You want to hang a 10 kg sign by using a pivot to attach the base of a 5.0 kg be

Question

You want to hang a 10 kg sign by using a pivot to attach the base of a 5.0 kg beam to a wall (see figure). You then attach a cable to the beam and to the wall in such a wall that the cable and beam are perpendicular to each other. The beam is 2.0 m long and makes an angle of 37degree with the vertical. You hang the sign from the end of the beam attached to the cable. What must be the minimum tensile strength (minimum amount of tension it can sustain) if it is not to snap? Determine the horizontal and vertical components of the force exerted by the pivot on the beam.

Explanation / Answer

Sum moments about the hinge point to zero
Let T be the unknown tension and the minimum tensile strength needed for the cable.
Let CCW be the positive moment
Let L be the length of the beam

T[L] - 5.0(9.81)[(L/2)sin37] - 10.0(9.81)[Lsin37] = 0

as L is common to all terms, it divides out

T - 5.0(9.81)[(1/2)sin37] - 10.0(9.81)[sin37] = 0
T - 2.5(9.81)[sin37] - 10.0(9.81)[sin37] = 0
T = (2.5 + 10.0)(9.81)sin37
T = 73.8
T = 74 N       (answer a)

Part B

Fx = 0

Fx = Tcos37

Fx = 73.8*cos37

Fx = 58.94 N -- Horizontal Force

Fy = 0

Fy + Tsin37 = 5.0(9.81) + 10.0(9.81)

Fy = 102.74 N -- Vertical Force

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