PHY 151: Lecture Forces of Friction 5.9 Newton’s Second Law

PHY 151: Lecture 5 The Laws of Motion 5.8 Forces of Friction

Forces of Friction - 1 When an object is in contact with a surface, there is a force acting on the object Component of this force that is perpendicular to the surface is called the normal force When the object moves or attempts to move along the surface, there is also a component of force that is parallel to the surface This parallel component is called the frictional force, or friction

Forces of Friction - 2 Friction is proportional to the normal force –ƒ s  µ s n and ƒ k = µ k n μ is the coefficient of friction –These equations relate the magnitudes of the forces; they are not vector equations –For static friction, the equals sign is valid only at impeding motion, the surfaces are on the verge of slipping –Use the inequality for static friction if the surfaces are not on the verge of slipping

Forces of Friction - 3 The coefficient of friction depends on the surfaces in contact The force of static friction is generally greater than the force of kinetic friction The direction of the frictional force is opposite the direction of motion and parallel to the surfaces in contact The coefficients of friction are nearly independent of the area of contact

Static Friction Magnitude When two surfaces are not in relative motion, the force of friction is called static friction The magnitude f s of the static frictional force can have any value from zero up to a maximum value of f s Max, depending on the applied force In this equation,  s is the coefficient of static friction, and n is the magnitude of the normal force

Static Friction Static friction acts to keep the object from moving As long as the object is not moving, ƒ s = F If increases, so does If decreases, so does ƒ s  µ s n –Remember, the equality holds when the surfaces are on the verge of slipping

Kinetic Friction Magnitude When two surfaces are in relative motion, the force of friction is called kinetic friction Magnitude of the kinetic frictional force is given by f k f k =  k n In this equation,  k is the coefficient of kinetic friction, and n is the magnitude of the normal force

Kinetic Friction The force of kinetic friction acts when the object is in motion Although µ k can vary with speed, we shall neglect any such variations ƒ k = µ k n

Friction Force Direction Static Friction  The direction is opposite to the direction in which the object would move if there was no friction Kinetic Friction  The direction is opposite to the direction in which the object is actually moving

Some Coefficients of Friction

Friction in Newton’s Laws Problems Friction is a force, so it simply is included in the in Newton’s Laws The rules of friction allow you to determine the direction and magnitude of the force of friction

Friction Force Example - 1 In moving a 35.0-kg desk from one side of a classroom to the other, a professor finds that a horizontal force of 275 N is necessary to set the desk in motion, and a force of 195 N is necessary to keep it in motion at a constant speed What are the coefficients of (a) static and (b) kinetic friction between the desk and the floor? (a)  Static friction  f s =  s N   s = f s /N = f s /mg = 275/(35)(9.8) = (b)  Kinetic friction  f k =  k N   k = f k /N = f k /mg = 195/(35)(9.8) = 0.569

Friction Force Example - 2 An object has an initial velocity of 20 m/s Object slides on surface with  k = 0.8 How far does the object slide until it stops?  Vertical F = N – mg = ma = 0 N = mg  Horizontal Friction = -  k N = -  k mg F = -  k mg = ma a = -  k g  Stopping distance v f 2 = v i 2 + 2ax (we will soon learn this formula) 0 = v i 2 – 2  k gx x = v i 2 /2  k g = 20 2 /2/(.8)/9.8 x = 25.5 m

Friction Force Example - 3  k = 0.4 No friction  F = 400 = ma = 50a  a = 8 m/s 2 With friction Vertical  F = ma = 0  -mg + N = 0  N = mg = 50(9.8) = 490 N Horizontal  f k =  k N = 490(0.4) = 196 N  F = P – f k = ma  400 – 196 = 50a  a = 204/50 = 4.08 m/s 2

Friction Force Example - 4  k = 0.4 No friction Horizontal  P x = 400cos36.9 = 320  P y = 400sin36.9 = 240  F = P x = 320 = ma = 50a  a = 320/50 = 6.4 m/s 2 With friction Vertical  F = -mg + N + P y = N = 0  N = 490 – 240 = 250 Horizontal  f k =  k N = 0.4(250) = 100  F = -f k + P x = = ma = 50a  a = 220/50 = 4.4 m/s 2

