Chapter 5

1. Chapter 5 Applying Newton’s Laws

2. Goals for Chapter 5 • To use and apply Newton’s First Law • To use and apply Newton’s Second Law • To study friction and fluid resistance • To consider forces in circular motion

3. Introduction • It’s not hard to state Newton’s laws well; how should we apply them to everyday situations? • We can start with forces balanced, nothing moving (statics). • Next, we’ll study unbalanced forces causing motion (dynamics). • We’ll end Chapter 5 considering the role of forces in circular motion.

4. Using Newton’s First Law when forces are in equilibrium

5. 5.1 using Newton's 1st law: particles in equilibrium • A body is in equilibrium when it is at rest or moving with constant velocity in an inertial frame of reference.

6. Example 5.1: tension in a massless rope • A gymnast with mass mG = 50.0 kg suspends herself from the lower end of a hanging rope. The upper end of the rope is attached to the gymnasium ceiling. • What is the gymnast’s weight? • What force (magnitude and direction) does the rope exert on her? • What is the tension at the top of the rope? Assume that the mass of the rope itself is negligible. 490 N 490 N 490 N

7. Example 5.2: tension in a rope with mass • Suppose that in example 5.1, the weight of the rope is not negligible but is 120 N. find the tension at each end of the rope. • Lower end of the rope: 490 N • Upper end of the rope: 610 N

8. Example 5.3: two-dimensional equilibrium • A car’s engine with weight w hangs from a chain that is linked at ring O to two other chains, one fastened to the ceiling and the other to the wall. Find the tension in each of the three chains in terms of w, the weights of the ring and chains are negligible. Engine:T1 = w T3 = 1.155w T2 = 0.577w

9. Example 5.4: an inclined plane • A car of weight w rests on a slanted ramp leading to a car-transporter trailer. Only a cable running from the trailer to the car prevents the car from rolling backward off the ramp. Find the tension in the cable and the force that the tracks exert on the car’s tires.

10. Example 5.5 – tension over a frictionless pulley • Blocks of granite are to be hauled up a 15o slope out of a quarry and dirt is to be dumped into the quarry to fill up old holes to simplify the process, you design a system in which a granite block on a cart with steel wheels (weight w1, including both block and cart) is pulled uphill on steel rails by a dirt-filled bucket (weight w2, including both dirt and bucket) dropping vertically into the quarry. How must the weights w1 and w2 be related in order for the system to move constant speed? Ignore friction in the pulley and wheels and the weight of the cable. Block and cart: w2=0.26w1

11. Test your understanding 5.1 • A traffic light of weight w hangs from two lightweight cables, one on each side of the light. Each cable hands at a 45o angle from the horizontal. What is the tension in each cable? • w/2 • w/√2 • w • w√2 • 2w T T 45o 45o w

12. Caution: ma doesn’t belong in free-body diagrams. 5.2 using Newton's 2nd law: dynamics of particles

13. Example 5.6 Straight-line motion with a constant force • An ice boat is at rest on a perfectly frictionless horizontal surface. A wind is blowing (along the direction of the runners) so that 4.0 s after the iceboat is release, it attains a velocity of 6.0 m/s. what constant horizontal force Fw does the wind exert on the iceboat? The mass of iceboat and rider is 200 kg. Fw = 300 N

14. Example 5.7 Straight-line motion with friction • Suppose a constant horizontal friction force with magnitude 100 N opposes the motion of the iceboat in Example 5.6. In this case, what constant force FW must the wind exert on the iceboat to cause the same constant x-acceleration ax=1.5 m/s2? Fw = 400 N

15. Example 5.8 Tension in an elevator cable • An elevator and its load have a total mass of 800 kg. The elevator is originally moving downward at 10.0 m/s; it slows to a stop with constant acceleration in a distance of 25.0 m. Find the tension T in the supporting cable while the elevator is being brought to rest. T = 9440 N

16. Example 5.9 Apparent weight in an accelerating elevator • A 50.0 kg woman stands on a bathroom scale while riding in the elevator in Example 5.8. What is the reading on the scale? n = 390 N

17. Apparent weight and apparent weightlessness • When a passenger with mass m rides in an elevator with y-acceleration ay, a scale shows the passenger’s apparent weight to be n = m∙(g + ay) • When the elevator is accelerating upward, ay is positive and n is greater than the passenger’s weight w = mg. when the elevator is accelerating downward, ay is negative and n is less than the weight. • When ay= g, the elevator is in free fall, n = 0 and the passenger seems to be weightless. • Similarly, an astronaut orbiting the earth in a spacecraft experiences apparent weightlessness. • In each case, the person is not truly weightless because there is still a gravitational force acting. But the person's sensation in this free-fall condition are exactly the same as though the person were in outer space with no gravitational force at all.

