Physics for Scientists and Engineers. Chapter 6 Dynamics I: Motion Along a Line

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1 Physics for Scientists and Engineers Chapter 6 Dynamics I: Motion Along a Line Spring, 008 Ho Jung Paik

2 Applications of Newton s Law Objects can be modeled as particles Masses of strings or ropes are negligible When a rope attached to an object is pulling it, the magnitude of that force,, is the tension in the rope Interested only in the external forces acting on the object Can neglect internal reaction forces 7-eb-08 Paik p.

3 Objects in Equilibrium If the acceleration of an object is zero, the object is said to be in equilibrium Mathematically, the net force acting on the object is zero Apply Newton s first Law in component form: r 0 x 0, y 0 7-eb-08 Paik p. 3

4 Static Equilibrium, Example A traffic light is suspended with three ropes as shown in the figure. ind the tension in the three ropes. he traffic light does not move, so the net force must be zero. Static equilibrium Need two free-body diagrams: one for the light and the other for the knot. 7-eb-08 Paik p. 4

5 Example, cont Light: y 0: 3 mg 0 3 mg Knot : x 0: cos 37 + cos y 0: sin 37 + sin mg 0.60 mg 7-eb-08 Paik p. 5

6 Dynamic Equilibrium, Example A car with a weight of 5,000 N is being towed up a 0 slope at constant velocity. riction is negligible. he tow rope is rated at 6000 N. Will it break? Know: θ 0, w 5,000 N ind: 7-eb-08 Paik p. 6

7 Example, cont Apply Newton s nd Law (a 0) r 0 x wsinθ 0 y n wcosθ 0 rom the x equation, wsinθ (5,000 N)sin N 7-eb-08 Paik p. 7 0 Since < 6,000 N, the rope won t break. Note that y equation is not needed. Check solution at θ 0 and 90.

8 Objects Experiencing a Net orce If an object experiences an acceleration, there must be a nonzero net force acting on it Draw a free-body diagram Apply Newton s Second Law in component form: r x r ma ma x, y ma y 7-eb-08 Paik p. 8

9 Crate Pulled on loor, Example 3 A crate weighing 490 N is pulled with a rope on a frictionless floor. he tension in the rope is 0 N. What is the resulting acceleration of the crate? orces acting on the crate: ension: 0 N Gravitational force: g 490 N Normal force n exerted by the floor 7-eb-08 Paik p. 9

10 Example 3, cont Apply Newton s nd Law in component form: Solve for the unknown a x : a x m 0 N m/s 490 N/9.80 m/s If a constant and v x0 0, then v a t (0.400 m/s ) t xt x 7-eb-08 Paik p. 0

Car on Incline, Example 4 A car is at rest on top of a hill.

0 m hill with an incline angle of 30.0. orces acting on the

straight down Choose the coordinate system with x // the

11 Car on Incline, Example 4 A car is at rest on top of a hill. ind the car s velocity at the bottom of the 30.0 m hill with an incline angle of orces acting on the object: Normal force n the plane Gravitational force g straight down Choose the coordinate system with x // the incline, y the incline Replace the force of gravity with its components 7-eb-08 Paik p.

12 Example 4, cont Apply Newton s nd Law in component form: a x x y 0 mg sin 30 max n mg cos30 0 Solve for the unknown a x : 0 g sin m/s a x? he velocity at the bottom of the slope is obtained from v xf v + a Δx, v xi x xf 0+ (4.90m/s )30.0 m 7.m/s 7-eb-08 Paik p.

13 Weight & Apparent Weight, Weight is the force exerted by the earth on mass m w mg, where g 9.80 m/s Consider a scale and weight at rest When the weight comes to rest with respect to the spring, sp w he pointer is calibrated to read sp w at equilibrium (a 0) 7-eb-08 Paik p. 3

14 Weight & Apparent Weight, Consider an object hanging from a scale in an elevator moving up with acceleration a sp he y component of nd Law: y sp sp mg m( g + a) ma he spring dial comes to rest with the force sp m(g+a) > w m (g+a) is the apparent weight 7-eb-08 Paik p. 4

15 Weight & Apparent Weight, 3 Now consider the elevator accelerating downward sp he y component of nd Law: mg m( a) y sp sp m( g a) he scale reads sp m(g -a) < w If a g (freefall), the apparent weight is 0 weightless 7-eb-08 Paik p. 5

16 Multiple Weights/Objects orces acting on the objects: ension (same along the same string) Gravitational force Normal force and friction Objects connected by a (non-stretch) string have the same magnitude of acceleration Draw the free-body diagrams Apply Newton s Laws 7-eb-08 Paik p. 6

17 7-eb-08 Paik p. 7 Example 5 Check if the answer makes sense. m m a 0 m << m a g, m >> m a g g m m m m a a g m a g m a g m a m m g m a g m m a m g m y y ) ( ) ( ) ( ) ( : ) ( : An object of mass m is connected to a light string that passes over a frictionless pulley and is fastened to another object of mass m. ind the acceleration of the objects.

