Friction and drag

Suppose that you want to move a heavy file cabinet which is standing in the middle of your office into a corner. You push on it, but nothing happens. What is going on?

You exert a force, but there is no acceleration. The net force must be zero.

Which force of equal magnitude points in a direction opposite to the direction of the force you are applying?

The force of static friction (f_s) cancels the applied force when the cabinet is at rest while you are pushing on it.

You push harder. Eventually the cabinet breaks away and starts accelerating. But you have to keep on pushing just to keep it moving with a constant velocity. When you stop pushing, it quickly slows down and comes to rest. Why?

While the cabinet is moving the force of kinetic friction (f_k) opposes the applied force. When it is moving with constant velocity, the two forces exactly cancel.

Where do these frictional forces come from?

Frictional forces are intermolecular forces. These forces act between the molecules of two different surfaces that are in close contact with each other. On a microscopic scale, most surfaces are rough. Even surfaces that look perfectly smooth to the naked eye show many projections and dents under a microscope. The intermolecular forces are strongest where these projections and dents interlock resulting in close contact. The component of the intermolecular force normal to the surfaces provides the normal force which prevents objects from passing through each other and the component parallel to the surface is responsible for the frictional force.

Assume a cabinet is resting on the floor. Nobody is pushing on it. The intermolecular forces are normal to the surface. The force of gravity (red arrow) is balanced by the normal force (black arrow).

Now assume that you are pushing against the cabinet. The cabinet is not moving, but the surface molecules are displaced by microscopic amounts. This results in intermolecular forces which have a component tangential to the surface (the force of static friction). This tangential component opposes the applied force. The net force on the cabinet is zero. The harder you push the greater is the microscopic displacement of the surface molecules and the greater is the tangential component of the intermolecular forces.

When you push hard enough, some of the projections on the surfaces will break off, i.e. some of the surface molecules will be completely displaced. The horizontal component of the intermolecular forces diminishes and no longer completely opposes the applied force. The cabinet accelerates. But while the horizontal component has diminished, it has not vanished. It is now called the force of kinetic friction. For the cabinet to keep accelerating, you have to push with a force greater in magnitude than the force of kinetic friction. To keep it going with constant velocity you have to push with a force equal in magnitude to the force of kinetic friction. If you stop pushing, the force of kinetic friction will produce an acceleration in the opposite direction of the velocity, and the cabinet will slow down and stop.

The frictional force always acts between two surfaces, and opposes the relative motion of the two surfaces.

The maximum force of static friction between two surfaces is roughly proportional to the magnitude of the force pressing the two surfaces together. The proportional constant is called the coefficient of static friction m_s. The magnitude of the force of static friction is always smaller than or equal to m_sN, We write f_s£ m_sN, where f_s is the magnitude of the frictional force and N is the magnitude of the force pressing the surfaces together. For the cabinet and the floor, N is the weight of the cabinet. The coefficient of static friction is a number (no units). The rougher the surface, the greater is the coefficient of static friction. As long as the applied force has a magnitude smaller than m_sN, the force of static friction f_s has the same magnitude as the applied force, but points in the opposite direction.

The magnitude of the force of kinetic friction acting on an object is f_k= m_kN, where m_k is the coefficient of kinetic friction. For most surfaces, m_k is less than m_s.

If you put your file cabinet on a small cart with wheels, you can greatly reduce the force you need to apply to move the cabinet with constant speed. Wheels do not slide across the floor, they roll across the floor. At any instant only a very small portion of the surface of a wheel is in contact with the floor. If you push on the cart, static friction prevents this portion from sliding. As the wheel rolls, a new portion of the wheel is lowered onto the floor, while the portion previously in contact with the floor is raised. Pure rolling involves only static friction, not kinetic friction.

Problems:

A racing car accelerates uniformly from 0 to 80mi/h in 8s. The magnitude of the force that accelerates the car is approximately equal to the magnitude of the frictional force between the tires and the road. If the tires do not spin, determine the minimum coefficient of static friction between the tires and the road.

Solution:

The magnitude of the acceleration is a = (80mile/h)/8s = (35.76m/s)/8s = 4.47m/s².
The magnitude of the force accelerating the car is f_s= ma £ m_sN. The normal force is N = mg.
We therefore have ma £ m_smg, or a £ m_sg, m_s³ a/g = 4.47/9.8 = 0.46.

A woman at an airport is towing her 20kg suitcase at constant speed by pulling on a strap at an angle of q above the horizontal. She pulls on the strap with a 35N force, and the frictional force on the suitcase is 20N.
(a) Draw a free body diagram of the suitcase.
(b) What angle does the strap make with the horizontal?
(c) What normal force does the ground exert on the suitcase?

