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

Step-by-step 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) * 6.84 m + 316 N * (6.84)/(2) m + 608 N * 6.84 m`

`-F_(T) * 0.81915 * 6.84 m + 316 N * (6.84)/(2) m + 608 N * 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.