Chapter 5 Applying Newton s Laws 5.1 Using Newton s First Law First Law. Abodyactedonbynonetforce,i.e. F i =0 i has a constant velocity (which may be zero) and zero acceleration. Example 5.1. Agymnastwithmassm G =50kg suspends herself from the lower end of a hanging rope of negligible mass. The upper end of the rope is attached to the gymnasium ceiling. (a) What is the gymnasts s weight? (b) What force (magnitude and direction) does the rope exert on her? (c) What is the tension at the top of the rope? Step 1: Choose coordinate system. Let s choose x-axis to point upward. Step 2: Draw free-body diagrams. 49
CHAPTER 5. APPLYING NEWTON S LAWS 50 Step 3: Apply Newton s Laws. (a) What is the gymnasts s weight? Although weight is a force (due to gravity near the surface of Earth) and thus a vector, one usually speaks of weight as a magnitude of force vector given in general by Eq. (4.16). For the gymnast it is w =9.8 m/s 2 50 kg =490N. (5.1) If we choose a coordinate system such that x-axis point upward, then w earth on gymnast =( 490 N) î. (5.2) (b) What force (magnitude and direction) does the rope exert on her? Draw a free-body diagram for the gymnast. There are only two forces action the gymnast: weight w earth on gymnast and tension T rope on gymnast and due to first law they must balance each other or using Eq. (5.2) whose magnitude is w earth on gymnast + T rope on gymnast =0 (5.3) T rope on gymnast =(+490N) î. (5.4) T rope on gymnast =490N (5.5) and the direction is pointing upward. (c) What is the tension at the top of the rope? Draw a free-body diagram for the rope. There are again only two forces action on the rope anddueto first law they must also balance each other T gymnast on rope + T ceiling on rope =0. (5.6) We can now use the third law and Eq. (5.4) tofind T gymnast on rope = T rope on gymnast =( 490 N) î. (5.7) By combining (5.6) and(5.7) we get T ceiling on rope =(+490N) î (5.8) whose magnitude is T ceiling on rope =+490N. (5.9) Example 5.2. Find the tension at each end of rope in Example 5.1 if the weight of the rope is 120 N.
CHAPTER 5. APPLYING NEWTON S LAWS 51 Step 1: Coordinate system. Was already chosen in Example 5.1. Step 2: Draw a free-body diagram for the gymnast and the rope. Step 3: Apply Newton s Laws. Nothing changed for the gymnast s diagram and thus we can still follow the same logic and conclude that T gymnast on rope =( 490 N) î. (5.10) What is however different for the rope diagram is that there is yet another force acting on the rope - the weight - and thus the first law implies where T gymnast on rope + T ceiling on rope + w earth on rope =0. (5.11) w earth on rope =( 120 N) î. (5.12) By combining together Eqs. (5.10), (5.11) and(5.12) we get T ceiling on rope = T gymnast on rope w earth on rope = ( 490 N) î ( 120 N) î =(+610N) î. (5.13) Example 5.4. Acarofweightw rests on a slanted ramp attached to a trailer. (See figure below. Angle α is given. ) Only a cable running from the trailer to the car prevents the car from rolling off the ramp. (The car brakes are off and its transmission is neutral.) Find the tension in the car and the force that the ramp exerts on the car s tires. Step 1: Coordinate system. To proceed we must choose a 2D coordinate
CHAPTER 5. APPLYING NEWTON S LAWS 52 system with x-axis pointing to the right and y-axis pointing up (note that this is intentionally different from the choice made in your textbook). Step 2: Draw a free-body diagram for the car. Step 3: Apply Newton s laws. From the first law we have Then Eq. (5.14) canbewrittenincomponentsas w + T + n =0. (5.14) (0,w)+(T cos α, T sin α)+( n sin α, n cos α) =0 (5.15) or for each component separately T cos α n sin α = 0 w + T sin α + n cos α = 0. (5.16) We now have two equations with two unknowns which can be easilysolved and thus w + n sin2 α cos α T = n sin α cos α + n cos α = 0. (5.17) n = w cos α T = w sin α. (5.18)
