Question

Knowing that the coefficient of static friction between the tires and the road is 0.80 for the
car as shown in figure 2. At this car, suddenly brakes are applied when it was moving with
a velocity of 360 inches/sec and it skidded to rest in 240 inches. Determine the magnitude of the normal
reaction and of the friction force at each wheel as the car skidded to rest.

Answer 1

Answer:

bhai pata nahi ye coefficient mein nahi janta

Answer 2

Given:

μ = 0.80

v = 360 inches/s

s = 240 inches.

To find:

1) normal reaction N =?

2) friction force F =?

Explanation:

1)

Normal reaction at the car is equal to the weight of the car.

∴  N = W

But weight is also the product of mass & acceleration due to gravity.

∴  W = mg

⇒ N = W =mg

   N = m(9.81)

   N = 9.81m N

∴ The magnitude of the normal reaction of the car is 9.81 times the mass of the car.

2)

By the third law of kinematics

  $\displaystyle v^2 = u^2 +2as$

Here u = 0

∴ $(360)^2 = 0 + 2a(240)$

 a = 270 inch/$s^2$

Now, a frictional force is the product of mass & acceleration.

∴ F = ma

 F  = 270m N

∴ The required frictional force is 270 times the mass of the car.