* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Notes - UMD Physics
Survey
Document related concepts
Coriolis force wikipedia , lookup
Hunting oscillation wikipedia , lookup
Frictional contact mechanics wikipedia , lookup
Seismometer wikipedia , lookup
Equations of motion wikipedia , lookup
Modified Newtonian dynamics wikipedia , lookup
Jerk (physics) wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Classical mechanics wikipedia , lookup
Rigid body dynamics wikipedia , lookup
Fictitious force wikipedia , lookup
Centrifugal force wikipedia , lookup
Classical central-force problem wikipedia , lookup
Transcript
Physics for Scientists and Engineers Chapter 6 Dynamics I: Motion Along a Line Spring, 2008 Ho Jung Paik 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, T, is the tension in the rope Interested only in the external forces acting on the object Can neglect internal reaction forces 27-Feb-08 Paik p. 2 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 ∑F =0 ∑F = 0, ∑F = 0 x 27-Feb-08 y Paik p. 3 Static Equilibrium, Example 1 A traffic light is suspended with three ropes as shown in the figure. Find the tension in the three ropes. The 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. 27-Feb-08 Paik p. 4 Example 1, cont Light: ∑Fy = 0: T3 − mg = 0 ⇒ T3 = mg Knot : ∑Fx = 0: −T1 cos 37˚+T2 cos 53˚= 0 ⇒ T1 = 0.754 T2 ∑Fy = 0: T1 sin 37˚+T2 sin 53˚−T3 = 0 ⇒ T2 = 0.798 mg ⇒ T1 = 0.602 mg 27-Feb-08 Paik p. 5 Dynamic Equilibrium, Example 2 A car with a weight of 15,000 N is being towed up a 20° slope at constant velocity. Friction is negligible. The tow rope is rated at 6000 N. Will it break? Know: θ = 20°, w = 15,000 N Find: T 27-Feb-08 Paik p. 6 Example 2, cont Apply Newton’s 2nd Law (a = 0) r ∑F = 0 ∑ F = T − w sin θ = 0 x ∑ F = n − w cos θ = 0 y From the x equation, T = w sin θ = (15,000 N) sin 20 0 = 5130 N Since T < 6,000 N, the rope won’t break. Note that y equation is not needed. Check solution at θ = 0° and 90°. 27-Feb-08 Paik p. 7 Objects Experiencing a Net Force 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 r ∑ F = ma ∑F x 27-Feb-08 = ma x , ∑F y Paik = ma y p. 8 Crate Pulled on Floor, Example 3 A crate weighing 490 N is pulled with a rope on a frictionless floor. The tension in the rope is 20 N. What is the resulting acceleration of the crate? Forces acting on the crate: Tension: T = 20 N Gravitational force: Fg = 490 N Normal force n exerted by the floor 27-Feb-08 Paik p. 9 Example 3, cont Apply Newton’s 2nd Law in component form: Solve for the unknown ax: T 20 N 2 ax = = = 0 . 400 m/s m 490 N/9.80 m/s 2 If a = constant and vx0 = 0, then vxt = a x t = (0.400 m/s 2 )t 27-Feb-08 Paik p. 10 Car on Incline, Example 4 A car is at rest on top of a hill. Find the car’s velocity at the bottom of the 30.0 m hill with an incline angle of 30.0˚. Forces acting on the object: ⊥ Normal force n the plane Gravitational force Fg straight down Choose the coordinate system with x // the incline, y⊥ the incline Replace the force of gravity with its components 27-Feb-08 Paik p. 11 Example 4, cont Apply Newton’s 2nd Law in component form: 0 F = mg sin 30 = max ∑ x ∑F y = n − mg cos 30° = 0 ax = ? Solve for the unknown ax: a x = g sin 30 0 = 4.90 m/s 2 The velocity