CHAPTER 5. APPLYING NEWTON S LAWS 53 5.2 Using Newton s Second Law Second Law. If a net external force acts on a body, the body accelerates. The direction of acceleration is the same as the direction of the net force. The mass of the body times the acceleration vector of the body equals to the net force vector, i.e. F i = m a (5.19) i Example 5.6. An iceboat is at rest on a frictionless horizontal surface. Due to the blowing wind, 4.0 saftertheiceboatisreleased,itismovingtothe right at 6.0 m/s. What constant horizontal force F W does the wind exert on the iceboat? The combined mass of iceboat and rider is 200 kg. Step 1: Coordinate system: We can choose a 1D coordinate system with x-axis pointing in the direction of wind and origin at the place wherethe boat was released. Step 2: Initial Conditions are Step 3: Final conditions are t 0 = 0s x 0 = 0m v 0x = 0m/s (5.20) v(4.0 s) = 6.0 m/s (5.21)
CHAPTER 5. APPLYING NEWTON S LAWS 54 and from equation of motions with constant acceleration we get v x (t) = v 0x + a x t v x (4.0 s) = 0m/s + a x 4.0 s. (5.22) By equating Eqs. (5.21) and(5.22) wearriveattheexpressionforacceleration due to wind a x =1.5 m/s 2 (5.23) but since the mass of the boat is the second law implies m =200kg (5.24) F W =(200kg) ( 1.5 m/s 2) î =(300N) î. (5.25) Example 5.10. Atobogganloadedwithphysicsstudents(totalweightw) slides down a snow-covered hill that slopes at a constant angle α. Thetoboggan is well waxed, so there is virtually no friction. (a) what is its acceleration? (b) What is normal force? Step 1: Coordinate system. This time let s work with a coordinate suggested in the textbook: x-axis pointing in the direction of the slope and y-axis normal to the slope. Exact location of origin is not important. Step 2: Free-body diagram. Draw a free-body diagram for the toboggan. Step 3: Apply Newton s Laws. From the second law w + n = m a (5.26)
CHAPTER 5. APPLYING NEWTON S LAWS 55 where mass can be obtained from the weight m = w g. (5.27) The second law can also be written in components as (w sin α, w cos α)+(0,n)=( w g a x, 0) (5.28) or as two separate equations w sin α = w g a x w cos α + n = 0. (5.29) Then the acceleration is and normal force is a = g sin αî (5.30) n = w cos αĵ. (5.31) 5.3 Friction forces Friction coefficients. There two types of contact forces between macroscopic objects. One is the normal force which is perpendicular to the contact surface and another one is parallel to the contact surface: n normal force always perpendicular to the contact surface f friction force is always parallel to the contact surface. Both forces arise due to microscopic (electromagnetic) interaction between molecules, but we shall only study their macroscopic properties. If there is a motion along the surface of contact, then these two forces are related to each other by the so-called coefficient of kinetic friction or µ k = f k n (5.32) f k = µ k n. (5.33) (Note that the friction coefficients are dimensionless, i.e. have no units.) If there is no motion along the surface of contact, then the friction force is bounded from above f s (f s ) max = µ s n. (5.34)
CHAPTER 5. APPLYING NEWTON S LAWS 56 It turns out that µ s >µ k. (5.35) and so a larger force T must be applied to an object at rest for it to start moving, i.e. fs max,butthenasmallerforcewillbesufficienttocontinue motion, i.e. f k,since f k <fs max. (5.36) Here is a plot of frictional force as a function of applied force The exact values of both coefficients depend on the materials inconstant, e.g. Material µ s µ k steel on steel 0.74 0.57 ice on steel 0.03 0.015 dry rubber on dry concrete 1.0 0.8 wet rubber on wet concrete 0.3 0.25 There is also rolling friction which is typically much smaller. For steel wheels on steel rails it is 0.002 0.003. Example 5.13. You want to move a 500 Ncrateacrossalevelfloor. To start the crate moving, you have to pull with a 230 Nhorizontalforce.Once the crate starts to move, you can keep it moving at constant velocity with only 200 Nforce.Whatarethecoefficientofstaticandkineticfriction?
