Mod 5

Module - 5: Force & Friction Introduction: The science of mechanics is based on three natural laws relating force and motion. These were clearly stated for the first time by Sir Isaac Newton [1642 – 1727] and were published in 1686 in his Philosophiae Naturalis Principia Mathematica. Newton’s three laws relate an object’s acceleration to its mass and the forces acting on it. A modern wording of Newton’s laws follows:

I. Newton's First Law of Motion: Every object continues to be at rest or in a state of uniform motion unless acted on by an external force. This we recognize as essentially Galileo's concept of inertia, and this is often termed simply the "Law of Inertia". To say that something is moving always implies a specific frame of reference. An inertial frame of reference is one in which Newton’s first law of motion holds.

II. Newton's Second Law of Motion: Newton's second law of motion explains how an object will change velocity if it is pushed or pulled upon. The rate of change of momentum of a body is directly proportional to the applied force acting on the body. Firstly, this law states that if you do place a force on an object, it will accelerate, i.e., change its velocity, and it will change its velocity in the direction of the force. It accelerates in the direction………….. That you push it. Secondly, this acceleration is directly proportional to the force. For example, if you are pushing on an object, causing it to accelerate, and then you push, say, three times harder, the acceleration will be three times greater. If you push twice as hard………….. It accelerates twice as much.

F∝a

Thirdly, this acceleration is inversely proportional to the mass of the object. For example, if you are pushing equally on two objects, and one of the objects has five times more mass than the other, it will accelerate at one fifth the acceleration of the other. If it gets twice the mass…………….. It accelerates half as much.

a = F/m

III. Newton's Third Law of Motion: The word force is used to describe the interaction between two objects. When two objects interact, they exert force on each other. Newton’s third law states that these forces are equal in magnitude and opposite in direction. For example, if you push on a wall, it will push back on you as hard as you are pushing on it. If you push on it………………… It pushes on you. The full power of Newton’s second law emerges when it is combined with the force laws that describe the interactions of objects. For example, Newton’s law for gravitation, gives the gravitational force exerted by one object on another in terms of the distance between the objects and the masses of each. This, combined with Newton’s second law, enables us to calculate the orbits of planets around the sun, the motion of the moon, and variations with altitude of g, the acceleration due to gravity.

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The fundamental forces in nature: All the different forces observed in nature can be explained in terms of four basic interactions that occur between elementary particles: 1. The gravitational force 2. The electromagnetic force 3. The strong nuclear force 4. The weak nuclear force The everyday forces that we observe between macroscopic objects are due to either the gravitational force or the electromagnetic force. Forces may be placed into two broad categories, based on whether the force resulted from the contact or non-contact of the two interacting objects. Action at a distance: The fundamental forces of gravity and electromagnetism act between particles that are separated in space. This creates a philosophical problem referred to as action at a distance. Contact forces: Many forces we encounter are exerted by objects in direct contact. These forces are electromagnetic in origin and are exerted between the molecules of each object. Normal force: Consider a book on a table. The weight of the book pulls it downward, pressing it against the molecules in the table’s surface, which resist compression and exert a force upward on the book. Such a force, perpendicular to the surface, is called a normal force. Frictional force: Objects in contact can also exert forces on each other that are parallel to the surfaces in contact. The parallel component of a contact force is called a frictional force. Static friction: Friction is a complicated, incompletely understood phenomenon that arises due to the bonding of molecules between two surfaces that are in close contact. This bonding is the same as the molecular bonding that holds an object together. When you apply a small horizontal force to a large box resting on the floor, the box may not move because of the force of static friction,

 f s′ exerted by the floor on the box,

balances the force you are applying. The force of static friction, which opposes the applied force, can adjust from zero to some maximum force f s, max depending on how hard you push. You might expect f s, max to be proportional to the area of contact between the two surfaces, but this is not the case. To a good approximation, f s, max is independent of the area of contact and is simply proportional to the normal force exerted by one surface on the other: f s, max = µ s Fn where, µs is called the coefficient of static friction, a dimensionless quantity that depends on the nature of the surfaces in contact. If you exert a horizontal force smaller than f s, max on the box, the frictional force will just balance this horizontal force. In general, we can write f s ≤ µ s Fn Kinetic friction: If you push the box hard enough, it will slide across the floor. When the box is sliding, molecular bonds are continually being formed and ruptured, and small pieces of the surfaces are being broken off. The result is a force of kinetic friction,

 f k that opposes the motion. To keep the box sliding with constant

velocity, you must exert a force on the box that is equal in magnitude and opposite in direction to the force of kinetic friction exerted by the floor.

