A block placed on a horizontal surface is being pushed by a force F making an angle `theta` with the vertical as shown in Fig. 7.39 . The coefficient of friction between block and surface is `mu` . (a) Find the force required to slide the block with uniform velocity on the floor . (b) Show that if `theta` is smaller than a certain

angle `theta_(0)` , the block cannot be made to slide across the floor , no matter how great the force be .

Explanation

(a) Force acting on the block are shown in Fig.7.39
For vertical equilibrium of the block ,
` R = F "cos" theta + "mg" " " …….`(i)
While for horizontal motion ,
`F "sin" theta - mu R = ma`
or `" " F "sin" theta = mu R " " ["as" v` = constant a = 0]
Substituting R from Eqn. (i) in the above ,
`F "sin" theta = mu ( F "cos" theta + "mg")`
i.e., `" " F = (mu "mg")/(("sin" theta - mu "cos" theta))`
(b) As friction is present , for motion F must be + ve . This is possible only if
`"sin" theta - mu "cos" theta gt 0 `
i.e., `" " "tan" theta gt mu `
or `" " theta gt tan^(-1) (mu)`
or `" " theta gt theta_(0) " " "with" " " theta_(0) = tan^(-1)(mu)`
So , for angle `theta lt theta_(0)[= tan^(-1) (mu)]` no motion will take place however great the force F be .