## Numerical Problems Based On Concept of Friction

**Solution:**
Given : m=2kg,µ=0.8

Friction force = µmg=0.8*2*9.8=15.68 N

**Solution:**
Applied force =5N

Object is at rest.

Friction force =5N

**Solution:**
Given : m=10kg,µ=0.3

frictional force=µmg=0.3*10*9.8=29.4 N

**Solution:**
Given : m=40kg,µ=0.45

frictional force =µmg =0.45*40*9.8=176.4 N

**Solution:**
Given : F = 200N at 30° with horizontal

Fx = 200*Cos30 = 173.2 N.

Fy = 200*sin30 = 100 N.

Wt. of the object =M*g = 30kg * 9.8 = 294 N acting downwards

Reaction of surface on box N = mg-Fy =(30*10)-100=200N

Force of kinetic friction(Fk) = µ*N = 0.2*200=40N

**Solution:**
Limiting friction, f = μmg=0.3×50×10==150 N

Applied force, F = 50N

As Applied force < Limiting frictional force,

so force of friction = 50 N

**Solution:**
In this problem frictional force will act as retarding force.

Frictional force acting on block mass M = µMg

Retardation =-µMg/M=-µg

Initial velocity (u)=v ,Final velocity(v) =0

Let shortest distance that object will cover= s.

Accoding to equation of motion

v²=u²+2as

0=v²+2(-µg)s

v²=2(µg)s

s =v²/2(µg)

shortest distance that object will cover= s = v²/2(µg)

Let after time t object will come at rest.

Accoding to equation of motion

v=u+at

0=u+(-µg)t

t=u/µg

**Solution:**
Let acceleration of block be a .

Force applied on the block F= 40N

Friction force acting on the block f= 0.2*5*10=10N

Friction force is opposite of applied force.

Net force acting on the block=40-10=30N

Therefore a=30/5= 6 m/s²

**Solution:**
static friction = µmg

50=µ*4*9.8

µ=50/4*9.8=1.28

**Solution:**
Let tension in string T.
To move m1,

Tention T> µm1g (Limiting friction on m1 mass) or T>0.9*7*10

T>63

Calculation of T :

T=m2g

T=5*10

T=50

Here we see T< Limiting friction on m1 mass.

Hence m1 will not move.

To move m1 :

T>63

m2g>63

m2>6.3 kg

**Solution:**
The minimum force required to prevent slipping is equal to the minimum force that will prevent the block from sliding down .

Fmin = 10gsin(45°)−10gcos(45°)×0.5

Fmin = 10*9.8*0.707−10*9.8*0.707*0.5=34.20 N

The maximum force required on block to motion in upward direction

Fmax = 10gsin(45°)+10gcos(45°)×0.5

Fmax = 10*9.8*0.707+10*9.8*0.707*0.5= 103.3 N

**Solution:**

Given,

Weight of the block (force due to gravity) = 200 N

Force required to start the motion (static friction) = 80 N

Force required to keep the block moving at a constant velocity (kinetic friction) = 40 N

We can calculate the coefficients of friction as follows:

Coefficient of static friction (μs):

μs = (Force of static friction) / (Normal force)

μs = 80 N / 200 N

μs = 0.4

Coefficient of kinetic friction (μk):

μk = (Force of kinetic friction) / (Normal force)

μk = 40 N / 200 N

μk = 0.2

So, the coefficients of friction for the block and the surface are μs = 0.4 and μk = 0.2, respectively.

**Solution:**

Given,

The total work done = 270J

The force (F) = 50N

The angle (θ) = 30 degrees

The height (h) = 4m.

Coefficient of friction (μ)=?

Total work done = Work against gravity (W) + Work against friction(w)

Calculate the work against gravity(W):

W = 50sin30°*(4/sin30°)

W = 50N *1/2 * 8

W = 25*8

W = 200J

Calculate the work against friction(w):w

w= Total work done - Work done against gravity

w = 270J - 200 J

w = 70

Calculation of force of friction

F1=μN=μ50cos30°=μ50*(1.732/2)

Work against friction(w)=F1*(4/sin30°)

=F1*8

=μ50*(1.732/2)*8

=346.4*μ

∴346.4*μ =70

μ =70/346.4

μ = 0.20

So, the coefficient of friction on the incline is approximately 0.20.

**Solution:**

Maximum static friction acting between the crate and the truck =F1

F1=0.4*48*9.8=188.16 N

Net force acting on the crate due to motion of trruck=F2

F2=48Xa (where a=acceleration of truck)

Crate is not moving so F1 and F2 are balanced forces.

48Xa=188.16

a=188.16/48 =3.92 m/s2

**Solution:**

Maximum static friction force that can be exerted on an object=0.4*40*9.8=156.96N

Therefore, the maximum pushing force that can be applied without moving the box is approximately 156.96 Newtons.

**Solution:**

Given,

m=5kg

μ=0.1

u=5m/s

v=0

Let time required to stop the box=t

The normal force acting on box=N=mg

N= 5 kg × 9.81 m/s^2 ≈ 49.05 N

Force of Kinetic Friction of box =μN= 0.1 × 49.05 N ≈ 4.905 N

This force will oppose the motion of object.

accelration =4.905/5=0.981 m/s^2 (This will be negative)

Equation of motion

a=(v-u)/t

-0.981=-5/t

t=5/0.981=5.1 second

So, it will take approximately 5.1 seconds for the box to come to a stop.