Friction Force Example - 5  k = 0.4 No friction Horizontal  P x = 400cos36.9 = 320  P y = -400sin36.9 = -240  F = P x = 320 = ma = 50a  a = 320/50 = 6.4 m/s 2 With friction Vertical  F = -mg + N + P y = N = 0  N = = 730 Horizontal  f k =  k N = 0.4(730) = 292  F = -f k + P x = = ma = 50a  a = 28/50 = 0.56 m/s 2

Friction Force Summarize Examples 3, 4, 5 Example 4 with friction has greater a than Example 3 because box lifted up decreasing friction Example 5 with friction has lower a than Example 4 because box is pushed down increasing friction Note: a is acceleration Ex 3 a m/s 2 Horz Ex 4 a m/s 2 Pull Up Ex 5 a m/s 2 Push Down No Friction Friction

PHY 151: Lecture 5 The Laws of Motion 5.9 Problems Using Newton’s Second Law

Problems Using Newton’s Second Law Assumptions –Objects can be modeled as particles –Interested only in external forces acting on the object Can neglect internal forces –Initially dealing with frictionless surfaces –Masses of strings or ropes are negligible The force the rope exerts is away from the object and parallel to the rope When a rope attached to an object is pulling it, the magnitude of that force is the tension in the rope

Particle in Equilibrium If the acceleration of an object that can be modeled as a particle is zero, the object is said to be in equilibrium Mathematically, the net force acting on the object is zero

Particle Not in Equilibrium If the acceleration of an object is not zero, the object is said to be not in equilibrium Mathematically, the net force acting on the object is zero

Equilibrium Example – 1a You and two friends find three ropes tied together with a single knot and decide to have a three-way tug-of-war Alice pulls to the west with 100 N of force Bob pulls to the south with 200 N How hard, and in which direction, should you pull to keep the knot from moving?

Equilibrium Example – 1b   F x = T 1x + T 2x + T 3x   F y = T 1y + T 2y + T 3y  T 1x = - T 1 T 2x = 0T 3x = T 3 cos   T 1y = 0T 2y = - T 2 T 3y = T 3 sin   -T 1 + T 3 cos  = 0  -T 2 + T 3 sin  = 0  T 1 = T 3 cos   T 2 = T 3 sin  T 2 / T 1 = tan   = tan -1 (T 2 /T 1 )  = tan -1 (200/100)= T 3 = T 1 /cos  T 3  = 100/cos(63.4) T 3 = 223 N

Normal Force - 1 Normal force is one component of the force that a surface exerts on an object with which it is in contact This is component that is perpendicular to the surface Depending on the physical situation, the Normal Force can be vertical, horizontal, or some other direction For example: –Block exerts a force on table by pressing down on it –Consistent with the third law, table exerts an oppositely directed force of equal magnitude on block –This reaction force is the normal force

Normal Force - 2 The normal force is not always equal to the gravitational force of the object For example, in this case may also be less than

Normal Force Apparent Weight There are situations in which a scale does not give correct weight In such situations, the reading on the scale gives only the “apparent” weight, rather than the gravitational force or “true” weight The apparent weight is the force that the object exerts on the scale with which it is in contact Consider a person on a scale in an elevator:  If the elevator is not moving or moving with constant velocity, the scale registers the true weight  If the elevator is accelerating, apparent weight and the true weight are not equal  When the elevator accelerates upward, the apparent weight is greater than the true weight

Apparent Weight Example - 1 Consider a person on a scale in an elevator Normal force of scale on person is N in up direction Weight of gravity, mg, on person is in down direction Newton’s second law gives  F = N – mg = ma  N = ma + mg “a” can be either + for up acceleration or – for down acceleration N is the force the scale exerts on the person By Newton’s third law, N is the force the person exerts on the scale, the apparent weight

