Question

A body of mass 5 kg is placed on a rough horizontal surface. if coefficient of friction is 1/√3 find what pulling force should act on the body at an angle 30 degree to the horizontal so that the body just begins to move?

Answer 1

`f_"max" = 22.0` `"N"`

Explanation:

I'll assume the given coefficient is that of static friction.

The maximum horizontal force `f_"max"` that can be applied to the object (which is what we're trying to find) is given by

`f_"max" = mu_sn`

where

• `mu_s` is the coefficient of static friction (`1/sqrt3`)

• `n` is the magnitude of the normal force

`bb(METHOD` `bb(1`:

To find this, we recognize that the pulling force makes a `30^"o"` angle with the horizontal, so the vertical component of this applied force is

`F_"applied-y" = Fsin30^"o"` = `F/2 = (f_"max")/2`

The net vertical force `sumF_y` is

`sumF_y = mg - (f_"max")/2 =n`

Rearranging:

`color(blue)(f_"max" = 2(mg-n)`

From the first equation:

`color(blue)(f_"max" = 1/sqrt3n = n/sqrt3`

So, setting the two equations for `f_"max"` (in blue) equal to each other:

`2(mg - n) = n/sqrt3`

`n = 38.1` `"N"` `[mg = (5color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)]`

So,

`f_"max" = (38.1color(white)(l)"N")/(sqrt3) = color(red)(22.0` `color(red)("N"`

`bb(METHOD)` `bb(2`:

Method 1 was more of an indirect solution; we can also find `f_"max"` directly by rearranging the two blue equations to solve for `n`, and then setting them equal to each other:

`color(blue)(n = mg - (f_"max")/2`

`color(blue)(n = (f_"max")/(1/sqrt3) = sqrt3f_"max"`

Thus,

`mg - (f_"max")/2 = sqrt3f_"max"`

Knowing that `mg = (5color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)`,

`f_"max" = color(red)(22.0` `color(red)("N"`