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The coefficient of kinetic friction µ k is defined as the ratio of magnitudes of the kinetic frictional force f and the normal force Fn:

k

f k =µ k Fn where µ k depends on the nature of the surfaces in contact. Experimentally, it is found that µ k is less than µ s and is approximately constant for speeds ranging from about 1 cm/s to several meters per second.

The plot of the frictional force vs. the applied force illustrates some of the features of the frictional force. Note that the frictional force equals the applied force (in magnitude) until it reaches the maximum possible value µsN. Then the object begins to move as the applied force exceeds the maximum frictional force. When the object is moving the frictional force is kinetic and roughly constant at the value µkN which is below the maximum static friction force.

Examples of kinetic friction: ♦

Sliding friction is when two objects are rubbing against each other. Putting a book flat on a desk and moving it around is an example of sliding friction.

♦

Rolling friction occurs when the two objects are moving relative to each other and one "rolls" on the other (like a car's wheels on the ground). The coefficient of rolling friction is typically denoted as μ r.

♦

Fluid friction is the friction between a solid object as it moves through a liquid or a gas. The drag of air on an airplane or of water on a swimmer are two examples of fluid friction.

Contact Forces Frictional Force Tensional Force Normal Force Air Resistance Force Applied Force Spring Force

Action-at-a-Distance Forces Gravitational Force Electrical Force Magnetic Force

Causes of friction: Friction is caused by the roughness of the materials rubbing against each other, deformations in the materials, and a molecular attraction between molecules of two surfaces. 1. Surfaces not completely smooth: Most friction results because the surfaces of materials being rubbed together are not completely smooth. If you looked at what seems to be a smooth surface under a microscope, you would see bumps, hills and valleys that would interfere with sliding motion. Of course, the rougher the surface, the more is the friction. If both surfaces become ultra-smooth and flat, the friction from surface roughness becomes negligible, but then friction from molecular attraction comes into play, often becoming greater than the normal friction.

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2. Deformations: Soft materials will deform when under pressure. This also increased the resistance to motion. For example, when you stand on a rug, you sink in slightly, which causes resistance when you try to drag your feet along the rug's surface. Another example is how rubber tires flatten out at the area on contact with the road. When materials deform, you must "plow" through to move, thus creating a resistive force. 3. Molecular attraction: There is another factor in friction, and that is stickiness caused by molecular attraction. This was mentioned above where surfaces are so smooth that the materials stick together due to molecular forces. Soft rubber is an example of a material that can have this type of friction. This factor is usually seen in rolling friction. The stickiness will create a resistance to any motion. Although this force is the smallest, it still can be a factor when the other causes of friction are low.

Reducing Friction: A common way to reduce friction is by using a lubricant, such as oil, that is placed between the two surfaces, often dramatically lessening the coefficient of friction. The science of friction and lubrication is called tribology. Superlubricity, a recently-discovered effect, has been observed in graphite: it is the substantial decrease of friction between two sliding objects, approaching zero levels - a very small amount of frictional energy would be dissipated due to electronic and/or atomic vibrations. Lubricants to overcome friction need not always be thin, turbulent fluids or powdery solids such as graphite and talc; acoustic lubrication actually uses sound as a lubricant.

Component of the force acting on the object (Strings, pulleys, and inclines):

Figure: Block sliding down an incline

Figure: Block dragged over a horizontal surface

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Figure: Block suspended by a string Figure: Block suspended by three strings

Figure: Block sliding over a smooth table, pulled by a second block

Figure: An Atwood machine

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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 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)DρAv2. Here A is the cross sectional area of the falling object in a plane perpendicular to its velocity, ρ is the density of the fluid (liquid or gas), 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)DρAv2 = mdv/dt or dv/dt = g - DρAv2/(2m) The object again reaches a terminal speed vt. We find vt without solving the differential equation by simply setting dv/dt = 0.

