Newton's Laws and Frictional Forces: Equilibrium and Dynamics, Assignments of Physics

Download Newton's Laws and Frictional Forces: Equilibrium and Dynamics and more Assignments Physics in PDF only on Docsity! Chapter 5 Applying ewton’s Laws • Newton’s 1st Law – Translational equilibrium • Newton’s 2nd and 3rd Laws – Dynamics of particles • Frictional forces • Circular motion Translational equilibrium • A body is in translational equilibrium when it is at rest or it moves with constant linear velocity: notice that, if the velocity is constant, we can find an inertial frame where the object is at rest and one can describe its dynamics using Newton’s Laws. • By Newton’s 1st Law, the state is determined by the condition In component form: (However, the largest dimensionality of our problems will be 2D...) 0F =∑
0 0 0 x y z F F F  =  =  = ∑ ∑ ∑ ( ) ( ) ( )2 3 23.0 m s 0.20 m sv t t t= + Problem: Dynamics based on a time dependent velocity: A person of mass m = 72 kg stands on a bathroom scale in an elevator in a tall building. The elevator starts from rest and travels upward with a speed that varies with time according to When t = 4.0 s, what is the reading of the bathroom scale? Frictional forces • Types of friction: - kinetic and static friction - fluid resistance (or drag) Kinetic friction • Appears at the surface of contact between two surfaces moving relative to each other. Direction: in the surface of contact (perpendicular on the normal force), against the relative motion Magnitude: proportional to the magnitude of the normal. The coefficient of proportionality µk is a property characterizing the surfaces in contact: • It is a good approximation to assume that the kinetic friction does not depend on the relative velocity of the surfaces in contact, or on their area. k kf µ= coefficient of kinetic friction Problems: 1. Deceleration by friction: A car moves on a surface with coefficient of kinetic friction µk. The driver pushes the brakes. a) What is the acceleration a of the car? b) Knowing the the length of the skid marks is d, how fast was the car moving at the instant when the driver pushed the brakes? 2. Acceleration by a net force including friction: A box of mass m is pulled by a cord under tension T on a rough, flat surface with coefficients of kinetic friction µk. The box accelerates in the direction of T. What is the acceleration a of the box? kf  T W a m kf  W m homework related HW Problem: Incline with friction: A block with mass m1 is places on an inclined plane with slope angle θ and is connected to a second hanging block of mass m2 by a cord passing over a small, frictionless pulley. The coefficient of kinetic friction is µk. Suppose that the mass m1 is larger enough than m2 such that it moves down the incline. a) Find the acceleration of the masses in the kinetic regime. b) Find the tension in the cord. c) Assuming that the mass m1 is released from rest on top of the ramp, calculate its velocity at the bottom knowing that the length of the ramp is d. m1 m2 µk θ homework related HW d Fluid resistance (drag) • Is the resistance opposed by a fluid against the motion of an object submerged in it. • Fluid dynamics is a rather complex field of physics so here we’ll model the drag only very approximately. Ex: The water resistance on a submersible, or the air drag on an airplane Direction: against the direction of object’s motion Magnitude: depends on the speed of the object. Then, using a simple model, , where is a constant. • For small objects moving with low speed (Stokes Drag or Viscous Resistance): • For objects moving with high speed (Rayleigh Drag or Drag Equation): ˆn nf c v v= −
ˆf kvv= − ⇒
nc 2 ˆf Dv v= − ⇒
f kv= 2f Dv= Exercise: An object falling through a fluid – Terminal velocity. A skydiver jumps from rest and falls vertically downward. Calculate the time dependence of his velocity. 2 ˆ small ˆ high kvv v f Dv v v − =  −
w mg=

v w f ma+ =


mg f ma− = Terminal velocity: velocity reached asymptotically by the falling object, when a = 0: 2 tmg Dv= ⇒ tmg kv= ⇒ t mg v k = t mg v D = ( ) ( )1 kt mtv t v e−= − ( ) ( )tanhtv t v t gD m=