PHYS 210 Chapter 6, Spring 2012

Chapter 6 Chapter 6 Force and Motion II Force and Motion II Forces of Friction Forces of Friction  When an object is in motion on a When an object is in motion on a surface or through a viscous medium, surface or through a viscous medium, there will be a resistance to the there will be a resistance to the motion motion  This is due to the interactions between This is due to the interactions between the object and its environment the object and its environment  This is resistance is called the This is resistance is called the force of force of friction friction More About Friction More About Friction  Friction is proportional to the normal force Friction is proportional to the normal force  The force of static friction is generally The force of static friction is generally greater than the force of kinetic friction greater than the force of kinetic friction  The coefficient of friction (µ) depends on The coefficient of friction (µ) depends on the surfaces in contact the surfaces in contact  The direction of the frictional force is The direction of the frictional force is opposite the direction of motion opposite the direction of motion  The coefficients of friction are nearly The coefficients of friction are nearly independent of the area of contact independent of the area of contact Static Friction, ƒ Static Friction, ƒss  Static friction acts to Static friction acts to keep the object from keep the object from moving moving  If F increases, so does ƒ If F increases, so does ƒss  If F decreases, so does If F decreases, so does ƒƒss  ƒƒss   µ µ nn Kinetic Friction Kinetic Friction  The force of The force of kinetic friction kinetic friction acts when the acts when the object is in object is in motion motion  ƒƒkk = µ = µ nn Example: Example:  A Block of mass m A Block of mass m11 on a on a rough, horizontal surface rough, horizontal surface is connected to a ball of is connected to a ball of mass m mass m22 by a lightweight by a lightweight cord over a lightweight cord over a lightweight frictionless pulley. A force frictionless pulley. A force of magnitude F at an angle of magnitude F at an angle  with the horizontal with the horizontal is applied to the block as shown. The coefficient is applied to the block as shown. The coefficient of Kinetic friction between block and surface is of Kinetic friction between block and surface is kk. Determine the acceleration of the objects. . Determine the acceleration of the objects. Solution: Solution: Free Body diagrams: Free Body diagrams: Newton’s 2 Newton’s 2nd nd Law: Law: X-axis: F X-axis: F cos cos – T – f – T – fkk = m = m11a a (1) (1) Y-axis: F Y-axis: F sin sin + n – m + n – m11g = 0 g = 0 (2) (2) (2) (2)   n = m n = m11g – F g – F sin sin ffkk = = kkn = n = kk((m m11g – F g – F sin sin)) (1) (1)   T = Fcos T = Fcos - - kk((m m11g – F g – F sin sin) - ) - m m11aa But Also, But Also, T = m2a – m2g Solution Cont’d: Solution Cont’d:  T = T T = T   Fcos Fcos - - kk((m m11g – F sin g – F sin) - ) - m m11aa = = m2a – m2g Knowing the value of a one can easily deduce the value of T = m2a – m2g. 2 1 1 k 2 k m m μ m ) μ sin ) - g(m F(cos       a Connected Objects Connected Objects  Apply Newton’s Laws Apply Newton’s Laws separately to each separately to each object object  The acceleration of The acceleration of both objects will be both objects will be the same the same  The tension is the The tension is the same in each diagram same in each diagram  Solve the Solve the simultaneous simultaneous equations equations More About Connected More About Connected Objects Objects  Treating the system as one object Treating the system as one object allows an alternative method or a allows an alternative method or a check check  Use only external forces Use only external forces  Not the tension – it’s internal Not the tension – it’s internal  The mass is the mass of the system The mass is the mass of the system  Doesn’t tell you anything about any Doesn’t tell you anything about any internal forces internal forces Newton’s Law applied to Newton’s Law applied to uniform circular motion: uniform circular motion:  A particle moving with uniform speed v in a circular A particle moving with uniform speed v in a circular path of radius r, experiences a centripetal path of radius r, experiences a centripetal acceleration acceleration aacc that has a magnitude: that has a magnitude: aacc is directed toward the is directed toward the center of the circle and is center of the circle and is always always  to v. to v. r v2 c  | a | Newton’s Law applied to Newton’s Law applied to uniform circular motion Cont’d: uniform circular motion Cont’d:  If we apply Newton’s law along the radial If we apply Newton’s law along the radial direction, direction, F acts toward the center F acts toward the center of the circle. of the circle. r m v m F 2   c  a Newton’s Law applied to Newton s Law applied to uniform circular motion Cont’d uniform circular motion Cont’d Ex: Ex:  The ball is put in a The ball is put in a circular motion on a circular motion on a frictionless table. frictionless table. If the string breaks at If the string breaks at some point how the some point how the ball moves afterward? ball moves afterward? Tangential ( Tangential (aatt) and Radial ( ) and Radial (aaN N) ) Acceleration Acceleration  In the case when the speed changes along In the case when the speed changes along a circular path the total Acceleration vector a circular path the total Acceleration vector is given by: is given by: aa = = aatt + + aaN N Where, Where, |a |aN N| = v | = v22/r (directed toward the center of the circle) /r (directed toward the center of the circle) |a |att| = (directed tangent to the circle) | = (directed tangent to the circle) dt d|v| Tangential ( Tangential (aatt) and Radial ( ) and Radial (aaN N) ) Acceleration Cont’d: Acceleration Cont’d: at aN a aa = = aatt + + aaNN ||aaNN| = v | = v22/r/r dt ||aatt| = d|v| | = Example Example The above figure shows the path of a park ride that travels at constant speed through five circular arcs of radii R0, 2R0, and 3R0. Rank the arcs according to the magnitude of the centripetal force on a rider traveling in the arcs, greatest first. 