A force of $19.6\, N$ when applied parallel to the surface just moves a body of mass $10 \,kg$ kept on a horizontal surface. If a $5\, kg$ mass is kept on the first mass, the force applied parallel to the surface to just move the combined body is........ $N.$
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(a) ${F_l} \propto R$ $\therefore $ ${F_l} \propto m$ i.e. limiting friction depends upon the mass of body. So, $\frac{{({F_l})\,\prime }}{{({F_l})}} = \frac{{m'}}{m} = \frac{{10 + 5}}{{10}}$
==>$({F_l})\,\prime = \frac{3}{2} \times {F_l} = \frac{3}{2} \times 19.6 = 29.4\;N$
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