A uniform beam of length L and mass M has its lower end pivoted at P on the floo

Question

A uniform beam of length L and mass M has its lower end pivoted at P on the floor, making an angle 6 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. Write an expression for the component P_y of the force exerted by the pivot on the beam. Write an expression for the tension T in the horizontal cable AB Write an expression for the x component P_x of the force exerted by the pivot on the beam, in terms of T. What is the tension in the horizontal cable, in newtons, if the mass of the beam is 49 kg, the length of the beam is 13 m, and the angle is 42 degree ?

Explanation / Answer

From force equilibrium;

Py = Mg + Mg = 2Mg

Torque equilibrium from pivot,

MgL/2 * cos (theta) + 3MgL/4 cos (thata) - T L sin( theta) = 0

T = 5Mg /4 * cot(theta)

From horizontal equilibrium

Px -T = 0 So Px = 5Mg cot ( theta) /4

T = 5 * 49 * 9.8 * cot ( 42) / 4

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