Question

$\begin{array}{c}\text{A block of mass}10\text{kg is placed on the rough}\\ \text{horizontal surface. A pulling force}F\text{on the rough}\\ \text{block which makes an angle}\theta \text{above the horizontal.}\\ \text{If coefficient of friction between block and surface is}\\ \end{array}$ $\begin{array}{c}\frac{4}{3},\text{then minimum value of force required to just}\\ \text{move the block is}(g=10m/s^2)\end{array}$

• A. $80N$
• B. $160N$
• C. $60N$
• D. $120N$

Answer 1

The correct option is A $80N$

AT minimum force to just move it,static friction will maximum (i.e $\mu N$)

$F\cos \theta =\mu N$

$F\sin \theta +N=\mu g$

$\Rightarrow$ $N=\mu g-F\sin \theta .....\left(ii\right)$

From (i) and (ii) we get

$F(\cos \theta +\mu \sin \theta )=\mu Mg$

$\Rightarrow$ $f=\frac{\mu Mg}{\cos \theta +\mu \sin \theta }$

For $F$ to be minimum $(\cos \theta +\mu\sin \theta)$ should be maximum

Max value of $\cos \theta +\mu \sin \theta =\sqrt{1+{\mu }^2}$

So, $F_{min}=\frac{\mu Mg}{\sqrt{1+{\mu }^2}}$

$=\frac{\left(\frac{4}{3}\right)\left(0\right)\left(0\right)}{\sqrt{1+{\left(\frac{4}{3}\right)}^2}}$

$F_{min}=80N$