A uniform beam of length L = 2.8 m and mass M = 21 kg has its lower end fixed to

Question

A uniform beam of length L = 2.8 m and mass M = 21 kg has its lower end fixed to pivot at a point P on the floor, making an angle 0 = 11 degree as shown in the diagram. A horizontal cable is attached at its upper end B to a point A on a wall. A box of the same mass M as the beam is suspended from a rope that is attached to the beam one-fourth L from its upper end. What is the y-component P_y of the force, in newtons, exerted by the pivot on the beam? P_y = Write an expression for the tension T in the horizontal cable AB. What is the x-component P_k of the force, in newtons, exerted by the pivot on the beam?

Explanation / Answer

According to the question provided we need to find the following values.

Part a) P(y) = 2 * M * 9.81 = 2 * 21 kg * 9.81 = 412.02 Newton =====ANSWER)

Part b) T = 5/4 * M *g cotan( theta) ========================ANSWER)

Part c) Px = 5/4*M*9.81*1/tan(t*3.14159/180) = 1.25 * 21 * 9.81 * 1 / tan ( 11 * 3.14159/180 ) = 1324.788113 Newton =============================================ANSWER)

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