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:
Applied force, F = ma =1× 5=5N
As Applied force < Limiting frictional force,
so force of friction = 5 N
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:
∴Friction force acting on box=35N
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.
