Subject: Mathematics — Differential Calculus
Question: A load of weight $P$ lying on a horizontal plane must
be displaced by a force $F$ applied to it. The frictional force is
directly proportional to the force pressing the load against the plane
and is directed against the displacing force. The proportionality factor
(co-efficient of friction) is equal to $K$.
At what angle $\phi$ to the horizontal must the force $F$ be applied
so that its value turns out to be the least possible? Determine the
least value of the displacing force.
Answer: $F = \frac{KP}{\sqrt{1 + K^{2}}}$
Solution: Frictional force
$$\text{Frictional force} = KN$$
$$\text{Frictional force} = Kmg = KP$$
[IMAGE] Net horizontal force
$$\text{Net horizontal force} = K\left( mg - F\sin\phi \right) = F\cos\phi$$
$$\text{Net horizontal force} = K\left( P - F\sin\phi \right) = F\cos\phi$$
$$\frac{KP}{K\sin\phi + cos\phi} = F$$
$$F(\phi) = \frac{KP}{K\sin\phi + cos\phi}$$
$$F'(\phi) = \frac{- KP\left( K\cos\phi - sin\phi \right)}{\left( K\sin\phi + cos\phi \right)^{2}}$$
$$F'(\phi) = 0$$
\[Condition for least value of force\]
$$K\cos\phi - sin\phi = 0$$
or,
$$\tan\phi = K$$
$$\phi = \tan^{- 1}K$$
[IMAGE] $$F = \frac{KP}{\frac{K^{2}}{\sqrt{1 + K^{2}}} + \frac{1}{\sqrt{1 + K^{2}}}}$$
$$F_{\min} = \frac{KP}{\sqrt{1 + K^{2}}}$$
$F = \frac{KP}{\sqrt{1 + K^{2}}}$ (least force)
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