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 theta 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 y-component P_y of the force exerted by the pivot on the beam. Expression: P_y = Select from the variables below to write your expression. Note that all variables may not be required. cocotan(theta), cos(alpha), cos(phi), cos(theta), cotan(theta), sin(alpha), sin(phi), sin(theta), tan(theta), alpha, d, g, M, T, y Write an expression for the x-component P_x of the force exerted by the pivot on the beam, in terms of T. Expression: P_x = ___ Select from the variables below to write your expression. Note that all variables may not be required. cotan(theta), tan(theta), alpha, beta, theta, cos(theta), d, g, h, j, k, M, P, sin(theta), T What is the tension in the horizontal cable, in newtons, if the mass of the beam is 48 kg, the length of the beam is 15 m, and the angle is 42 degree?

Explanation / Answer

a) As vertical forces must balance

Py = 2 * M * g = 19.6 M

b) Sum of the moments about the pivot

T L sin theta = M g (3L/4) cos theta + Mg (L/2) cos theta

T = 5/4 * Mg / tan theta = 5/4 * M g cotan theta

c) Px = T  = 5/4 * Mg / tan theta

d) T  =  5/4 * Mg / tan theta

= 5 * 48 * 9.8 / 4 tan 42

= 653 N

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