at the bottom of the slope is obtained from vxf2 = vxi2 + 2ax Δx, vxf = 0 + 2(4.90 m/s2 )30.0 m = 17.1 m/s 27-Feb-08 Paik p. 12 Weight & Apparent Weight, 1 Weight is the force exerted by the earth on mass m w = mg, where g = 9.80 m/s2 Consider a scale and weight at rest When the weight comes to rest with respect to the spring, Fsp = w The pointer is calibrated to read Fsp = w at equilibrium (a = 0) 27-Feb-08 Paik p. 13 Weight & Apparent Weight, 2 Consider an object hanging from a scale in an elevator moving up with acceleration a T = Fsp The y component of 2nd Law: ∑ Fy = Fsp − mg = ma ∴ Fsp = m( g + a ) The spring dial comes to rest with the force Fsp= m (g+a) > w m (g+a) is the apparent weight 27-Feb-08 Paik p. 14 Weight & Apparent Weight, 3 Now consider the elevator accelerating downward T = Fsp The y component of 2nd Law: ∑F y = Fsp − mg = m(−a) ∴Fsp = m( g − a) The scale reads Fsp= m(g -a) < w If a =g (freefall), the apparent weight is 0 ⇒ “weightless” 27-Feb-08 Paik p. 15 Multiple Weights/Objects Forces acting on the objects: Tension (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 27-Feb-08 Paik p. 16 Example 5 An object of mass m1 is connected to a light string that passes over a frictionless pulley and is fastened to another object of mass m2. Find the acceleration of the objects. ∑F : ∑F m1 : y = T − m1 g = m1a ⇒ T = m1 ( g + a ) m2 y = T − m2 g = m2 (−a ) ⇒ T = m2 ( g − a ) m2 − m1 ⇒ m1 ( g + a) = m2 ( g − a) ⇒ a = g m2 + m1 Check if the answer makes sense. m1 = m2 ⇒ a = 0 m1 << m2 ⇒ a ≈ g, m1 >> m2 ⇒ a ≈ −g 27-Feb-08 Paik p. 17 Multiple Pulleys, Example 6 Assume the block is accelerated upward, i.e. a > 0. Let F = 20.0 N and M = 5.00 kg. Find the acceleration and all the T’s. All the pulleys and strings are considered massless and there is no friction in the pulleys ⇒ F = T1 = T3 = T2 Note that top pulley doesn’t move. The bottom pulley and mass M have the same acceleration. 27-Feb-08 Paik p. 18 Example 6, cont M: Plower ∑ F = T − Mg = Ma ⇒ : ∑ F = T + T − T = 0a y 5 y 2 . 3 T5 = M ( g + a ) 5 ⇒ T2 + T3 = T5 = M ( g + a ) ⇒ T2 = T3 = 12 M ( g + a ) T1 Pupper : ∑ Fy = T4 − T1 − T2 − T3 = 0 ⇒ T1 = T3 = 12 M ( g + a ), T4 = 32 M ( g + a ) S: ∑F y = T1 − F = 0 ⇒ F = T1 = 12 M ( g + a ) Therefore, 2F 2(20 N) a= −g = − 9.8 m/s 2 = −1.8 m/s 2 , a <0! M 5 kg T1 = T2 = T3 = F = 20 N, T4 = 3F = 60 N, T5 = 2 F = 40 N 27-Feb-08 Paik F p. 19 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. Find the acceleration of the objects and the tension in the string. ∑F = 9 kg : ∑ F ∑F n m1 = 5 kg : x = T = m1a, m2 y = T − m2 g = m2 (− a ) y = n − m1 g = 0 m2 9.00 ⇒a= g= × 9.80 m/s 2 = 6.30 m/s 2 m1 + m2 5.00 + 9.00 T = 5.00 kg × 6.30 m/s 2 = 31.5 N 27-Feb-08 Paik a T m1g T a m2g p. 20 Ball and Cube, Example 8 Two objects are connected by a light string that passes over a frictionless pulley. The incline is frictionless, and m1 = 2.00 kg, m2 = 6.00 kg, and θ = 55.0°. Draw free-body diagrams of both objects. Find (a) the acceleration of the objects, (b) the tension in the string, and (c) the speed of each object 2.00 s after being released from rest. 27-Feb-08 Paik p. 21 Example 8, cont y (a) m1 = 2.00 kg : a ∑ Fy = T − m1 g = m1a T m2 = 6.00 kg : a ∑ Fx = m2 g sin θ − T = m2 