How does a lubricant help in reducing friction ?

Question: How does a lubricant help in reducing friction ?

Explanation : When a lubricant is added to a machine , it spreads between the two surfaces rubbing each other . On spreading of lubricant , the irregularities present on the surfaces are filled and it forms a thin layer between the surfaces in contact . As a result of it , the contact between the two hard surfaces is replaced by the contact between the hard surfaces and lubricant layer . Due to it the force of friction is reduced considerably .

Blocks A and B of masses m1 and m2 are rest on the fixed wedge. The coefficient of friction between the wedge and the block is μ = 0.5 . If there is no motion then

Q: Blocks A and B of masses m1 and m2 are rest on the fixed wedge. The coefficient of friction between the wedge and the block is μ = 0.5 . If there is no motion then

Numerical

(a) $ \frac{2 – \sqrt{3}}{1 + 2 \sqrt{3}} \le \frac{m_1}{m_2} \le \frac{2 + \sqrt{3}}{2 \sqrt{3}-1} $

(b) $ \frac{\sqrt{3}-1}{1 + 2 \sqrt{3}} \le \frac{m_1}{m_2} \le \frac{2 + \sqrt{3}}{2 \sqrt{3}-1} $

(c) $ \frac{m_1}{m_2} \le \sqrt{3} $

(d) $ \frac{m_1}{m_2} \ge \sqrt{3} $

Click to See Answer :
Ans: (a)

 

A turn of radius 20 m is banked for the vehicle of mass 200 kg going at a speed of 10 m/s. Find the direction and magnitude…

Q: A turn of radius 20 m is banked for the vehicle of mass 200 kg going at a speed of 10 m/s. Find the direction and magnitude of frictional force acting on a vehicle if it moves with a speed a) 5 m/s b) 15 m/s Assume that friction is sufficient to prevent slipping (g = 10m/s2)

Click to See Solution :
Sol: v = 10 m/s

$\large tan\theta = \frac{v^2}{r g} $

$\large tan\theta = \frac{(10)^2}{20 \times 10 } = \frac{1}{2}$

Now, as speed is decreased, force of friction f acts upwards.

$\large N sin\theta – f cos\theta = \frac{m v^2}{r}$ ;

$\large N cos\theta + f sin\theta = m g $ ;

Substituting $\large tan\theta = \frac{1}{2} $ , v = 5 m/s, m = 200 kg and r = 20 m,

We get , f = 300√5 N

 

A car is driven round a curved path of radius 18 m without the danger of skidding. The coefficient of friction between…

Q: A car is driven round a curved path of radius 18 m without the danger of skidding. The coefficient of friction between the tyres of the car and the surface of the curved path is 0.2. What is the maximum speed of the car for safe driving? (g = 10 ms-2)

Click to See Solution :
Sol: Maximum speed is

$\large v = \sqrt{\mu_s g r}$

$\large v = \sqrt{0.2 \times 10 \times 18}$

$\large v = \sqrt{36} = 6 m/s$

 

Two cars of masses m1 and m2 are moving in circles of radii r1 and r2 respectively. Their speeds are such that…

Q: Two cars of masses m1 and m2 are moving in circles of radii r1 and r2 respectively. Their speeds are such that they make complete circle in the same time t . The ratio of their centripetal acceleration is

Click to See Solution :
Sol: As their time period of revolution is same, angular speed is also same centripetal acceleration is

a = ω2 r ;

$\large \frac{a_1}{a_2} = \frac{\omega^2 r_1}{\omega^2 r_2} = \frac{r_1}{r_2}$