A unifrom cube of side "`b` " and mass M rests on a rough horizontal table.A horizontal force F is applied normal to one of the face to a point,at a height `(3b)/4` above the base. What should be the coefficient of friction (`mu` ) between cube and table so that is will tip about an edge before it starts slipping?
A) `mu >` `2/3`
B) `mu >` `1/3`
C) `mu >` `3/2`
D) ` ` none
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A uniform cube of side b and mass M rests on a rough horizontal table. A horizontal force F is applied normal to one of the face to a point, at a height `(3*b)/4` above the base.
About the bottom right edge, there is a torque created due to two forces, one of them is the weight of the the cube. This is equal to `F*r*sin theta` = `(m*g)*(b/2)*sqrt 2*(1/sqrt 2)` = `(m*g*b)/2`
The other is due to the force F being applied at a height `(3*b)/4` . The torque in this case is `F*(3*b)/4` .
The net torque is `F*(3*b)/4 - m*g*b/2` . If the box topples over before it starts to slide, `F*(3*b)/4 - M*g*b/2 > 0` and `F < mu*M*g`
`F*3/2 > M*g`
=>` F > (2*M*g)/3`
As `mu*M*g > F`
`mu*M*g > (2*M*g)/3`
=> `mu > 2/3`
The coefficient of friction should be greater than `2/3` .
The correct answer is option A
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