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 at an angle `theta` with the horizontal as shown in Fig. 7.40 . For vertical equilibrium ,

`R + F "sin" theta = "mg" ` , R = mg - F sin `theta` ……. (i)
while for horizontal motion
`F "cos" theta ge f_(L)`
i.e., `" " F "cos" theta ge mu R " " ["as" f_(L) = muR] " " ......`(ii)
Substituting value of R from Eqn. (i) in (ii) , we get
`F "cos" theta ge mu ("mg" - F "sin" theta)`
i .e., `" " F ge (mu "mg")/(("cos" theta + mu "sin" theta)) " " .......` (iii)
For the force F to be minimum (`"cos" theta + mu "sin" theta`) must be maximum , i.e.,
`(d)/(d theta) ("cos"theta + mu "sin" theta ) = 0 `
or `" " -"sin"theta + mu "cos" theta = 0 `
i.e., `" " "tan" theta = mu " " ......` (iv)
So that , `" " "sin" theta = (mu)/(sqrt(1 + mu^(2))` and `"cos" theta = (1)/(sqrt(1 + mu^(2)))`
Substituting these values in Eqn . (iii) ,
`F ge (mu "mg")/((1)/(sqrt(1 + mu^(2))) + (mu^(2))/(sqrt(1 + mu^(2))))`
i.e., `" " F ge (mu "mg")/(sqrt(1 + mu^(2)))`
So that , `F_("min") = (mu "mg")/(sqrt(1 + mu^(2)))` with `theta = tan^(-1) (mu)`