CHAPTER 5. APPLYING NEWTON S LAWS 57 Step 1: Choose coordinate system. Step 2: Draw free-body diagrams. Step 3: Apply Newton s Laws. For the static case the First Law implies n + w + T + (f s ) max =0 (5.37) or and thus nĵ wĵ + T î (f s) max î =0 (5.38) n = w By combining with Eq. (5.34) weget µ s = (f s) max n For the kinetic case the First Law implies T = (f s ) max. (5.39) = T n = 230N =0.46. (5.40) 500N n + w + T + f k =0 (5.41) or and thus nĵ wĵ + T î f kî =0 (5.42) n = w T = f k (5.43)
CHAPTER 5. APPLYING NEWTON S LAWS 58 By combining with Eq. (5.32) weget µ k = f k n = T n = 200N =0.40. (5.44) 500N Example 5.16. Atobogganloadedwithphysicsstudents(fromExample 5.10) slides down a snow-covered hill. The wax has worn off, so there is a nonzero coefficient of kinetic friction µ k. The slope has just the right angle α to make the toboggan slide with constant velocity. Find the angle in terms of w and µ k. Step 1: Coordinate system. Let s choose x-axis to point in the direction of the slope and y-axis normal to the slope. Exact location of origin is (once again) not important. Step 2: Draw a free body diagram. Step 3: Apply Newton s Laws. For the kinetic case the First Law implies n + w + f k =0 (5.45) or and thus nĵ +(wsin α) î (w cos α) ĵ f kî =0 (5.46) By combining with Eq. (5.32) weget n = w cos α w sin α = f k (5.47) µ k = f k n = w sin α =tanα (5.48) w cos α
CHAPTER 5. APPLYING NEWTON S LAWS 59 and so α =arctanµ k. (5.49) Fluid (air) resistance. The (magnitude of) force of fluid resistance depends on the velocity, f = { kv for "small" velocities Dv 2 for "large" velocities (5.50) where the coefficients depend on many factors: type of fluid, shape of object, etc. We can apply the second law (in the vertical direction) for a falling object to get mg (kv) =ma (5.51) for small velocities. This acceleration will keep accelerating the object until the (so-called terminal) velocity is v t = mg k (5.52) in which case according to Eq. (5.51) itwillstartmovingwithconstant velocity, i.e. a =0. Similarly in the regimes where the large velocities approximation of resistance force of Eq. (5.50) isvalid,thesecondlawimplies and thus the terminal velocity is mg (Dv 2 )=ma. (5.53) v t = mg D. (5.54) Example 5.18. For a human body falling through air in a spread-eagle position, the numerical value of the constant D in Eq. (5.6) is about 0.25 kg/m. Find the terminal speed for a 50 kg skydiver. Step 1: Coordinate system. Let s choose the x-axis to point upward. Step 2: Draw a free body diagram.
CHAPTER 5. APPLYING NEWTON S LAWS 60 Step 3: Apply Newton s laws. We have already used the First Law toderive Eq. (5.54) andthuswecanjustuseittogettheterminalvelocity mg v t = D = (50 kg)(9.8 m/s 2 ) =44m/s. (5.55) 0.25 kg/m 5.4 Dynamics of circular motion We have already looked at a circular motion with constant speed in Section 3.4 and derived the following relations a = v2 R (5.56)
CHAPTER 5. APPLYING NEWTON S LAWS 61 and where a = 4π2 R T 2 (5.57) a magnitude of acceleration v constant speed R radius of circular path T period of motion. (5.58) An object in such a motion experiences a constant (in magnitude) acceleration and thus according to Second Law the (magnitude) of net force must be F net = ma = m v2 R = R m4π2 T. (5.59) 2 Example 5.19. Asledwithamassof25.0 kg rests on a horizontal sheet of essentially frictionless ice. It is attached by a 5.00 mropetoapostsetin the ice. Once given a push, the sled revolves uniformly in a circle around the post. If the sled makes five complete revolutions every minute, find the force F exerted on it by the rope. Step 1: Coordinate system. Let s choose a moving coordinate system with perpendicular direction towards center and parallel direction in the direction of motion of the sled. Step 2: Draw a free body diagram.
CHAPTER 5. APPLYING NEWTON S LAWS 62 Step 3: Apply Newton s Laws. From the Second Law we got Eq. (5.59) which implies that F net = m 4π2 R T where T = 60 s 5 =12s m = 25.0 kg R = 5.00 m. and thus F net =34.3 N.