In fig .1 2-40,one end of a uniform beam of weight 222n is hinged to a wall ; the other end is sup -ported by a wire that makes angles 6 =30.0with both wall and beam .find (a)thetension in the wire and the ( b)horizontaland (c)verticalcompo -nents of the force of the hinge on the beam .

  (a) 74√3 N

  (b) 37√3 N

  (c) 111 N

Step-by-step explanation:

(a) The moment about the hinge produced by the beam is the product of the weight of the beam and the distance of its center from the wall. The tension in the wire counteracts that moment.

The tension in the wire acts at a horizontal distance from the wall that is twice the distance to the beam's center, so the tension's vertical component is only half the weight of the beam.

Since the wire is at a 30° angle to the wall, the horizontal component of the tension is 1/√3 times the vertical component. Altogether, the tension in the wire is (2/√3) times half the beam's weight, or 74√3 N.

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(b) The horizontal force at the hinge counteracts the horizontal component of the tension in the wire, so is 111/√3 N = 37√3 N.

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(c) The vertical component of the force at the hinge is half the beam weight, so is 111 N.

the nearest power is 1.651.

means 7482 to the 1.651th power is almost equal to 2489287