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A block placed on a horizontal surface is being pushed by a force`F` making an angle `0` with the verticle If the friction coefficient is `mu` how much force is needed to get the block just started Discuss the situation when tan ` theta lt mu `. |
Answer» In limitting equilibrium force of friction `f = mu R` In equilibrium along the horizontal `F sin theta = mu R` and along the verticle `F cos theta + mg = R = (F sin theta )/(mu` `:. Mg = F(( sin theta )/(mu) - cos theta)` or `F = (mu mg)/(sin theta - mu cos theta)` If tan `theta lt mu` `(sin theta - mu cos theta lt theta)` `:. F` is negative So for angles less angles less than `tan^(-1) mu` on cannot push the block ahed howsoever large the force may be . |
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