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

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

Grade:11

4 Answers

Chetan Mandayam Nayakar
312 Points
13 years ago

let the normal reactions be NA and NB. NA+NB=W. (NA)*x=(NB)*(d-x)

manu saxena
46 Points
13 years ago

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

Sneha Singh
11 Points
6 years ago
N1+N2=W=>N2=W-N1...........(1)Anticlockwise momentum=Clockwise momentum N1x = N2(d-x)putting value of N2 from (1)N1x=(W-N1)(d-x)N1x=Wd-Wx-N1d+N1xN1x-N1x+N1d=W (d-x)0+N1d=W (d-x)N1=W(d-x)/d
Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the solution to your problem.

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

Thanks and Regards

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