A particle of mass m rests on a horizontal floor with which it has a coefficient of static friction `mu`. It is desired to make the body move by applying the minimum possible force F. Find the magnitude of F and the direction in which it has to be applied.

Explanation

Let the force F be applied on the body at an angle `theta` with the horizontal as shown in The force of friction `f = mu R` acts along the surface of contact to the left as shown For horizontal equilibrium
` F cos theta = f = mu R `
For vertical equilibrium
` R + F sin theta = mg `
` R = mg - F sin theta `
Put in (i) ` F cos theta = mu (mg - F sin theta ) `
` = mu mg - mu F sin theta `
` F ( cos theta + mu sin theta) = mu mg `
` F = (mu mg )/(cos theta + mu sin theta )`
For force F to be minimum the denominator should be maximum for which
`(d)/(d theta) (cos theta + mu sin theta) = 0 `
` - sin theta + mu cos theta = 0 or mu cos theta = sin theta `
` mu = (sin theta)/(cos theta) = tan theta `
` sin theta = (mu)/(sqrt(mu^(2) + 1)) `
` cos theta = (1)/(sqrt(mu^(2 + 1) `
From (ii) ` F_(min) = (mu mg)/(1/(sqrt(mu^(2) + 1 ))+ mu^(2)/sqrt(mu^(2 + 1)`
` F_(min) = (mumg)/sqrt(mu^(2) + 1 `
Angle which this force makes with horizontal is `theta = tan^(-1) mu`

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