a m2 g sin θ − m1 g ⇒ a= = 3.57 m/s 2 m1 + m2 T n y’ x θ m1g m g cosθ 2 m2g sinθ m2g x’ (b) T = m1 (a + g ) = 26.7 N (c) v f = vi + at = 0 m/s + (3.57 m/s 2 )(2.00 s) θ = 55.0° = 7.14 m/s 27-Feb-08 Paik p. 22 Forces of Friction When an object is in contact with a surface, there is a resistance to its motion, the force of friction This is due to the interaction between the molecules on the mating surfaces Friction is proportional to the normal force Static friction: ƒs ≤ µsn and kinetic friction: ƒk= µkn These equations relate the magnitudes of the forces only Generally, µs > µk The coefficient of friction (µ) depends on the surfaces in contact but is nearly independent of the area of contact 27-Feb-08 Paik p. 23 Static Friction There is a maximum static frictional force, fs,max = µsn. If the applied force is greater, then the object slips. 27-Feb-08 Paik p. 24 Kinetic Friction The kinetic friction coefficient is less than the static friction coefficient, µk < µs. Therefore, fk = µkn = constant < fs,max. 27-Feb-08 Paik p. 25 Friction Forces vs. Applied Force 27-Feb-08 Paik p. 26 Rolling Friction Kinetic friction is operable for a wheel sliding on a surface Rolling friction is for a non-sliding, rolling wheel fr = µrn and points opposite to the motion µr < µk 27-Feb-08 Paik p. 27 Some Coefficients of Friction 27-Feb-08 Paik p. 28 Friction 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 Friction is a force, so it is simply included in the ΣF in Newton’s Laws Continue with the solution as with any Newton’s Law problem 27-Feb-08 Paik p. 29 Static Friction - Measuring µs The 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 = mg sin θ − f x s, max = 0, ∑ F = n − mg cos θ = 0 y ⇒ f s , max = mg sin θ = μs n = μs mg cos θ ⇒ μs = tan θ 27-Feb-08 Paik p. 30 Sliding Down a Plane, Example 9 If µs = 0.30 and θ = 30°, will the cube slide? Will not slide if fs ≥ mg sinθ = 0.5mg Since fs ≤ µsn and n = mg cosθ, fs ≤ µsmg cos 30° = 0.30 x 0.866mg = 0.26mg ⇒ Will slide 27-Feb-08 Paik p. 31 Truck with a Crate, Example 10 A truck is carrying a crate up a 10° hill. The coefficient of static friction between the truck bed and the crate is μs = 0.35. Find the maximum acceleration that the truck can reach before the crate begins to slip backward. n The crate will not slip if its acceleration is equal to that of the truck. For it to be accelerated, some force must act on it. One of these is the static friction force fs. This force must be directed upward to keep the crate from sliding backwards. As the acceleration of the truck increases, so must fs. Once fs, max = µsn is reached, the crate will start sliding. 27-Feb-08 Paik p. 32 Example 10, cont Applying Newton' s law to the crate : ΣFx = − mg sin θ + μ s n = ma max ΣFy = − mg cos θ + n = 0 n ⇒ n = mg cos θ − mg sin θ + μ s mg cos θ = ma max ⇒ a max = g ( μ s cos θ − sin θ ) Substituting θ = 10° and μ s = 0.35, a max = 1.68 m/s 2 27-Feb-08 Paik p. 33 Hockey Puck, Example 11 Consider a puck is hit and given a speed v = 20.0 m/s. How far will it go if µk = 0.1? 1) x : −fk = ma y : n − mg = 0 a = −fk/m = −µkn/m = −µkg 2) Use vf2 = vi2 + 2aΔx vi = 20.0 m/s, vf = 0, Δx = ? Δx = −vi2 /2a = −400/(−2 x 0.1 x 9.80) = 204 m 27-Feb-08 Paik p. 34 Pulling at an Angle, Example 12 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. The optimum angle must be somewhere in between. 