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 a distance x from a. What is the normal reaction on a?
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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
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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