Question

A motor van weighing 4400 kg rounds a level curve of radius 200 m on unbanked road at 60 km / hr. What should be minimum value coefficient of friction to prevent skidding? A what angle the road should be banked for this velocity ? (Ans : 0.1417, 8° 4')

Answer 1

Answer:

co-efficient of friction = 3.11

Explanation:

Given,

Mass of motor van,m = 4400 kg

radius of circular path,r = 200 m

speed of the circular path,v = 60 km/hr

                                            $=\dfrac{60\times 1000}{60\times 60}$ m/s

                                            = 16.66 m/s

Hence, the centrifugal fiorce on the van will be given by,

$F\ =\ \dfrac{m.v^2}{r}$

   $=\ \dfrac{4400\times 16.66^2}{200}$

  $=\ 6106.22\ N$

Since, we have to calculate the value of co-efficient of friction to prevent friction.

So, centrifugal force = frictional force

$=>\ 6106.22 N\ =\ \mu\times r\times g$

$=>\ \mu\ =\ 3.11$

Hence, the value of co-efficient of friction will be 3.11.

Answer 2

Answer:

Co-efficient of friction = 0.1417

and,

Angle of banking = 8.06 degrees

Explanation:

In the question,

Mass of the van = 4400 kg

Radius of the curve, r = 200 m

Speed of the vehicle, v = 60 km/hr = 16.66 m/s

Let, us say the co-efficient of friction is given by μ.

So,

To prevent the skidding we need to put the centripetal force and Force of friction equal.

So,

Force of friction, f = μN

and,

Centripetal force is,

$F=\frac{mv^{2}}{r}$

So,

Also,

Normal, N = mg

So,

$\frac{mv^{2}}{r}=\mu N\\\frac{(4400)(16.66)^{2}}{200}=\mu (4400\times 9.8)\\\mu = 0.1417$

Therefore, the co-efficient of friction is 0.1417.

Now,

Also,

$v=\sqrt{Rgtan\theta}\\16.666=\sqrt{200\times 9.8\times tan\theta}\\tan\theta = 0.1417\\\theta=8.06\ degree$

Therefore, the angle is also given by 8.06 degrees.