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Type of Force and its Symbol

Description of Force

Applied Force F app

An applied force is a force which is applied to an object by another object or by a person. If a person is pushing a desk across the room, then there is an applied force acting upon the desk. The applied force is the force exerted on the desk by the person.

The force of gravity is the force with which the earth, moon, or other massive body attracts an object towards itself. By definition, this is the weight of the object. All objects upon earth Gravity Force (also experience a force of gravity which is directed "downward" towards the center of the earth. known as Weight) The force of gravity on an object on earth is always equal to the weight of the object as given by the equation: F grav F grav = m * g where: g = acceleration of gravity = 9.8 m/s2 (on Earth) and m = mass (in kg)

Normal Force F norm

The normal force is the support force exerted upon an object which is in contact with another stable object. For example, if a book is resting upon a surface, then the surface is exerting an upward force upon the book in order to support the weight of the book. On occasion, a normal force is exerted horizontally between two objects which are in contact with each other. The friction force is the force exerted by a surface as an object moves across it or makes an effort to move across it. The friction force opposes the motion of the object. For example, if a book moves across the surface of a desk, the desk exerts a friction force in the direction opposite to the motion of the book.

Friction Force F frict

Friction results when two surfaces are pressed together closely, causing attractive intermolecular forces between the molecules of the two different surfaces. As such, friction depends upon the nature of the two surfaces and upon the degree to which they are pressed together. The friction force can be calculated using the equation:

Air resistance is a special type of frictional force which acts upon objects as they travel Air Resistance Force through the air. Like all frictional forces, the force of air resistance always opposes the motion of the object. This force will frequently be ignored due to its negligible magnitude. It is Fair most noticeable for objects which travel at high speeds (e.g., a skydiver or a downhill skier) or for objects with large surface areas. Tensional Force F tens

Tension is the force which is transmitted through a string, rope, or wire when it is pulled tight by forces acting at each end. The tensional force is directed along the wire and pulls equally on the objects on either end of the wire.

Spring Force F spring

The spring force is the force exerted by a compressed or stretched spring upon any object which is attached to it. This force acts to restores the object, which compresses or stretches a spring, to its rest or equilibrium position. For most springs (specifically, for those said to obey "Hooke's Law"), the magnitude of the force is directly proportional to the amount of stretch or compression

Consider the simple experiment from Figure 1.

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Figure 1. On the table there is a box. The gravitational force mg is compensated by reaction force FR as Newton’s third law requires. We apply a force F to it, which is measured by the force gauge G. We slowly increase the force and the box does not move. This means that frictional force Ff is increasing as well and is all the time equal to F. At some value of the pulling force the box starts moving. For a moment it will accelerate and we can decrease the pulling force F to the value resulting in motion with constant velocity. Motion with constant velocity means that the net force acting on box is ZERO. This means that while the box is moving the force of smaller value is required to compensate the frictional force. The conclusion is, there are two types of frictional forces: static frictional force FS and kinetic frictional force FK The frictional force in the case of solids moving against each other are defined by Equations FS = μS mg (1) FK = μK mg (2) where μS and μK are coefficients of static and kinetic friction. Their values depend on certain properties of the moving bodies and on the quality of surfaces which touch each other. This is obvious to every child. We cannot slide on our shoes on asphalt, but can easily do it on ice. The term surface means much more than the surface of a table, floor, road or any other surface from our daily life. Later on we will discuss the frictional forces between molecules in liquid or gas and in such cases the surface of molecules plays a role in the formulas defining these frictional forces. A very important field related to friction is motion of solid objects in the air or in water. In this case the frictional force is called drag force. In spite of a quite different name the drag force is also a frictional force and only the mechanism which creates this type of friction is very different from the one creating the friction between two solid objects. More details concerning frictional forces will be presented in problems, as this physics tutorial is a kind of classroom to help you understand physics through solving numerical examples.

