Question

A uniform 6.84 m long horizontal beam that weighs 316 N is attached to a wall by a pin connection that allows the beam to rotate. Its far end is supported by a cable that makes an angle of 55◦ with the horizontal, and a 608 N person is standing 1.9 m from the pin. 1.9 m 6.84 m 608 N 316 N R FT 55◦ Find the force FT in the cable by assuming that the origin of our coordinate system is at the rod’s center of mass.

Answer 1

Explanation:

Let us assume that moment about the pin and then setting it equal to zero as the rod is in equilibrium is as follows.

          Moment = Force × Leverage

        $-F_{T} Sin 55^{o} \times 6.84 m + 316 N \times \frac{6.84}{2} m + 608 N \times 6.84 m$

        $-F_{T} \times 0.81915 \times 6.84 m + 316 N \times \frac{6.84}{2} m + 608 N \times 6.84 m$ = 0

                  $F_{T} = 935.11 N$

Therefore, we can conclude that the force ($F_{T}$) in the cable by assuming that the origin of our coordinate system is at the rod’s center of mass is 935.11 N.