now equate the torque about the center of mass and resolve the forces in the vertical direction to get ur answer
a rod of weight, w is supportd by two parallel knife edges A and B and is in equilibrium in a horizontal position. the knives are at a distance of d from each other. the centre of mass of the rod is at a distance x from A. the normal reaction on A is ........ ??
and on B is............ ?
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4 Answers
Mani Pal Singh
·2009-03-29 12:01:09
Kalyan Pilla
·2009-03-29 12:10:57
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
questions
·2009-03-29 12:30:41
yeah exactly but the answer what BT depicts is
on A its w(d-x)
and on B its wx
guys !!
Kalyan Pilla
·2009-03-29 12:36:45
That means there is no ~/d term there.
Jus' check if w is density[7][7][333]