?

    Notes:
    the sign is 1.7 m from the wall
    the CG of the beam is located .75 m from the wall and .95 m from the sign
    at the CG of the beam the wire is attached with the other end attached to the wall
    the angle that the beam makes with the wire is 50 degrees

    2. Relevant equations

    in order to achieve static equilibrium

    SIGMA torque = 0
    SIGMA F_y = 0
    SIGMA F_x = 0

    3. The attempt at a solution

    subscripts
    F_g is the force of gravity
    F_h is the force of the hinge
    F_T is the force of tension
    _x was added on to forces to indicate a x component
    _y was added on to forces to indicate a y component
    _beam was added on to forces to indicate that a force exerted on the beam
    _sign was added on to forces to indicated that a force exerted on the sign

    SIGMA F_y = F_h_y + F_T_y - F_g_sign - F_g_beam = 0
    SIGMA F_y = F_h_y + F_T sin THETA - F_g_sign - F_g_beam = 0

    SIGMA F_x = F_h_x - F_T_x = 0
    SIGMA F_x = F_h_x - F_T cos THETA= 0

    PP at hinge
    SIGMA torque = F_T_y * r_3 - F_g_beam * r_2 - F_g_sign * r_1 = 0
    SIGMA torque = F_T sin THETA * r_3 - F_g_beam * r_2 - F_g_sign * r_1 = 0

    add F_g_sign * r_1 to both sides
    SIGMA torque = F_T sin THETA * r_3 - F_g_beam * r_2 = F_g_sign * r_1

    add F_g_beam * r_2 to both sides
    SIGMA torque = F_T sin THETA * r_3 = F_g_sign * r_1 + F_g_beam * r_2

    divide both sides by sin THETA * r_3
    SIGMA torque = F_T = ( F_g_sign * r_1 + F_g_beam * r_2 ) / sin THETA * r_3

    plug and chug
    SIGMA torque = F_T = ( 90 N(.95 m + .75 m) + 55 N * .75 m ) / (.75 m) sin 50
    = 198.4 N

    we know this
    SIGMA F_x = F_h_x - F_T cos THETA= 0

    add F_T cos THETA to both sides
    SIGMA F_x = F_h_x = F_T cos THETA

    plug chug
    SIGMA F_x = F_h_x
    = 198.4 N cos 50
    = 130 N
    rounded to two sig figs

    we know this
    SIGMA F_y = F_h_y + F_T sin THETA - F_g_sign - F_g_beam = 0

    add F_g_sign to both sides
    SIGMA F_y = F_h_y + F_T sin THETA - F_g_beam = F_g_sign

    add F_g_beam to both sides
    SIGMA F_y = F_h_y + F_T sin THETA = F_g_sign + F_g_beam

    subtract F_T sin THETA from both sides
    SIGMA F_y = F_h_y = F_g_sign + F_g_beam - F_T sin THETA

    plug chug
    SIGMA F_y = F_h_y = 90 N + 55 N - 198.4 N sin 50
    = - 7.0 N

    What gives? I should all of my steps... ALL of them... So could you please show me were I went wrong?
     
  2. jcsd
  3. Jan 5, 2010 #2
    yah so i'm really not sure what I did wrong here
     
  4. Jan 5, 2010 #3
    sorry i still dont get it
     
  5. Jan 5, 2010 #4

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    When you sum moments about the hinge, the moment from he tension force is F_T_x times r3, not F_T_y times r3, you got your sin and cos mixed up. I didn't check the rest of your work.
     
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