Atwood machine - testing Newton’s Laws of motion:

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The Rev. George Atwood (1746-1807) developed a simple construction known as the Atwood machine which gives students an exceptionally simple but excellent tool for testing and verifying Newton’s Laws of Motion. The idea of the Atwood Machine is explained on Fig.1

A pulley is mounted about 2.0m from the ground on a stand with a scale on it to measure the positions of attached objects. Two weights are attached to both ends of a long string which passes over the pulley. In the free body diagram of the Atwood's machine, T is the tension in the string, m is the smaller mass, M is the larger mass, and g is gravity’s acceleration. Assuming that the pulley and the string are massless and the string doesn't stretch, and that there is no friction, the resultant force on M is the difference between the tension and mg . The net force on M is the difference between the tension and Mg. The resultant force exerted on both masses is F = Mg - mg = (M - m)g (1) and this force is accelerating both masses at a = F/(M+m) (2) Substituting (1) into (2) we get

(3) If the difference M-m is relatively small, acceleration a is easy to measure with a simple stopwatch. If we assume the knowledge of gravitational acceleration g, the Atwood Machine allows us to check Newton’s Second Law of Motion. When M = m we can verify Newton’s First Law of Motion. If we assumed Newton’s Second Law of Motion as true, we can then determine, from Equation (3), the value of gravitational acceleration g. The applications of the Atwood Machine are quite universal, especially for educational purposes. Endless generations of students, including myself have learned Newton’s Laws of Motion with the Atwood Machine. Some of the { Problems link do problems } included in this Chapter are classical examples of the application of the Atwood Machine idea. At the Atwood Machine web page you can see photos of many practical realizations of the equipment described above. You can also read there some historical information if you are interested in such matters.

Solved problems: 9

1. A woman pulls a loaded sled of mass m = 75 kg along a horizontal surface at constant velocity. The coefficient of kinetic friction µ k between the runners and the snow is 0.10, and the angle θ is 42 o. (i) What is the tension T in the rope? (ii) What is the normal force with which the snow pushes vertically upward on the sled?

FN

Solution:

FT

[i]

Ffk

θ

x-component of the forces:

x

FT cosθ – Ffk = 0, ………[1] y-component of the forces:

mg FT sin θ + FN - mg = 0, ………..[2]

We know that the force of kinetic friction Ffk can be written as Ffk = µk FN, …..[3] Combining [1] & [3], we get FN = FT cosθ/µk, …..[4] Combining [2] & [4], we get FT [µk sin θ + cos θ] = µk mg Or,

FT =

µ k mg 0.1x75 x9.8 = = 90.73 N µ k sin θ + cos θ 0.1x sin 42 0 + cos 42 0

[ii]

FT cos θ 90.73 x cos 42 0 = = 674.29 N Normal force = Fn = µk 0.1 2. A Crate of dilled pickles with mass m1 = 14 kg moves along a plane that makes an angle of θ = 30o with the horizontal. That crate is connected to a crate of pickled dills with mass m 2 = 14 kg by a taut, mass-less cord that runs around a frictionless, mass-less pulley. The hanging crate of dills descends with constant velocity. (i) What are the magnitude and direction of the frictional force exerted on m1 by the plane? (ii) What is µ k ? Solution: [i]

FN

FT

X - component of the forces acting on the sliding block:

FT − F fk − mg sin θ = 0,.....[1] Y - component of the forces acting on the hanging block:

mgsinθ Ffk mg

FT − mg = 0, or , FT = mg ,....[2] Using [2] in [1], we get

F fk = mg[1 − sin θ ] = 14 x9.8 x[1 − sin 30 0 ] = 68.6 N [ii]

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mgcosθ

FT m mg

Y - component of the forces acting on the sliding block:

FN = mg cos θ ,.....[3] We also know that

F fk = µ k FN , or , µ k =

F fk FN

=

F fk mg cos θ

=

68.6 = 0.58 14 x9.8 x cos 30 0

3. A girl of mass mg sits on a toboggan of mass m t, which in turn sits on a frozen pond assumed to be frictionless. The toboggan is pulled with a horizontal force F. The coefficients of static and sliding friction between the girl and toboggan are µ s and µ k . (i) Find the maximum value of F for which the girl will not slide relative to the toboggan. Find the acceleration of the toboggan and girl when F is greater than this value. Solution:

FN

FN

FT

FT Ffk

θ

Ffk

θ

x Girlmg

x Toboggan

mg

[i] The only force accelerating the girl forward is the frictional force exerted by the toboggan on the girl. X - component of the forces acting on the girl and the toboggan can be written as:

  f = mG a G    F − f ′ = mt a t    or , F = f ′ + mt at Using

  f = f ′ , we get        F = f ′ + m t a t = f + m t a t = m G a G + mt a t

If

   aG = at = a = the common acceleration, then we can write   F = [ mG + mt ] a

Now

    f f ,max = µ s FN 1 = µ s mG g = mG a max   ∴ a max = µ s g The maximum force for which the girl will not slide is then