18. Example 5.10 Acceleration down a hill • A toboggan loaded with vacationing students (total weight w) slides down a long, snow-covered slope. The hill slopes at a constant angle α, and the toboggan is so well waxed that there is virtually no friction. What is its acceleration? a = g sinα n = mg cosα

19. Watch for this common error—Figure 5.13 • A good “road sign” is to be sure that the normal force comes out perpendicular to the surface.

20. Example 5.11 Two bodies with same acceleration • You push a 1.00 kg food tray through the cafeteria line with a constant 9.0 N force. As the tray moves, it pushes on a 0.50 kg carton of milk. The tray and carton slide on a horizontal surface that is so greasy that friction can be neglected. Find the acceleration of the tray and carton and the horizontal force that the tray exerts on the carton. a = 6.0 m/s2 F T on C = 3.0 N

21. Double Trouble (a.k.a., Two Body Problems) • Two body-problems can typically be approached using one of two basic approaches. • One approach is the system analysis, the two objects are considered to be a single object moving (or accelerating) together as a whole. • Another approach is the individual object analysis, either one of the two objects is isolated and considered as a separate, independent object. Sometimes, you need to combine both approaches to solve a problem

22. Example 5.12 Two bodies with the same magnitude of acceleration • An air-track glider with mass m1 moving on a level, frictionless air track in the physics lab. The glider is connected to a lab weight with mass m2 by a light, flexible, non stretching string that passes over a small frictionless pulley. Find the acceleration of each body and the tension in the string.

23. Test Your Understanding 5.2 • Suppose you hold the glider in Example 5.12 so that it and the weight are initially at rest. You give the glider a push to the left and then release it. The string remains taut as the glider moves to the left, comes instantaneously to rest, than moves to the right. At the instant the glider has zero velocity, what is the tension in the string? • Greater than in Example 5.12 • The same as in Example 5.12 • Less than in Example 5.12, but greater than zero • zero

24. 5.3 Frictional forces, kinetic and static • Friction can keep an object from moving or slow its motion. • Microscopic imperfections cause non ideal motion. The friction and normal forces are really components of a single contact force. • The direction of friction force is opposite of the direction of motion. • Lubricant are used to reduce friction.

25. Kinetic Friction • The kind of friction that acts when a body slides over a surface is called kinetic friction forcefk. The magnitude of the kinetic friction force can be represented by the equation: μkis a constant called the coefficient of kinetic friction. μkdepends on the types of surface. Because it is a ratio of two force magnitudes, μkis a pure number, without units. The equation is only an approximate representation of a complex phenomenon. As a box slides over the floor, bonds between the two surfaces form and break, and the total number of such bonds varies; hence the kinetic friction force is not perfectly constant.

26. Static Friction • Friction forces may also act when there is no relative motion. If you try to slide a box across the floor, the box may not move at all because the floor exerts an equal and opposite friction force on the box. This is called a static friction force fs. Both equations of friction are relationships between magnitudes, not vector relationships.

27. Coefficients of friction

28. Applied force is proportional until the object moves—Figure 5.19 • Notice the transition between static and kinetic friction.

29. Example 5.13 Friction in horizontal motion • You are trying to move a 500 N crate across a level floor. To start the crate moving, you have to pull with a 230 N horizontal force. Once the crate “breaks loose” and starts to move, you can keep it moving at constant velocity with only 200 N. What are the coefficients of static and kinetic friction?

30. 5.14 Static friction can be less than the maximum • In Example 5.13, what is the friction force if the crate is at rest on the surface and a horizontal force of 50 N is applied to it?

31. Example 5.15 Minimizing kinetic friction • In Example 5.13, suppose you try to move the crate by tying a rope around it and pulling upward on the rope at an angle of 30o above the horizontal. How hard do you have to pull to keep the crate moving with constant velocity? Is this easier or harder than pulling horizontally? Assume w = 500 N and µk = 0.40. T = 188 N The tension required is less. n = 406 N

32. Example 5.16 Toboggan ride with friction I • Let’s go back to Example 5.10. the wax has worn off and there is now a nonzero coefficient of kinetic friction µk. The slope has just the right angle to make the toboggan slide with constant speed. Derive an expression for the slope angle in terms of w and µk.

33. Example 5.17 Toboggan ride with friction II • The same toboggan with same coefficient of friction as in example 5.16 accelerates down a steeper hill. Derive and expression for the acceleration in terms g, a, µk, and w.

34. Rolling friction • It’s a lot easier to move a loaded filing cabinet across a horizontal floor using a cart with wheels than to slide it. • We can define a coefficient of rolling friction μr , the tractive resistance. Typical values of μr are 0.002 to 0.003 for steel wheels on steel rails and 0.01 to 0.02 for rubber tires on concrete.

35. Example 5.18 Motion with rolling friction • A typical car weighs about 12,000 N. if the coefficient of rolling friction is µr = 0.015, what horizontal force is needed to make the car move with constant speed on a level road? Neglect air resistance.

