Question

A 100 kg uniform beam is attached to a vertical wall at one end and is supported by a cable at the other end. Calculate the magnitude of the vertical component of the force that the wall exerts on the left end of the beam if the angle between the cable and horizontal is θ = 43°. The angle between the horizontal and the beam is 30 degrees.

Answer 1

let tension in the cable to be T N.

let length of the beam is L.

let magnitude of vertical component of force that the wall exerts on the left end of the beam is F N.

balancing force in vertical direction:

T*sin(43)+F=weight of the beam

==>0.682*T+F=100*9.8=980

==>0.682*T+F=980...(1)

perpendicular distance between weight and the left end=(L/2)*cos(30)=0.433*L

perpendicular distance between horizontal component of tension (i.e. T*cos(43)=0.73135*T) and left end of the beam=L*sin(30)=0.5*L

perpendicular distance between vertical component of tension (i.e. T*sin(43)=0.682*T) and left end of the beam=L*cos(30)=0.866*L

now balancing torque about left end:

100*9.8*0.433*L=0.73135*T*0.5*L+0.682*T*0.866*L

==>424.34=0.36567*T+0.591*T

==>0.9566*T=424.34

==>T=443.6 N

hence vertical compoennt of the force exerted by the left end of the beam, as obtained from equation 1:

F=980-0.682*T=980-0.682*443.6=677.46 N