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?

1 Answer
Jul 19, 2017

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"