36. Fluid Resistance and Terminal speed • Fluid resistance is the force that a fluid (a gas or liquid) exerts on a body moving through. The direction of the fluid resistance force acting on body is always opposite the direction of the body’s velocity relative to the fluid. • The magnitude of the fluid resistance force usually increase with speed of the body through the fluid. • For very low speed, the magnitude f of the fluid resistance force can be described using the following equation Fluid resistance at low speed Where k is a proportionality constant that depends on the shape and size of the body and the properties of the fluid. The unit of k is N∙s/m or kg/s

37. Air drag • In motion through air at the speed of a tossed tennis ball or faster, the resisting force is approximately proportional to v2 rather than v. It is then called air drag or simply drag. Airplanes, falling raindrops, and bicyclists all experience air drag. The air drag on a typical car is negligible at low speeds but comparable to or greater than rolling friction at highway speeds. Fluid resistance at high speed The value of D depends on the shape and size of the body and on the density of the air. The units of the constant D are N∙s2/m2 or kg/m

38. Because of the effects of fluid resistance, an object falling in a fluid does not have a constant acceleration. • To find acceleration at a point of time, we need to use Newton’s second law.

39. Terminal velocity – slow moving object • When the rock first starts to move, vy = 0, the resisting force is zero, and the initial acceleration is ay = g. As the speed increases, the air resistance force also increases, until finally it is equal in magnitude to the weight. At this time mg – k∙vy = 0, the acceleration becomes zero, and there is no further increase in speed. The final speed vt, called the terminal speed is given by mg – k∙vt = 0 or

40. Terminal velocity – fast moving object • When the rock first starts to move, vy = 0, the resisting force is zero, and the initial acceleration is ay = g. As the speed increases, the air resistance force also increases, until finally it is equal in magnitude to the weight. At this time mg – D∙vy2 = 0, the acceleration becomes zero, and there is no further increase in speed. The final speed vt, called the terminal speed is given by mg – D∙vy2 = 0, or ∑Fy = mg + (-Dvy2) = may

41. Graphs of the motion of a body falling with fluid resistance proportional to the speed.

42. 5.19 Determine the terminal speed of a sky diver • For a human body falling through air in a spread-eagle position, the numerical value of the constant D is about 0.25 km/m. Find the terminal speed for a lightweight 50 kg skydiver. vt= 44 m/s

43. Test Your Understanding 5.3 • Consider a box that is placed on different surfaces. • In which situation(s) is there no friction force acting on the box? • In which situations(s) is there a static friction force acting on the box? • In which situation(s) is there a kinetic friction force on the box? • The box is at rest on a rough horizontal surface. • The box is at rest on a rough tilted surface. • The box is on the rough-surfaced flat bed of a truck, the tuck is moving at a constant velocity on a straight, level road, and the box remains in the same place in the middle of the truck bed. • The box is on the rough-surfaced flat bed of a truck, the truck is speeding up on a straight, level road, and the box remains in the same place in the middle of the truck bed. • The box is on the rough surfaced flat bed of a truck, the truck is climbing a hill, the box is sliding toward the back of the truck. • i, iii • ii, iv • v

44. The dynamics of uniform circular motion • In uniform circular motion, both the acceleration and force are centripetal. • Cut the cord restraining an object in such motion and observe the object’s behavior without the inward force.

45. Equations in Uniform Circular Motion

46. Example 5.20 Force in uniform circular motion • A sled with a mss of 25.0 kg rests on a horizontal sheet of essentially frictionless ice. It is attached by a 5.00 m rope to a post set in the ice. Once given a push, the sled revolves uniformly in a circle around the post. If the sled make five complete revolutions every minute, find the force F exerted on it by the rope.

47. Example 5.21 The conical pendulum • An inventor proposes to make a pendulum clock using a pendulum bob with mass m at the end of a thin wire of length L. instead of swinging back and for, the bob moves in a horizontal circle with constant speed v, with the wire making a constant angle β with the vertical direction. This system is called a conical pendulum because the suspending wire traces out a cone. Find the tension F in the wire and the period T in terms of β, g and L.

48. Example 5.22 Rounding a flat curve • A sports car is rounding a flat, unbanked curve with radius R. if the coefficient of static friction between tires and roads is µs, what is the maximum speed vmax at which the driver can take the curve without sliding?

49. Example 5.3 Rounding a banked curve • For a car traveling at a certain speed, it is possible to bank a curve at just the right angle so that no friction at all is needed to maintain the car’s turning radius. Then a car can safely round the curve even on wet ice. Your engineering firm plans to rebuild the curve in Example 5.22 so that a car moving at speed v can safely make the turn even with no friction. At what angle β should the curve be banked?

50. Banked curves and the flight of airplanes • Example 5.23 also apply to an airplane when it makes a turn in level flight. • Lcosβ = mg • Lsinβ=mv2/R • tanβ = v2/gR