A mass of 6kg is suspended by a rope oflength 2m from a celling a force of 50N in the horizontaldirection is applied at themidpoint of the rope as shown in the figure what is the angle the rope makes with the vertical is equalibrium take g=10 ms^(-2)neglect mass of the rope

A mass of 6kg is suspended by a rope oflength 2m from a celling a forc

1.	A mass of 6kg is suspended by a rope oflength 2m from a celling a force of 50N in the horizontaldirection is applied at themidpoint of the rope as shown in the figure what is the angle the rope makes with the vertical is equalibrium take g=10 ms^(-2)neglect mass of the rope
Answer» Solution :Givenm= 6KG length = 2 m force=50 N g=10`MS^(-2)` Asshownin thediagramthere are threeforcesactingon the midpointp of the ropeSupposethe ropemakesan angle`theta` with theverticel inequilibriumresolvingthe forceshorizontally and VERTICALLY we get `T_(1) SIN theta=T_(3) = 50 N` `T_(1) cos thetaT_(2)= 6kgwt = 60N` Dividing (1) by (2) we get tan `theta ` `=(5)/(6) ` (or) `thetatan ^(-1) ((5)/(6))` `=tan^(-1)(0.833)` `=theta= 39 .8^(@)`