Question

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 1

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Q.) 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



Ans.) By Equating forces .



N1+N2=W


Equating Torques about A 



Wx=N2d



=>N2=Wx/d



So N1=W[1- (x/d)]



.°. Normal reaction on A is W[1- (x/d)]



And , on B is Wx/d .


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Answer 2

Answer:Na=w(d-x)/d

Explanation:

w=Na+Nb

Nb=w-Na

Now balancing torque about COM,

i.e. clockwise momentum=clockwise momentum

NaX=Nb(d-x)

NaX=(w-Na)(d-x)

NaX=wd-wx-Nad+NaX

Nad=wd-wx

Na=w(d-x)/d

Hope it helps