Question

Find the weight WB so as to have its impending motion down the plane. Take weight of block A as 2kN. The pin connected rod AB is initially is in horizontal position. Refer Figure 13. Coefficient of friction = 0.25 for all surfaces.

Answer 1

   

Consider block A	Considering block B,
ΣFx=0 FAB + RAcos(60) + 0.25RAcos(30) =0 FAB + 0.7165RA =0 ……(1) ΣFy=0 ΣFy=0 -2 + RAsin(60) – 0.25RAsin(30) =0 0.741RA =2 RA = 2.699 kN …..(2) FAB = -0.7165RA FAB = -1.934 kN ……(3)	ΣFx=0 -FAB – RBcos(45) + 0.25RBcos(45) =0 FAB = -0.53RB RB = 3.649 kN ΣFy=0  -WB + RBsin(45) + 0.25RBsin(45) WB = 3.225 kN

The weight of the block B is 3.225 kN.