Friction Moves the Car and Friction Stops the Car

It doesn't matter how beautiful your theory is, it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong.

– Richard Feynman

 
When you think of friction in detail you will find that it is an ordinary force with extraordinary features. Its direction is always self adjusting and is opposite to the direction of the force which tries to move a body. The magnitude of frictional force is self-adjusting up to the limiting value.
Have you ever thought how exactly a car moves forward? The engine of the car does not exert any forward force. It just spins the wheels of the car. The portion of the wheel (tyre) in contact with the ground moves backwards. The frictional force between the wheel and the ground is therefore directed forwards. It is this frictional force which moves the car forward.
When you apply the brakes, the braking mechanism does not apply any backward force. It just stops the spin of the wheels. The forward frictional force ceases. But the car will continue to move forward due to inertia. The wheel will then slide forward. Immediately, a backward frictional force is called into play and the car stops after traveling some distance. Now, consider the following questions:(1) If μ is the coefficient of friction between the road and the tyre of a car, the minimum time in which the car can cover a distance ‘s’, starting from rest is
(a) directly proportional to μ
(b) inversely proportional to μ
(c) inversely proportional to √μ
(d) directly proportional to √μ (e) independent of μThe correct option is not (b) but is (c). We have s = 0 + ½ at². Since the driving force is the frictional force μmg, acceleration a = μg. . Therefore s = ½ μg t² and hence t α 1/√μ. The correct option is (c).(2) If μ is the coefficient of friction between the road and the tyre of a car moving with a velocity ‘v’, the minimum distance in which it can be stopped is
(a) directly proportional to μ
(b) inversely proportional to μ
(c) inversely proportional to √μ
(d) directly proportional to √μ
(e) independent of μThe correct option is (b) which you can obtain using the equation, 0 = v² – 2as, on substituting for a = μg. Thus , stopping distance, s = v²/2μg.
Note this equation. It says that the stopping distance of any vehicle is directly proportional to the square of the speed of the vehicle.(3) A cube of mass 2kg is kept pressed against a vertical wall by applying a horizontal force of 60N. The coefficient of friction between the cube and the wall is 0.5. What is the frictional force between the cube and the wall?
(a) 60N
(b) 30N
(c) 19.6N
(d) 9.8N
(e) zeroDon’t be tempted to chose option (b) as the answer. Option (b) gives the maximum possible frictional force μR, where R is the normal reaction. Here friction is just sufficient to balance the weight of the cube (mg). So the correct option is (c).Certain simple questions may confuse you occasionally. Consider the following question:A wooden block of mass 3kg is resting on an inclined plane of inclination 30˚. If the coefficient of static friction between the block and the plane is 0.8, the frictional force between the block and the plane is (g=10m/s²)
(a) 20.8 N
(b) 24N
(c) 30N
(d) 12.5N
(e) 15NYou may be inclined to calculate the force as μmg cosθ, which yields the value20.8N, which is not the correct answer. Remember, μmg cosθ is the maximum possible frictional force (limiting friction). Since the block is resting on the inclined plane, the frictional force required is just enough to balance the component of the weight along the plane, which is mg sinθ = 15N. The correct option therefore is (e). Since frictional force is self adjusting (up to the limiting value), the value gets automatically adjusted to 15N in the present case.
Now consider the following MCQ:A heavy uniform chain is lying on a horizontal table with a certain portion hanging over one edge. If the coefficient of static friction between the chain and the table is 0.25, what is the maximum fraction of the length of the chain that can hang over the edge of the table?
(a) 10% (b) 20% (c) 33% (d) 45% (e) 50%If ‘L’ is the full length of the chain and ‘l’ the maximum length that can overhang, we have, mlg = μm(L-l)g where m is the mass per unit length (linear density) of the chain. The weight of the hanging portion tries to displace the chain while the frictional force between the table and the remaining portion tries to restore the chain. We have equated their magnitudes in the limiting case. From this we get l = 0.25(L-l) so that l/L = 0.2. The fraction written as percentage is 20%.An insect inside a hemispherical bowlHere is an interesting problem:An insect is trapped in a hemispherical bowl of radius R kept on a horizontal table. Up to what height can it crawl if the angle of friction is θ?
(a) Rcosθ
(b) Rsinθ
(c) R(1-cosθ)
(d) R(1-sinθ)
(e) RtanθAs the insect crawls upwards, the component of its weight along the surface of the bowl increases and the opposing frictional force also increases. When the insect reaches the maximum possible height, the frictional force is maximum and is equal to mg sinθ where θ is the angle of friction. This means that the inclination (with the horizontal) of the tangent to the hemispherical surface at the limiting position of the insect is equal to the angle of friction,θ. The angle between the vertical radius (of the hemispherical surface) and the radius drawn from the limiting position of the insect also is equal to θ, as ascertained from simple geometry.
Therefore, the maximum height attained by the insect is R-Rcosθ = R(1-cosθ).