A rod of weight W is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position . The knives are at a distance d from each other . The centre of mass of the rod is at distance x from A . The normal reaction on A is

A rod of weight W is supported by two parallel knife edges A and B and

1.	A rod of weight W is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position . The knives are at a distance d from each other . The centre of mass of the rod is at distance x from A . The normal reaction on A is
Answer» `(W(d- x))/(x)` `(W( d - x))/(d)` `(WX)/(d)` `(Wd)/(x)` Solution :Given situation is shown in figures . `N_(1)` = Normal reaction on A `N_(2)` = Normal reaction on B W = WEIGHT of the rod In vertical equilibrium `N_(1) + N_(2) = W "" ….. (i)` Torque BALANCE about centre of MASS of the rod , `N_(1) x = N_(2) (d - x)` Putting value of `N_2` from equation (i) `N_(1)x = (W - N_1) (d - x)` `implies N_1 x = Wd - Wx - N_1 d + N_1 x implies N_1 d = W (d - x)` `therefore N_(1) = (W (d - x))/(d)`