Apparent Weight Example - 2 A man of mass 80 kg stands in an elevator accelerating up at 4 m/s 2 Calculate the Normal force on the man  Weight down = -mg = -80(9.8) = -784 N  F = ma  N – 784 = 80(4)  N = = 1004 N

Equilibrium Example – 2a What are tensions on each rope? Mass Upper:   F y = W y + N y + f y + T y + P y =0  W y =-(m U )g=-(10)(9.8)=-98  N y = 0f y = 0  T y = T U -T L P y = 0  T U – T L + 0 = 0  T U – T L = 98 N Mass Lower:   F y = W y + N y + f y + T y +P y =0  W y =-m L g = -(10)(9.8) = -98  N y = 0f y = 0  T y = T L P y = 0  T L +0 =0  T L = 98 N

Equilibrium Example – 2b What are tensions on each rope? T U – T L = 98 N T L = 98 N T u = = 196 N

Equilibrium Alternative Example – 2c What are tensions on each rope? Mass Upper and Mass Lower:   F y = W y + N y + f y + T y + P y =0  W y =-(m U +m L )g=-(20)(9.8)=-196  N y = 0f y = 0  T y = T U P y = 0  T U + 0 = 0  T U = 196 N Mass Lower:   F y = W y + N y + f y + T y +P y =0  W y =-m L g = -(10)(9.8) = -98  N y = 0f y = 0  T y = T L P y = 0  T L +0 =0  T L = 98 N

Equilibrium Example – 3a A car with a weight of 15,000 N is being towed up a 20 0 slope at constant velocity Friction is negligible The tow rope is rated at 6000 N maximum tension Will it break?

Equilibrium Example – 3b   F x = W x + N x + f x + T x + P x   F y = W y + N y + f y + T y + P y  W X = -wsin  W y = -wcos   N x = 0 N y = n  T x = T T y = 0  T – Wsin  = 0  N – Wcos  = 0 T = (15000)sin(20) = 5130 N Rope will not break

Not in Equilibrium Example - 1 Two forces act on an object of mass 4.00 kg One force is 40.0 N in the +x-direction The other force is 60.0 N in the + y-direction Find magnitude and direction of the acceleration of the object –x-direction   F x = ma x  40 = 4a x  40/4 = a x = 10 m/s 2 –y-direction   F y = ma y  60 = 4a y  60/4 = a y = 15 m/s 2  magnitude of a = sqrt( ) = 18 m/s 2   = tan -1 (15/10) = 56 0

Not in Equilibrium Example 2 A boy pulls a box of mass 30 kg with a force of 25 N at an angle of 30 0 above the horizontal (a)Ignoring friction, what is the acceleration of the box? (b)What is the normal force exerted on the box by the ground?   F x = W x + N x + (f k ) x + T x + P x = ma x   F y = W y + N y + (f k ) y + T y + P y = ma y  W x = 0W y = mg = 30(-9.8) = -294  N x = 0N y = N  T x = 0T y = 0  (f k ) x 0(f k ) y = 0  P x = 25 cos30 = 21.7P y = 25 sin30 = 12.5 (a) x-direction  = 30a x  a x = 21.7/30 = 0.72 m/s 2 (b) y-direction  N = N = 30a y = 0  N = newtons

Multiple Objects When two or more objects are connected or in contact, Newton’s laws may be applied to the system as a whole and/or to each individual object Whichever you use to solve the problem, the other approach can be used as a check

Not in Equilibrium Example – 3a The Atwood machine consists of two masses suspended from a fixed pulley m 1 = 0.55 kg m 2 = 0.80 kg (a) What is the acceleration of the system? (b) What is magnitude of tension in string?

