Question

Dont use vector method

5. The uniform 200 kg bar AB is raised in the vertical plane by the application of a constant couple M 3 kNm applied to the link at C. The mass of the links is small and may be neglected. If the bar starts from rest at 0-0, determine the magnitude of the force supported by the pin at A as the position 0 600 is passed. 0.5 1.75 0.75

Answer 1

Let R and H are the vertical and horizontal reaction at support C.

Total weight of bar EABF= 200 kg

Weight of portion EA= 200/3 kg acting at a distance 0.25 m from A.

Weight of portion AB= 350/3 kg acting at a distance 0.875 m from B.

Weight of portion BF= 50 kg acting at a distance 0.375 m from B.

Applied moment M= 3kNm = 3*1000/9.81= 305.8kgm

Using clockwise moment conservation about D,

-100/3(0.5+1.5Cos60 +1.75) -350/3(0.875+ 1.5Cos60 ) -50(1.5Cos60 - 0.375) + 305.8 + R*1.75=0

By solving the equation we get R= 1.45 kg.

Now consider joint C and T is tension in bar AC,

Force equilibrium in vertical direction,

TSin60 + R= 0

T = -R/Sin60 = -1.45/Sin60 = -1.67 kg

Negative sign of T indicates that the force in member AC is compressive in nature.