18 Multiple Pulleys, Example 6 Assume the block is accelerated upward, i.e. a > 0. Let 0.0 N and M 5.00 kg. ind the acceleration and all the s. All the pulleys and strings are considered massless and there is no friction in the pulleys 3 Note that top pulley doesn t move. he bottom pulley and mass M have the same acceleration. 7-eb-08 Paik p. 8

19 7-eb-08 Paik p N 60 N, 3 0 N,,.8 m/s 9.8 m/s 5 kg (0 N) herefore, ) ( 0 : ) ( ), ( 0 : ) ( ) ( 0 : ) ( : g M a a g M S a g M a g M P a g M a g M a P a g M Ma Mg M y y upper y lower y Example 6, cont a <0!

ind the acceleration of the objects and the tension in the

20 Example 7 A 5.00-kg object placed on a frictionless, horizontal table is connected to a cable that passes over a pulley and then is fastened to a hanging 9.00-kg object. ind the acceleration of the objects and the tension in the string. m m a 5 kg : 9 kg : m m + m 5.00 kg 6.30 m/s x y g m a, m g ( a) m/s m y 3.5 N n m g 6.30 m/s 7-eb-08 Paik p. 0 0 n m g a a m g

21 Ball and Cube, Example 8 wo objects are connected by a light string that passes over a frictionless pulley. he incline is frictionless, and m.00 kg, m 6.00 kg, and θ Draw free-body diagrams of both objects. ind (a) the acceleration of the objects, (b) the tension in the string, and (c) the speed of each object.00 s after being released from rest. 7-eb-08 Paik p.

4 m/s m g m g sinθ m g m + m + g) m a 6.

22 Example 8, cont y (a) m m (b) (c) v f.00 kg : 6.00 kg : m v y x a i m g sinθ ( a m/s m g m g sinθ m g m + m + g) m a 6.7 N m a at 0 m/s + (3.57 m/s 3.57 m/s )(.00 s) 7-eb-08 Paik p. a x θ m g m g cosθ a m g n y m g sinθ x θ 55. 0

23 orces of riction When an object is in contact with a surface, there is a resistance to its motion, the force of friction his is due to the interaction between the molecules on the mating surfaces riction is proportional to the normal force Static friction: ƒ s µ s n and kinetic friction: ƒ k µ k n hese equations relate the magnitudes of the forces only Generally, µ s > µ k he coefficient of friction (µ) depends on the surfaces in contact but is nearly independent of the area of contact 7-eb-08 Paik p. 3

24 Static riction here is a maximum static frictional force, f s,max µ s n. If the applied force is greater, then the object slips. 7-eb-08 Paik p. 4

25 Kinetic riction he kinetic friction coefficient is less than the static friction coefficient, µ k < µ s. herefore, k f µ k n constant < s,max f. 7-eb-08 Paik p. 5

26 riction orces vs. Applied orce 7-eb-08 Paik p. 6

27 Rolling riction Kinetic friction is operable for a wheel sliding on a surface Rolling friction is for a non-sliding, rolling wheel f r µ r n and points opposite to the motion µ r < µ k 7-eb-08 Paik p. 7

28 Some Coefficients of riction 7-eb-08 Paik p. 8

29 riction in Newton s Laws Draw the free-body diagram, including the force of kinetic friction Opposes the motion Is parallel to the surfaces in contact riction is a force, so it is simply included in the Σ in Newton s Laws Continue with the solution as with any Newton s Law problem 7-eb-08 Paik p. 9

30 Static riction - Measuring µ s he block tends to slide down the plane, so the friction acts up the plane Increase θ up to the instant slipping starts Apply Newton s Law: f x mg sinθ fs, max 0, s,max mg sinθ μ n μ mg cosθ s s y n mg cosθ 0 μ s tanθ 7-eb-08 Paik p. 30

31 Sliding Down a Plane, Example 9 If µ s 0.30 and θ 30, will the cube slide? Will not slide if f s mg sinθ 0.5mg Since f s µ s n and n mg cosθ, s f µ s mg cos x 0.866mg 0.6mg Will slide 7-eb-08 Paik p. 3

ruck with a Crate, Example 0 A truck is carrying a crate up a 0 hill. he coefficient of static friction between the truck bed and the crate is μ s 0.35.

32 ruck with a Crate, Example 0 A truck is carrying a crate up a 0 hill. he coefficient of static friction between the truck bed and the crate is μ s ind the maximum acceleration that the truck can reach before the crate begins to slip backward. n he crate will not slip if its acceleration is equal to that of the truck. or it to be accelerated, some force must act on it. One of these is the static friction force s f. his force must be directed upward to keep the crate from sliding backwards. As the acceleration of the truck increases, so must s f. Once s, f max µ s n is reached, the crate will start sliding. 7-eb-08 Paik p. 3

33 Example 0, cont Applying Newton's law to the crate : Σ Σ x y mg mg a max sinθ + μ n Substitutingθ 0 a n max mg cosθ mg sinθ + μ mg cosθ + n s g( μ cosθ sinθ ) s.68 m/s s 0 cosθ and ma μ max ma s max 0.35, n 7-eb-08 Paik p. 33

34 Hockey Puck, Example Consider a puck is hit and given a speed v 0.0 m/s. How far will it go if µ k 0.? ) x : f k ma y : n mg 0 a f k /m µ k n/m µ k g ) Use v f v i + aδx v i 0.0 m/s, v f 0, Δx? Δx v i /a 400/( x 0. x 9.80) 04 m 7-eb-08 Paik p. 34

Pulling at an Angle, Example What is the optimum angle for minimizing the pulling force that moves the block horizontally at constant velocity?