Solution:

	(a)
	(b) The suitcase moves with constant velocity, the net force on the suitcase is zero. F_x= 0 implies 35Ncosq = 20N, or cosq = 20/35, q = cos^-1(20/35) = 55.15^o.
	(c) F_y= 0 implies mg = n + 35Nsinq, 20kg´9.8m/s²= n + 35N´0.82, n = 167.3N.

Determine the stopping distance for a skier with a speed of 20m/s on a slope that makes an angle q with the horizontal. Assume m_k= 0.18 and q = 5^o.

Solution:

The frictional force is proportional to the normal force, which is mgcosq. We have f_k= m_kmgcosq = ma. The acceleration produced by this force is a = m_kgcosq = 0.18´ 9.8(m/s²)´cos(5^o) = 1.76m/s² in a direction opposite to the velocity. Gravity produces an acceleration a = gsinq = 9.8(m/s²)´ sin(5^o) = 0.85m/s² in the direction of the velocity. The total acceleration is 0.91m/s² in a direction opposite to the velocity. The acceleration is constant. We therefore have 0 - v₀= -at, t = v₀/a = (20/0.91)s = 21.98s for the time it takes the skier to come to a stop. In this time he travels a distance s = v₀t - (1/2)at². s = 219.8m.

Suppose you are driving a car along a highway at a high speed. Why should you avoid slamming on your brakes if you want to stop in the shortest distance? That is, why should you keep the wheels turning as you brake?

Solution:

When the wheels are turning, friction between the surfaces of the brake pads and the disks or drums of the wheels is responsible for decelerating the car. When the wheels are locked, friction between the tires and the road decelerates the car. The brake pads are designed for the job, and the coefficient of kinetic friction between the brake pads and the disk or drum is large.

Links:

	Frictional force
	Friction and inclined planes

Exercise (You can earn up to 5 points extra credit by completing this exercise.)

Drag

The frictional force acts between surfaces. But an object moving through a medium such as a gas or a liquid is also acted on by a resistive force. This force always points in a direction opposite to the direction of the velocity of the object, but, unlike friction, its magnitude depends on the speed of the object.

For objects moving through a liquid or for small objects moving slowly through a gas, the magnitude of the resistive force, R, is often proportional to the speed. We write

R = -bv.

The equation of motion for or a stone falling through water to the bottom of a lake (motion in one dimension, along a line) is

F = ma,

which leads to

mg - bv = mdv/dt

dv/dt = g - (b/m)v.

The solution to this differential equation is

v = (g/k) - [(kv₀+g)/k]e^-kt,

where k=b/m.

For a stone starting from rest (v₀= 0), v = (g/k)(1 - e^-kt).

At t = 0, e^-kt= 1 and the speed of the stone is v₀. As t becomes very large, e^-kt approaches zero, and v approaches v_t= g/k, the terminal speed. The stone reaches terminal speed when the resistive force R is equal in magnitude to the gravitational force mg.

We can find v_t without solving the differential equation by simply setting dv/dt = 0. When the speed does not change any longer, then the stone has reached terminal speed.

For objects moving at high speed through air the magnitude of the resistive force is often proportional to the square of the speed, and can be written as

R = (1/2)DrAv².

Here A is the cross sectional area of the falling object in a plane perpendicular to its velocity, r is the density of the air, and D is the drag coefficient, which depends on the shape of the object. For a spherical object D has a value of approximately 0.5.

For objects moving at high speed through air the equation of motion is

F = ma,

which leads for 1-dimensional motion in the vertical direction near the surface of the earth to

mg - (1/2)DrAv²= mdv/dt

dv/dt = g - DrAv²/(2m)

The object again reaches a terminal speed v_t. We find v_t without solving the differential equation by simply setting dv/dt = 0.

Problems:

On long journeys, jet aircraft usually fly at high altitudes of about 30000ft. What is the main advantage of flying at these altitudes from an economic viewpoint?

Solution:

The magnitude of the resistive force for an object moving with high speed through air increases as the density of the air increases. At high altitudes, the air density is lower and therefore the magnitude of the resistive force is smaller than at low altitudes.

A diver of mass 80 kg jumps from a slow-moving aircraft and reaches a terminal speed of 50m/s.
(a) What is the acceleration of the skydiver, when her speed is 30m/s?
(b) What is the drag force on the diver when her speed is 50 m/s?
(c) What is the drag force on the diver when her speed is 30 m/s?

Solution:
Given: m = 80kg, v_t= 50m/s.

	(a) Using we can find DrA. DrA = 2mg/v_t²= 0.6272kg/m. Therefore a = dv/dt = 9.8m/s²- (0.6272/160m)v². When v = 30m/s, a = 6.272m/s².
	(b) When v = v_t= 50m/s, the drag force is equal in magnitude to mg = 784N.
	(c) When v = 30m/s, the total force is equal in magnitude to ma = 80kg 6.272m/s²= 502N. The magnitude of the drag force is mg - ma = 282N.

Links:

	Skydiving
	Freefall lab

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