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Transcript
Chapter 5 Forces (Part the First) Forces “A push or a pull” Can cause acceleration Aristotelian (wrong) vs Newtonian (mostly right) mechanics Limitations of Newtonian mechanics Objects moving very fast (special relativity) Very small objects (quantum mechanics) Very small objects moving very fast (quantum field theory) 6/28/04 2 Forces Unit: kg m/s2 [Newtons] Force is a vector quantity Direction and magnitude The sum of the forces on an object is Fnet 6/28/04 3 The Four Fundamental Interactions 6/28/04 4 The Practical Forces of Nature Gravity Restoring forces Springs, Constraining forces Tension, etc. normal force Dissipative forces Friction, 6/28/04 air resistance 5 Newton’s 1st Law If there are no forces acting on an object, the velocity of the object will not change The “law of inertia” An object in motion will remain in motion, and object at rest will remain at rest, until acted upon by an outside force Then why do things seem to slow down when we stop pushing them? 6/28/04 6 Newton’s First Law vs Hollywood Is this possible? 6/28/04 7 Inertial Reference Frame A reference frame in which Newton’s laws hold Non-accelerating The earth is approximately an inertial reference frame Rotation, motion around the sun/galaxy/universe Coriolis effect 6/28/04 8 Newton’s 2nd Law The net force on a body is equal to the product of the body’s mass and the acceleration of the body Fnet ma Can break force into components Treat each direction independently Fnet , x max 6/28/04 Fnet , y may Fnet , z maz 9 What is Mass? Intrinsic property NOT dependent on an objects size, weight or density Proportionality constant between force and acceleration Difference between pushing a tennis ball and a bowling ball 6/28/04 10 Newton’s 3rd Law When object A exerts a force on object B, object B exerts a force on object A of equal magnitude and opposite direction “Action and reaction” A B FA on B FB on A 6/28/04 11 Forces: Gravity An object of mass m in freefall will experience a force: Fg m g Weight is equal to magnitude of Fg A peculiarity in Halliday, Resnick, and Walker W mg Note: Mass is intrinsic, while weight depends on where you are. They are NOT the same thing. 6/28/04 12 Demo: Newton’s Carts Two people push off each other The force is the same in magnitude, opposite in direction FA on B FB on A m A a A mB a B FB on A FA on B A 6/28/04 B 13 Demo: Newton’s Carts Double the mass on one cart. What happens? FA on B FB on A m A a A mB a B FB on A A 6/28/04 FA on B B 14 Forces: Normal “Normal” in the mathematical sense meaning perpendicular When an object pushes against a surface, the surface pushes back perpendicular to its surface with a force N N Fg 6/28/04 15 Forces: Friction Force that resists motion when an object slides over a surface Always opposite to the direction of motion We’ll discuss more next chapter v Ffr 6/28/04 16 Forces: Tension Force exerted by a rope Always points in the same direction as the rope Ropes are assumed to be massless unless otherwise specified v T 6/28/04 Ffr 17 Tension (Continued) T Frope Frope 1. A rope can only pull (cannot push) 2. Objects at the ends of a taut rope have the same magnitude velocity and acceleration 3. Frictionless and massless pulleys only change direction of rope, not the tension. T F 6/28/04 T F 18 Free Body Diagrams The first step in solving any force problem: 1. 2. 3. 