Not in Equilibrium Example - 3b (a) –Mass 1:   F y = W y + N y + (f k ) y + T y + P y = ma y  W y = -m 1 gN y = 0  T y = T(f k ) y = 0  P y = 0  -0.55g T + 0 = 0.55a y –Mass 2:  W y = -m 2 gN y = 0  T y = T(f k ) y = 0  P y = 0  -0.80g T + 0 = -0.80a y  Subtract equations: 0.25g = 1.35a y  a y = 1.8 m/s 2

Not in Equilibrium Example 3c (b)  -0.55g + T = 0.55a y  T = 0.55a y g  T = 0.55(1.8) (9.8) = 6.4 N

Not in Equilibrium Alternative Example - 3d The Atwood machine consists of two masses suspended from a fixed pulley m 1 = 0.55 kg m 2 = 0.80 kg (a) What is the acceleration of the system? (b) What is magnitude of tension in string?

Not in Equilibrium Alternative Example – 3e Lift masses m 1 and m 2 so that they are horizontal Gravity on m 1 becomes a pull to the left Gravity on m 2 becomes a pull to the right Treat two masses and rope between as a single system Tension in the rope is an Internal force and is ignored in Newton’s Second Law m1 Pull = m2g Pull = m1g a

Not in Equilibrium Alternative Example – 3f Mass 1 + Mass2:   F x = W x + N x + (f k ) x + T x + P x = ma x  W x = 0N x = 0  T x = 0(f k ) x = 0  P x = -m 1 g + m 2 g  -0.55g+0.80g = ( )a x  a x =.25(9.8) / 1.35 = 1.8 m/s 2 Mass 1:  W x = -m 1 gN x = 0  T x = T(f k ) x = 0  P x = 0  -0.55g +0 + T = 0.55(1.8)  T = (9.8) = 6.4 N

Not in Equilibrium Example – 4a Three blocks are pulled along a frictionless surface by a horizontal force of F = 18.0 N (a)What is the acceleration of the system? (b)What are tension forces in the strings?

Not in Equilibrium – Example 4b Mass 1:   F x = W x +N x +(f k ) x +T x +P x =m 1 a x  W x = 0N x = 0  T x = T 1 (f k ) x = 0  P x = 0  T = m 1 a x  T 1 = m 1 a x  T 1 = 1.0a x (Equation 1) Mass 2:   F x =W x +N x +(f k ) x +T x +P x =m 2 a x  W x = 0N x = 0  T x = -T 1 +T 2 (f k ) x = 0  P x = 0  T 1 + T = m 2 a x  -T 1 + T 2 = m 2 a x  -T 1 + T 2 = 2.0a x (Equation 2)

Not in Equilibrium Example – 4c Mass 3:   F x = W x N x +(f k ) x +T x +P x =m 3 a x  W x = 0N x = 0  T x = -T 2 (f k ) x = 0  P x = P  T 2 + P = m 3 a x  -T 2 + P = m 3 a x  -T = 3.0a x (Equation 3) Add Equations 1 and 2 T 2 = 3.0a x (Equation 4) Add Equations 3 and = 6.0a x a x = 18.0/6.0 = 3.0 m/s 2 Substitute into Equation 1 T 1 = 1.0(3.0) = 3.0 N Substitute into Equation 3 T 2 = 18.0 – 3.0(3.0) = 9.0 N

Not in Equilibrium Alternative Example – 4d Mass 1, Mass 2, and Mass 3:   F x = W x +N x +(f k ) x +T x +P x =ma x  W x = 0N x = 0  T x = 0(f k ) x = 0  P x = P  P = (m 1 +m 2 +m 3 )a x  18.0 = 6.0a x  a x =18.0 / 6.0=3.0(Equation 1) Mass 1:   F x =W x +N x +(f k ) x +T x +P x =m 1 a x  W x = 0N x = 0  T x = T 1 (f k ) x = 0  P x = 0  T = m 1 a x  T 1 = (1.0)(3.0) = 3.0 N

Not in Equilibrium Alternative Example – 4e Mass 3:   F x = W x + N x + (f k ) x + T x + P x =m 3 a x  W x = 0N x = 0  T x = -T 2 (f k ) x = 0  P x = P  – T 2 + P = m 3 a x  T 2 = P – m 3 a x  T 2 = 18 – (3.0)(3.0) = 9.0 N

