I already have the correct answer to this problem; however, I need it worked aga

Question

I already have the correct answer to this problem; however, I need it worked again step-by-step so I can learn how to do it. Please show work for every step. Thanks! A uniform beam of length L = 3.4 m and mass M = 47 kg has its lower end fixed to pivot at a point P on the floor, making an angle 0 = 27 degrees 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 = 922.1-this is the correct answer; I need to see step-by-step how to arrive at this answer) What is the x-component P_x of the force, in Newtons, exerted by the pivot on the beam? (P_x=1131-this is the correct answer; I need to see step-by-step how to arrive at this answer)

Explanation / Answer

forces acting on the system:

weight of the beam=M*g, in -ve y direction

weight of the block M=M*g, in -ve y direction

y component of the force exerted by pivot P, Py, in +ve y direction

tension in the cord, T, in -ve x direction

x component of the force exerted by pivot P on the beam, Px, in +ve x direction

balancing forces in y direction:

M*g+M*g=Py

==>Py=2*M*g=2*47*9.81=922.14 N

part b:

balancing force in x direction:

Px=T...(1)

balancing torqu about pivot P:

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

==>47*9.81*(3.4/2)*cos(27)+47*9.81*(3*3.4/4)*cos(27)=T*3.4*sin(27)

==>1745.9696=T*1.54356

==>T=1131.12 N

from equation 1, Px=T=1131.12 N

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