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   Fmax = [mG + mt ] a max = [ mG + mt ] µ s g When the force is greater than this value, then the force of kinetic friction acts on the girl and the toboggan. In this case, the horizontal force acting on the girl is that of kinetic friction. Therefore,

    f k = µ k FN 1 = µ k mG g = mG aG   ∴ aG = µ k g The horizontal component of the forces acting on the toboggan can be written as

   F − f k = mt a t     F − µ k mG g  F − fk ∴ at = = mt mt 

Remember when the girl slips on the toboggan, the net force on the girl is constant [ µ k mG g ], so the acceleration of the girl is constant.

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Problems 1. A block of mass m = 15 kg held by a cord on a frictionless inclined plane. What is the tension in the cord if θ = 27o? What force does the plane exert on the block?

2. Two blocks connected by a cord that passes over a mass-less frictionless pulley. Let m = 1.3 kg and M = 2.8 kg. Find the tension in the cord and the magnitude of the acceleration of the two blocks. 3. A sliding block of mass M = 3.3 kg. The block is free to move along horizontal frictionless surface. This block is connected by a cord that warps over a frictionless pulley to a second block, which is hanging of mass m = 2.1 kg. The cord and pulley has negligible mass compared to the blocks. The hanging block falls as the sliding block accelerates to the right. Find (i) the acceleration of the sliding block (ii) acceleration of the hanging block and (iii) the tension in the cord.

4. A sphere of mass 3.0x10-4 kg is suspended from a cord. A steady horizontal breeze pushes the sphere so that the cord makes a constant angle of 37 o with the vertical. Find (i) the magnitude of the push and (ii) the tension in the cord. 5. A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 24° above the horizontal. (i) If the coefficient of static friction is 0.47, what minimum force magnitude is required from the rope to start the crate moving? (ii) If the coefficient of kinetic friction = 0.29, what is the magnitude of the initial acceleration of the crate?

6. Two children are pulled on a sled over snow-covered ground. The sled, which is initially at rest, is pulled by a rope that makes an angle of 40o with the horizontal. The children have a combined mass of 45 kg and the sled has a mass of 5 kg. The coefficients of static and kinetic frictions are

µ s = 0.2 and µ k = 0.15. Find the frictional force exerted by the ground on the sled and the acceleration of the children and sled, starting from rest, if the tension in the rope is (i) 100 N and (ii) 140 N.

7. A car is traveling at 30 m/s along a horizontal road. The coefficients of friction between the road and the tires are µ s = 0.5 and µ k = 0.3. How far does the car travel before stopping if (i) the car is braked with antilock braking system so that the wheels do not slip, and (ii) the car is braked hard with no antilock braking system so that the wheels lock?

8. The mass m2 in Figure has been adjusted so that the block m1 is on the verge of sliding. [a] If m1 = 7 kg and m2 = 5 kg, what is the coefficient of static friction between the shelf and the block? [b] With a slight nudge, the blocks move with acceleration a. Find a if the coefficient of kinetic friction between the shelf and the block is µ k = 0.54.

9. A runaway baby buggy is sliding without friction across a frozen pond toward a hole in the ice. You race after the buggy on skates. As you grab it, you and the buggy are moving toward the hole at speed v0. The coefficient of friction between your skates and the ice as you turn out the blades to brake is µ k . D is the distance to the hole when you reach the buggy, M is the total mass of the buggy and m is your mass. [a] What is the least value of D such that you stop the buggy before it reaches the hole in the ice? [b] What force do you exert on the buggy?

10. A block rests on an inclined plane surface. The angle of inclination is increased until it reaches a critical angle Θ c , after which the block begins to slide. Find the coefficient of static friction µ s .

11. A woman at an airport is towing her 20kg suitcase at constant speed by pulling on a strap at an angle of θ 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.

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(b) What angle does the strap make with the horizontal? (c) What normal force does the ground exert on the suitcase?

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

13. 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. 14. 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?

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