Applications of Newton s Laws

Transcription

1 Chapter 5 Applications of Newton s Laws We shall apply Newton s laws of motion to a variety of situations in which several types of forces are involved: Gravitational force Tension Contact Forces (including friction) (5-1)

2 m g F g Gravitational force F g Also known as the weight of an object earth All objects near the surface of the earth are uniformly accelerated towards the center of the earth. The magnitude of the acceleration g = 9.8 m/s 2 From Newton s second law we get that F g = ma = mg (5-2)

3 A practical method of measuring F g Use a spring scale as shown in the figure. The spring has a pointer attached to it so that we can determine its elongation y-axis System = grapes. There are two forces acting on the grapes: The gravitational force mg and the spring force F s which is given by: F s = k y. Here y is the elongation of the spring The net force F ynet = F s - mg = ma y = 0 because the grapes are stationary. Thus mg = F s = k y (5.3)

4 crate floor T rope Tension It is the force exerted by a rope or a cable Tension has the following characteristics: Tension always pulls (never pushes) The direction of T is along the rope The magnitude T is constant even when T changes direction pulley T T (as in the case of the pulley) (5-4) crate pull

5 Normal Force (contact force) Characteristics of the normal force F N F N is perpendicular to the contact surface F N always pushes away from the contact surface (never pulls) F N crate F g floor (5-5)

6 The normal force and Newton s third law crate crate floor floor F N = F cf F fc System = crate The normal force on the crate exerted by the floor is labeled F cf System = floor The normal force on the floor exerted by the floor is labeled F fc Newton s third law: F cf + F fc = 0 F fc = - F cf = - F N The crate exerts a force F fc on the floor which is equal and opposite to F N (5-6)

7 Q: How can we measure F N? A: With a bathroom scale! The bathroom scale measures the normal force (and not necessarily mg unless the object on the scale does not accelerate) (5-7)

8 F as is the normal force exerted on the apple by the scale. apple scale y-axis F N = F as F N = F sa F sa is the normal force exerted on the scale by the apple These two forces are equal and opposite by virtue of Newton s third law System = apple F ynet = F N - mg F N F ynet = ma y = 0 (5-8) mg F N - mg = 0 Indeed: F N = mg

9 y-axis elevator What does the scale reads when the elevator accelerates upwards?.. System = man F ynet = F N - mg F ynet = ma (Newton s second law) F N F N - mg = ma F N = mg + ma a scale mg F N > mg The scale in this case gives a reading which is higher than the actual weight mg (5-9)

10 y-axis.. elevator What does the scale reads when the elevator accelerates downwards? System = man F ynet = F N - mg F ynet = - ma (Newton s second law) F N F N - mg = - ma F N = mg - ma a scale mg F N < mg The scale in this case gives a reading which is lower than the actual weight mg (5-10)

11 a m 2 a y Atwood s Machine It consists of two masses m 1 and m 2 (m 2 > m 1 ) connected with a rope that goes around a pulley. The rope and the pulley have negligible masses. m 1 The two masses are released and move under gravity which acts downwards. The heavier mass m 2 will move down and the lighter mass m 1 will move up. Determine the acceleration a of the two masses (5-11)

12 a a y y T a y T a m 1 g m 1 m 2 g m 2 m 2 ynet m 1 System= m 1 F = T mg = ma 1 1 T = mg eqs ma 1 System= m 2 F = T mg = ma ynet 2 2 T = mg 2 ma 2 eqs. 2 We combine eqs.1 and eqs.2 ( m2 m1) a = g ( m + m ) mg+ ma = mg ma Solve for a (5-12)

13 Friction Friction opposes motion or impending motion (5-13)

14 (5-14) motion crate f floor F N mg T rope Friction was first studied by Leonardo da Vinci He discovered that friction has the following characteristics: Friction opposes motion or impending motion Static and kinetic friction are independent of contact area The static and kinetic friction are proportional to F N 0 < fs <µ SFN Static friction f s µ s is a constant known as coefficient of static friction Kinetic friction f k k N k µ k is a constant known as coefficient of kinetic friction f = µ F

15 Leonardo da Vinci (self portrait) ( )

16 motion Consider the following experiment: crate f floor F N mg T rope We start pulling a heavy crate using a rope with a force T that increases with time. At some instant t m the crate will start to move We then plot the frictional force f as function of time. The static friction increases with time. At t m the static friction becomes maximum (µ s F N ) and the crate starts to move. The kinetic friction has a smaller constant value of µ k F N. (5-15) f Static friction t m m k < m s m s F N Motion starts kinetic friction t m k F N

17 y x Example (5-6),page 121 A mass m is at rest on a horizontal ramp. The ramp is slowly raised. Find the critical angle θ c at which the mass will start to slide F = F - mgcos θ = ma = 0 F = mg cos θ ynet N y N F = mgsinθ - f = ma = 0 mgsin θ = f = µ F xnet S x S S N From eqs.1 substitute F into eqs.2 mgsin θ = µ mg cos θ S N tanθ C = µ S ( eqs.1) ( eqs.2) (5-16)

18 Example (5-8) page 123. An automobile has a mass m = 1000 kg. The static friction with the road is µ s = 0.8 Determine the maximum forward acceleration (without spinning the wheels) F ynet = F N - mg = ma y = 0 F N = mg (eqs.1) F xnet = f s = ma a = f s /m a max = (f s ) max /m = µ s F N /m (eqs.2) From eqs.1 substitute F N into eqs.2 a max = µ s mg /m a max = µ s g = = 7.8 m/s 2

19 When the rope breaks the book does not stay on its circular path but instead flies along the tangent to the orbit Conclusion: For an object to be able to move on a circular orbit a force that points towards the center of the circle is required. This force is known as the centripetal force. When the centripetal force stops acting the object cannot stay on its circular orbit (5-18)

20 (5-19) m m m Consider an object of mass m moving on a circle of radius r with uniform speed v. The acceleration points towards the center of the orbit and has magnitude a = v 2 /r = ω 2 r From Newton s second law a force F = ma must act on the object. Thus the centripetal force F is given by the expression: F = 2 mv r The magnitude of F is constant, its direction is not. It always points towards the center of the circular orbit. or 2 F = mω r

21 C. y R Recipe for problems that involve uniform circular motion of an object of mass m on a circular orbit of radius R with speed v m v x Draw the force diagram for the object Choose one of the coordinate axes (the y-axis in this diagram) to point towards the orbit center C Determine F ynet Set F ynet = mv 2 /R or F ynet = mω 2 R (5-20)

22 Example (5-11) page 128 (5-21) x y A small dice sits 0.15 m from the center of a turntable. If µ s between the turntable and the dice is 0.55 what is the largest possible angular speed such that the dice will not slide off F ynet = F N - mg = ma y = 0 F N = mg (eqs.1) F xnet = f s = mω 2 r (eqs.2) ω 2 = f s /mr ω 2 max = (f s ) max /mr From eqs.1 we have: (f s ) max = µ s F N Thus ω 2 max = µ s mg/mr ω max = [µ s g/r] 1/2 = 6 rad/s

23 Example (5-12) page 129 y F ynet = F N = mv 2 /R A space station in the form of a hollow circular tube rotates around its axis. If the distance from the station outer wall to the center is 50 m what should be the speed v of the outer wall so that a scale will read what it would read on earth (eqs.1) Also F N = mg (eqs.2) If we combine the two equations we get: mv 2 /R = mg v = [Rg] 1/2 = [50 9.8] 1/2 = 22.1 m/s (5-22)