Not in Equilibrium Example - 5a Pulleys are frictionless There is no friction between mass m 3 and the table What is the acceleration of the system? m 1 =0.25 kg, m 2 = 0.50 kg, m 3 = 0.25 kg

Not in Equilibrium Example 5b Mass 1:   F =W y +N y +(f k ) y +T y +P y =m 1 a  W y = -m 1 gN y = 0  T y = T 1 (f k ) y = 0  P y = 0  -m 1 g T = m 1 a  -m 1 g+T 1 =m 1 a (Equation 1) Mass 2:   F =W y +N y +(f k ) y +T y +P y =m 2 a  W y = -m 2 gN y = 0  T y = T 2 (f k ) y = 0  P y = 0  -m 2 g T = m 2 (-a)  -m 2 g+ T 2 =-m 2 a (Equation 2)

Not in Equilibrium Example 5c Mass 3:   F x =W x +N x +(f k ) x +T x +P x =ma  W x = 0N x = 0  T x = -T 1 + T 2 (f k ) x = 0  P x = 0  – T 1 + T = m 3 a  – T 1 +T 2 =m 3 a (Equation 3) Subtract equation 2 from equation 1 (m 2 -m 1 )g-T 2 +T 1 =(m 1 +m 2 )a (Eq. 4) Add equations 3 and 4 (m 2 -m 1 )g=(m 1 +m 2 +m 3 )a a = (m 2 -m 1 )g/(m 1 +m 2 +m 3 ) a = ( )(9.8)/( ) a = 2.45 m/s 2

Not in Equilibrium Alternative Example – 5d Mass 1, Mass2, and Mass3   F x =W x +N x +(f k ) x +T x +P x =ma  W x = 0N x = 0  T x = 0(f k ) x = 0  P x = -m 1 g+m 2 g  m 1 g+m 2 g = ma  -m 1 g+m 2 g=ma  a = g(m 2 -m 1 )/(m 1 +m 2 +m 3 )  a=(9.8)( )/1=2.45 m/s 2 Mass 1:   F =W y +N y +(f k ) y +T y +P y =m 2 a  W y = 0N y = 0  T y = T 1 (f k ) y = 0  P y = -m 1 g  T 1 – m 1 g = m 1 a  -m 1 g+ T 1 =m 1 a  T 1 =m 1 a+m 1 g = (0.25)( )  T 1 = 3.1 N

Not in Equilibrium Alternative Example – 5e Mass 3:   F =W y +N y +(f k ) y +T y +P y =m 2 a  W y = 0N y = 0  T y = T 2 (f k ) y = 0  P y = -m 3 g  T 2 – m 3 g = -m 3 a  -m 3 g+ T 2 =-m 3 a  T 2 =-m 3 a+m 3 g = (0.25)( )  T 2 = 1.8 N

Not in Equilibrium Alternative Example – 6a Weight of the block on table is 422 N Weigh of the hanging block is 185 N Ignoring all friction and assume the pulley is massless (a) Find the acceleration of the two blocks (b) Find the tension in the cord –Mass of table block is m 1 = 422/9.8 =43.06 kg –Mass of hanging block is m2 = 185/9.8 = kg

Not in Equilibrium Alternative Example 6b Mass 1 and Mass2:   F x =W x +N x +(f k ) x +T x +P x =ma  W x = 0N x = 0  T x = 0(f k ) x = 0  P x = m 2 g  m 2 g = (m 1 + m 2 )a  m 2 g = (m 1 + m 2 )a  a = -g(m 2 /(m 1 +m 2 )  a=(9.8)(18.88)/(61.94)=2.99 m/s 2 Mass 1:   F x =W x +N x +(f k ) x +T x +P x =m 1 a  W x = 0N x = 0  T x = T(f k ) x = 0  P x = 0  T+0 = m 1 a  T = m 1 a  T = 43.06(2.99) = N