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"#