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Stuck here, help me understand: A rod of weight wis 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. T

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:

  • Option 1)

    \frac{wx}{d}

  • Option 2)

    \frac{wd}{x}

  • Option 3)

    \frac{w(d-x)}{x}

  • Option 4)

    \frac{w(d-x)}{d}

 
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By \ Equating \ forces . \\N_1+N_2=W \\ Equating \ Torques \ about \ A \\ Wx=N_2d =>\\N_2=Wx/d \\So N_1=W[1- (x/d)] \\.°. \ Normal \ reaction \ on \ A \ is \ W[1- (x/d)]


Option 1)

\frac{wx}{d}

Option 2)

\frac{wd}{x}

Option 3)

\frac{w(d-x)}{x}

Option 4)

\frac{w(d-x)}{d}

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