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

fmax=22.0 N

Explanation:

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

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

fmax=μsn

where

  • μs is the coefficient of static friction (13)

  • n is the magnitude of the normal force

METHOD 1:

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

Fapplied-y=Fsin30o = F2=fmax2

The net vertical force Fy is

Fy=mgfmax2=n

Rearranging:

fmax=2(mgn)

From the first equation:

fmax=13n=n3

So, setting the two equations for fmax (in blue) equal to each other:

2(mgn)=n3

n=38.1 N [mg=(5lkg)(9.81lm/s2)]

So,

fmax=38.1lN3=22.0 N

METHOD 2:

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

n=mgfmax2

n=fmax13=3fmax

Thus,

mgfmax2=3fmax

Knowing that mg=(5lkg)(9.81lm/s2),

fmax=22.0 N