423.15 Application: Application:  A ball of mass m= 0.500 Kg is attached to the A ball of mass m= 0.500 Kg is attached to the end of a cord 1.50 m long. The ball is whirled end of a cord 1.50 m long. The ball is whirled in a horizontal circle. in a horizontal circle. If the ball can withstand a maximum tension If the ball can withstand a maximum tension of 50.0 N, what is the maximum speed the of 50.0 N, what is the maximum speed the ball can attain before the cord breaks. ball can attain before the cord breaks. x Solution: Solution: Applying Newton’s Law: Maximum tension correspond to max v: r m v m F 2   c  a r m v T 2  x mg N T  m/s m T r V Max Max 12 2. 150 5.0 50 5.1       Example: Example:  A small object of mass A small object of mass m is suspended from a m is suspended from a string of length L. The string of length L. The object revolves with object revolves with constant speed v in a constant speed v in a horizontal circle of horizontal circle of radius r, Find an radius r, Find an expression for v as expression for v as function of T, r and m? function of T, r and m? Solution: Solution:  Applying Newton’s Applying Newton’s Law: Law: Projection along the Projection along the horizontal axis: horizontal axis: Projection along the Projection along the vertical axis: vertical axis: a   m T m g   r m v ) Tsin( 2   (1) mg Tcos( )   (2) r.g.tanθ v r g v tan 2      Dividing (1) by (2)  Example: Example:  A 1500-kg car moving on a A 1500-kg car moving on a flat, horizontal road flat, horizontal road negotiates a curve, (See negotiates a curve, (See Figure). If the radius of the Figure). If the radius of the curve is 35.0 m and the curve is 35.0 m and the coefficient of static friction coefficient of static friction between the tires and dry between the tires and dry pavement is 0.500, find the pavement is 0.500, find the maximum speed the car can maximum speed the car can have and still make the turn have and still make the turn successfully. successfully. Solution: Solution:  Force of friction enables Force of friction enables the car to remains in its the car to remains in its circular path. circular path. a    m n f m g s    Projection along the radius: μ mg μ n f s s s, max   r m v f 2 s  But, r mg m v μ 2max s   13.1m/s 9.80 35.0 0.500 μ .g.r v s max       More Applications: More Applications:  A small sphere of mass A small sphere of mass m is attached to the end m is attached to the end of a cord of length R and of a cord of length R and set into motion in a set into motion in a vertical circle about a vertical circle about a fixed point O. Determine fixed point O. Determine the tension in the cord the tension in the cord at any instant when the at any instant when the speed of the sphere is v speed of the sphere is v and the cord makes an and the cord makes an angle angle  with the vertical. with the vertical. Solution: Solution:  Note that along the path Note that along the path the speed is not the speed is not constant. constant. R m v T mgcos - 2    a   m T m g   Projection along the radius: Projection along the radius:             gcos R m v T 2 Application Cont’d Application Cont’d            gcos R m v T 2 What speed would the ball have as it passes over the top of the circle if the tension in the cord goes to zero instantaneously? At the top  = 180°  vtop  gR T = 0, (i.e. cord breaks)            g R m v T 2 Motion in the Presence of Motion in the Presence of Resistive Forces: Resistive Forces:  The medium through which the object The medium through which the object moves exerts a resistive force moves exerts a resistive force RR on it. on it.  The medium can be either a liquid or a gas. The medium can be either a liquid or a gas.  Some examples are the air resistance Some examples are the air resistance associated with moving vehicles (called also associated with moving vehicles (called also air Drag) and the viscous forces that act on air Drag) and the viscous forces that act on a object moving through a liquid. a object moving through a liquid. Resistive Forces: Resistive Forces:  The direction of the resistive force The direction of the resistive force RR is always is always opposite to the direction of motion. opposite to the direction of motion.  The Magnitude of The Magnitude of RR is proportional to the is proportional to the speed of the object: speed of the object: RR = - = -bb vv vv is the velocity of the object and is the velocity of the object and bb is a is a constant whose value depends on the constant whose value depends on the properties of the medium and on the shape properties of the medium and on the shape and dimensions of the object. and dimensions of the object. Terminal Speed Terminal Speed  When the Drag force ( upward air When the Drag force ( upward air resistance for example) equals the resistance for example) equals the driving force (downward force of driving force (downward force of gravity), the net force on the object is gravity), the net force on the object is zero. zero.  The constant speed of the object is The constant speed of the object is the the terminal speed terminal speed Application: Application:  Consider a small Consider a small sphere of mass m sphere of mass m released from rest released from rest in a liquid as shown in a liquid as shown in the adjacent in the adjacent figure, assuming figure, assuming that the buoyant that the buoyant force is negligible, force is negligible, describe the describe the motion of the motion of the sphere. sphere. Application Cont’d: Application Cont’d:  The forces acting The forces acting on the sphere are: on the sphere are: 1) Resistive Force: 1) Resistive Force: RR =- =-bb vv 2) The gravitational 2) The gravitational Force: Force: FFgg = m = mgg Application Cont’d: Application Cont’d:  Applying Newton’s law: Applying Newton’s law:  We can obtain the terminal We can obtain the terminal speed by putting dv/dt = 0. speed by putting dv/dt = 0. mg – mg – bb vvTT = 0 = 0 mg R y (+) b vT mg  dt mdv v mg-  b m v g dt dv   b  Application Cont’d: Application Cont’d:  The Equation The Equation is called a differential equation. is called a differential equation. m v g dt dv   b