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A uniform beam of length 1.0 m and mass 16 kg is attached to a wall by a cable that makes an angle of 30 degrees with the end of the beam, as shown in the figure. The beam is free to pivot at the point where it attaches to the wall. What is the tension in the cable?

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3 answers
  1. Sum moments around the beam end at the wall.

    mg*L/2-Tension*sin30=0

    solve for tension.

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    bobpursley
  2. error. Change the second term to Tension*length*sin30.

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    bobpursley
  3. us the formula for torque (t) the second condition (summation of torques is equal to zero)
    t of y component of the string(ty) - t of the weight of the beam(tw) = 0
    then ty=tw

    ty=rw/2 x Fw x sin(theta)

    ty= .5 x 16 (9.81) x sin 90

    now us the value of ty to solve for the Fy (tension force of the y component of the string)

    ty=r x Fy x sin 90

    use sohcahtoa to solve for the Fx (x component of string)

    use pythagorean to solve for the actual string

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