35 Pulling at an Angle, Example What is the optimum angle for minimizing the pulling force that moves the block horizontally at constant velocity? Pulling straight up minimizes the normal force and hence the frictional force. However, it does not move the block horizontally. Pulling parallel to the table maximizes the normal force and hence the frictional force. he optimum angle must be somewhere in between. 7-eb-08 Paik p. 35

36 x : y : n f f n herefore, Example, cont + cosθ ma + sinθ mg mg Substituting for Minimum occurs k sinθ μ ( mg θ f tan in the when x k 0 sinθ ) μ x 0 equation : d μkmg dθ μkmg μ sinθ + cosθ μ cosθ sinθ ( μ sinθ + cosθ ) minimizes the pulling force. 7-eb-08 Paik p. 36 k k k 0

Example 3 A van accelerates down a hill, going from

he acceleration is such that the string remains

37 Example 3 A van accelerates down a hill, going from rest to 30.0 m/s in 6.00 s. During the acceleration, a toy (m 0.00 kg) hangs by a string from the van's ceiling. he acceleration is such that the string remains perpendicular to the ceiling. Determine: (a) the angle and (b) the tension in the string. 7-eb-08 Paik p. 37

38 Example 3, cont Choose x-axis pointing down the slope. x : v f v i + at 30.0 m/s 0 + a(6.00 s) a 5.00 m/s Consider the forces on the toy. x : mg sinθ m (5.00 m/s ) θ 30.7 y : mg cosθ kg x 9.80 m/s xcos N 7-eb-08 Paik p. 38

spring scale reads 50 N. Pat's true weight is 30 N, and the chair weighs 60 N.

39 Example 4 Pat wants to reach an apple in a tree without climbing it. Sitting in a chair connected to a rope that passes over a frictionless pulley, Pat pulls on the loose end of the rope with a force that the spring scale reads 50 N. Pat's true weight is 30 N, and the chair weighs 60 N. (a) Draw free-body diagrams for Pat and the chair considered as separate systems, and another diagram for Pat and the chair considered as one system. (b) Show that the acceleration of the system is upward and find its magnitude. (c) ind the force Pat exerts on the chair. 7-eb-08 Paik p. 39

40 Ex 4, cont (a) ree-body diagrams: (b) Consider Pat and the chair as the system. Note that two ropes support the system and each rope has 50 N. y ma y x 50 N 480 N (480 N/9.80 m/s )a a m/s > 0. herefore, upward. (c) Now consider Pat as the system: y n + 30 N (30 N/9.80 m/s )a n 50 N + 30 N kg x m/s 83.3 N 7-eb-08 Paik p. 40

41 Motion with Resistive orces Motion can be through a medium Either a liquid or a gas Medium exerts a resistive drag force D on an object moving through the medium Magnitude of D depends on the medium Direction of D is opposite to the direction of motion Magnitude of D can depend on the speed in complex ways We will discuss only two cases: D v, D v 7-eb-08 Paik p. 4

dimensions of the object he negative sign indicates that D is opposite in direction to

42 Drag Proportional to v or objects moving at low speeds, the drag is approximately proportional to v : D bv b depends on the property of the medium, and on the shape and dimensions of the object he negative sign indicates that D is opposite in direction to v he equation of motion of the suspended mass: D mg m( a) bv mg m dv dt 7-eb-08 Paik p. 4 D

object: dv bv mg m 0 v v v mg b dt v(t) is found by solving

43 erminal Speed for D v Initially, v 0 D 0 and dv/dt g As t increases, D mg, a 0 and v v : the terminal speed of the object: dv bv mg m 0 v v v mg b dt v(t) is found by solving the differential equation: ( t / τ e ), τ m b / v 7-eb-08 Paik p. 43 t

44 Drag Proportional to v or objects moving at high speeds, the drag is approximately proportional to v : D (/)cρav c is a dimensionless empirical drag coefficient ρ is the density of the fluid A is the cross-sectional area of the object 7-eb-08 Paik p. 44

45 erminal Speed for D v Analysis of an object falling through a medium accounting for air resistance D mg a m( a), g cρa v m cρav he terminal speed will occur when the acceleration goes to zero mg Solving the equation gives v c ρ A mg ma D 7-eb-08 Paik p. 45

46 Some erminal Speeds 7-eb-08 Paik p. 46

Physics 101 Lecture 5 Newton`s Laws

Physics 101 Lecture 5 Newton`s Laws Dr. Ali ÖVGÜN EMU Physics Department The Laws of Motion q Newton s first law q Force q Mass q Newton s second law q Newton s third law qfrictional forces q Examples