27-Feb-08 Paik p. 35 Example 12, cont x : − f + F cos θ = max = 0 y : n + F sin θ − mg = 0 n = mg − F sin θ f = μ k (mg − F sin θ ) μ k mg Substituting for f in the x equation : F = μ k sin θ + cos θ μ k cos θ − sin θ dF Minimum occurs when = − μ k mg =0 2 dθ (μ k sin θ + cosθ ) Therefore, θ = tan −1 μ k minimizes the pulling force. 27-Feb-08 Paik p. 36 Example 13 A van accelerates down a hill, going from rest to 30.0 m/s in 6.00 s. During the acceleration, a toy (m = 0.100 kg) hangs by a string from the van's ceiling. The acceleration is such that the string remains perpendicular to the ceiling. Determine: (a) the angle and (b) the tension in the string. 27-Feb-08 Paik p. 37 Example 13, cont Choose x-axis pointing down the slope. x : vf = vi + at 30.0 m/s = 0 + a(6.00 s) ⇒ a = 5.00 m/s2 Consider the forces on the toy. x : mg sinθ = m (5.00 m/s2) ⇒ θ = 30.7° y : T – mg cosθ = 0 T = 0.100 kg x 9.80 m/s2 x cos 30.7° ⇒ T = 0.843 N 27-Feb-08 Paik p. 38 Example 14 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 250 N. Pat's true weight is 320 N, and the chair weighs 160 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) Find the force Pat exerts on the chair. 27-Feb-08 Paik p. 39 Ex 14, cont (a) Free-body diagrams: (b) Consider Pat and the chair as the system. Note that two ropes support the system and each rope has T = 250 N. ∑Fy = may ⇒ 2 x 250 N – 480 N = (480 N/9.80 m/s2)a ⇒ a = 0.408 m/s2 > 0. Therefore, upward. (c) Now consider Pat as the system: ∑Fy = n + T – 320 N = (320 N/9.80 m/s2)a ⇒ n = –250 N + 320 N + 32.7 kg x 0.408 m/s2 = 83.3 N 27-Feb-08 Paik p. 40 Motion with Resistive Forces Motion can be through a medium Medium exerts a resistive drag force D on an object moving through the medium Either a liquid or a gas 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 2 27-Feb-08 Paik p. 41 Drag Proportional to v For objects moving at low speeds, the drag is approximately proportional to v : D = −b v b depends on the property of the medium, and on the shape and dimensions of the object The negative sign indicates that D is opposite in direction to v The equation of motion of the suspended mass: D − mg = m(− a ) dv bv − mg = − m dt 27-Feb-08 Paik D p. 42 Terminal Speed for D ∝ v Initially, v = 0 ⇒ D = 0 and dv/dt = −g As t increases, D → mg, a → 0 and v → vT : the terminal speed of the object: dv bvT − mg = − m = 0 dt mg ⇒ vT = b v v (t) is found by solving the differential equation: v = vT (1 − e − t / τ ), τ = m / b 27-Feb-08 Paik t p. 43 Drag Proportional to v 2 For objects moving at high speeds, the drag is approximately proportional to v 2: D = (1/2)cρAv2 c is a dimensionless empirical “drag coefficient” ρ is the density of the fluid A is the cross-sectional area of the object 27-Feb-08 Paik p. 44 Terminal Speed for D ∝ v 2 Analysis of an object falling through a medium accounting for air resistance D D − mg = m(− a ), 12 cρAv 2 − mg = − ma ⎛ cρA ⎞ 2 ⇒ a = g −⎜ ⎟v ⎝ 2m ⎠ The terminal speed will occur when the acceleration goes to zero 2mg Solving the equation gives vT = cρA 27-Feb-08 Paik p. 45 Some Terminal Speeds 27-Feb-08 Paik p. 46