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
- W(d-x)x
- W(d-x)d
- Wxd
- Wdx
Answer
(2) W(d-x)d
Solution
Let NA and NB be the normal reactions at A and B. The weight W acts at the centre of mass, at distance x from A and (d-x) from B.
For rotational equilibrium, take torques about B:
NA d = W (d-x)
NA = W(d-x)d
(Check: NB = Wxd and NA + NB = W.)
Answer: (2)