If force `F` is gradually increased from zero to `50 N` , tehn (a) find the minimum contact force, (b) find the maximum value of angle of friction when `F = 30 N` , (d) find the angle of friction when the block is moving with constant velocity and ( e) sketch a graph between friction and applied force.

Explanation

`R` : Constant force , `phi` : angle of friction
`R = sqrt(N_(0)^(2) + f^(2)) , tan phi = (f)/(N)`
`N_(0) = 40, (f_(s))_(max) = mu_(s) N_(0) = N_(0) = 40 N`
`f_(k) = mu_(k) N_(0) = 0.8 xx 40 = 32 N`
(a) `R` is minimum when `F = 0`
`R = R_(min) = N_(0) = 40 N`
(b) `phi` is maximum when `F = (f_(s))_(max) = 40 N`
`tan phi_(max) = ((f_(s))_(max))/(N _(0)) = (40)/(40) = 1 rArr phi_(max) = 45^(@)`
(c ) When `F = 30 N , F lt (f_(s))_(max)` , the block is at rest
`F = f_(s) rArr F = 30 N`
` R = sqrt(N^(2) + f^(2)) = sqrt((40)^(2) + (30)^(2)) = 50 N`
`tan phi = (f)/(N) = (30)/(40) = (3)/(4) rArr phi = 37^(@)`

(d) `tan phi_(k) = (f_(k))/(N) = (32)/(40) = 0.8 rArr phi_(k) = tan^(-1) (0.8)`
(e) When `F lt (f_(s))_(max)` , the block is at rest
`f = f_(s) = f (f v//s F ` fraph will be an inclined line of slope `45^(@)`)
When `F = (f_(s))_(max)` , the block is about to move
When `F gt (f_(s))_(max)` , the block is in motion
`f = f_(k) = 32 n (f v//s F` graph will be straight line parallel to the `F`- axis)