Question

21 kg mass is located on the ground the static friction coefficient is 0.75 and the kinetic friction coefficient is 0.65 the mass pushes with 215 N of force an angle of 20° below the horizontal

Answer 1

GIVEN:

A block of 21 kg mass is placed onthe rough surface with

static friction coefficient $(\mu_s )=0.75$

kinetic friction coefficient $(\mu_k )=0.65$

applied force of 215 N at an angle $20\degree$ below the horizontal

DIAGRAM :

(fig.1) shows how force is applied on the block.

Now, we can make component of fore in x and y directions

as F₁ and F₂ respectively.

$\implies$ refer (fig. 2)

Values of component forces are as follows:

as, F=215 N and angle is $20\degree$

$\therefore F_1=215 cos(20\degree) =212.0339$

$F_2=215 Sin(20\degree) =73.5343$

$\implies refer (fig.3)$

now, normal force will increase and there is a +ve force along x direction.

Normal force

$N=mg +F_2$

$= (210+73.5343)$

$=283.5343$

$\implies refer(fig.4)$

When horizontal force F1 will work in x direction then box try to move and frictional force will come in action...

$\implies$ so first we find maximum friction force $(f_m)$ that can be applied by the box

$\boxed{f_m=f_s=\mu_sN}$

$=0.75*283.5343$

$\boxed{f_m=212.6507}$ ....................(1)

$\implies$ and force applied in x-deirection is

$\boxed{F_1=202.0339}$ ................................(2)

$\bigstar$ from (1) and (2) ,we get to know that maximum frictional force is more than horizontal force

$\therefore$ frictional force $f$ will be equal to $F_1$

i.e, 202.0399

$\bigstar$ it implies that the box will not move and remain static on its position.