Question

b Write an expression for the tension T in the horizontal cable AB $sig.gif?tid=4N89-35-A9-45-A5E1-16924$ 25% Part (c) Write an expression for the x-component Px of the force exerted by the pivot on the beam, in terms of T. $sig.gif?tid=4N89-35-A9-45-A5E1-16924$ 25% Part (d) What is the tension in the horizontal cable, in newtons, if the mass of the beam is 34 kg, the length of the beam is 11 m, and the angle is 28°? A uniform beam of length L and mass M has its lower end pivoted at P on the floor, making an angle θ with the floor. A horizontal cable is attached at its upper end B to a point A on a wall. A box of mass M is suspended from a rope that is attached to the beam one-fourth L from its upper end.

Answer 1

Answer OF Tlsin o-mg ( Jaso+mg (34) ago Tsino = mg eso + Ingaso. Tsino - (+%) ng aso s Tsino=(19) mg aso. T=(18) ng sino Ta lodo tano Pq = (1.25) mg sino @ Pa = T. 2: T= (1.25) mg Stone = 1.25 X 3469 x 9.8m/s? asas T =783.3 N sin 98