Chapter : 2. Force & Laws Of Motion
Applications of frictional forces
Applications of frictional forces
We would not be able to walk if there had been no friction between the soles of our shoes and the ground.
If there had been no friction, the wheels of a car will slip instead of rotating and stop moving. For that we have to increase the friction by making the tyres corrugated to get better grip of tyres on the road. Also, the friction is increased.
When brakes are applied, the vehicle stops due to the force of friction between the brake lining and the drum.
In the absence of friction, we cannot write on a blackboard with a chalk stick because the chalk stick will slide off the board without leaving any mark on the board.
Example : 11
Three blocks are tied together with strings as shown. They are lying on a horizontal frictionless table. These are pulled to the right with T3 = 12N force. If m1 = 1 kg, m2 = 2kg, and m3 = 3 kg, calculate the tension T1.
Solution.
m3a = T3 – T2
3a = 12 – T2...(1)
m2 a = T2 – T1
2a = T2 – T1...(2)
from (1) & (2)
5a = 12 – T1...(3)
m1 a = T1
a = T1...(4)
from (3) & (4)
5T1 = 12 – T1
T1 = 2N
We would not be able to walk if there had been no friction between the soles of our shoes and the ground.
If there had been no friction, the wheels of a car will slip instead of rotating and stop moving. For that we have to increase the friction by making the tyres corrugated to get better grip of tyres on the road. Also, the friction is increased.
When brakes are applied, the vehicle stops due to the force of friction between the brake lining and the drum.
In the absence of friction, we cannot write on a blackboard with a chalk stick because the chalk stick will slide off the board without leaving any mark on the board.
Example : 11
Three blocks are tied together with strings as shown. They are lying on a horizontal frictionless table. These are pulled to the right with T3 = 12N force. If m1 = 1 kg, m2 = 2kg, and m3 = 3 kg, calculate the tension T1.
Solution.
m3a = T3 – T2
3a = 12 – T2...(1)
m2 a = T2 – T1
2a = T2 – T1...(2)
from (1) & (2)
5a = 12 – T1...(3)
m1 a = T1
a = T1...(4)
from (3) & (4)
5T1 = 12 – T1
T1 = 2N
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