4. Sketch the object in question Draw an arrow for each force acting on the object Label each force Indicate the direction of acceleration off to the side (acceleration is NOT a force) F2 F1 F3 6/28/04 a 19 Net Force A free body diagram is used to calculate the net force on one object. Fnet= m a Note: The two equal forces in Newton’s Third Law are on different objects. They don’t appear on the same free body diagram. 6/28/04 20 How to Solve a Force Problem 1. 2. 3. 4. 5. 6/28/04 Draw a free body diagram Guess at forces of unknown magnitude or direction Break forces into components Sum of all the forces in one direction = mass * acceleration in that direction i.e. Fnet,x=m ax Repeat step 3 for each direction Solve for unknown quantities 21 Example: A Standing Person Say a man has a mass of 60 kg. Find the normal force exerted on him by the ground. 1) Draw a diagram N 2) Sum the forces in the y direction a=0 Fnet , y may N mg 0 N mg (60 kg)(9.8 m / s 2 ) 588 N Fg=mg 6/28/04 Note: We know the direction of the normal force from the diagram 22 Example: A Standing Person Now say our 60 kg man is jumping with a = 2 m/s2 1) Draw a diagram N 2) Sum the forces in the y direction Fy may a=2 m/s2 N mg ma N mg ma m( g a) N (60 kg) (9.8 m / s 2 ) (2 m / s 2 ) 708 N Fg=mg 6/28/04 Notice: The normal force is variable 23 Example: Ring on a Table Ring centered on round table and strings with weights are attached 0.5 kg How much force is needed to balance forces so the ring stays centered (zero acceleration)? 30° 1 kg 6/28/04 24 Solution: Draw a Free-Body Diagram A free body diagram shows all the forces operating on a single object F3,y F2 4.9 N We had to make an educated guess for where to put F3. If we guessed wrong, that’s OK. We will just pick up a minus sign F3 q 30° F3,x F1 9.8 N 6/28/04 25 Solution: Break Forces Into Components F3,y F1 : F3 F1, x 0 F1, y 9.8 N F2 4.9 N 30° F3,x F2 : F2, x (4.9 N ) cos 30 4.24 N F2, y (4.9 N ) sin 30 2.45 N 6/28/04 F1 9.8 N 26 Solution: Summing Forces In the y-direction: In the x-direction: Fnet , x max Fnet , y may F1, x F2, x F3, x 0 F1, y F2, y F3, y 0 0 4.24 N F3, x 0 9.8 N 2.45 N F3, y 0 F3, x 4.24 N F3, y 7.35 N F3,y F3 F2 4.9 N 6/28/04 30° F3,x F1=9.8 N 27 Solution: The magnitude of F3 is: F3 F3, x F3, y 2 F3,y F3 2 (4.24 N ) 2 (7.35 N ) 2 F2 4.9 N 30° q F3,x 8.5 N (0.87 kg) g The angle of F3 is: F1=9.8 N F3, y q tan F 3 , x 1 7.35 N 60 tan 4.24 N 1 6/28/04 Does it work? 28 Pig on Frictionless Inclined Plane N a Fg q Assume the pig has a mass of m and the angle of the ramp is q. Find N and a. Need to consider forces || and to plane… 6/28/04 29 Pig Forces: Break Into Components Acceleration: y ax a ay 0 N x a Normal Force: Nx 0 Ny N θ Fg,y θ Fg Gravity: Fg , x Fg sin q mg sin q Fg,x Fg , y Fg cosq mg cosq 6/28/04 30 Pig Forces: Sum Forces y-direction: Fnet , y may N y Fg , y 0 y N x a N mg cos q 0 N mg cosq x-direction: Fnet , x max θ Fg,y θ Fg Fg,x Fg , x max mg sin q ma 6/28/04 a g sin q 31 Pig on a Plane: Sanity Check N a q Fg N = mg cos θ θ= 0 θ = 90° 6/28/04 and a = g sin θ a=0 and N = mg a=g and N = 0 32 Cart on a Plane Now let’s add friction. A cart of mass m is stationary on an inclined plane of angle q. Find Ffr and N. N Ffr Fg q a=0 6/28/04 33 Cart on a Plane: Components Normal Force: N Nx 0 Ny N Gravity: Ffr Fg,y q Fg Fg,x Fg , x Fg sin q mg sin q Fg , y Fg cosq mg cosq Friction: F fr , x F fr Once again, use a “tilted” coordinate system F fr , y 0 6/28/04 34 Cart: Summing the Forces y-direction: Fnet , y may N Ffr N y Fg , y 0 N mg cos q 0 N mg cosq Fg,y q Fg Fg,x x-direction: Fnet , x max Fg , x F fr , x max F fr mg sin q 0 F fr mg sin q 6/28/04 35 Does it Work? If we replace the normal force and friction with another force, the cart should stay stationary after the ramp is removed. The mass of the cart is 1 kg and the angle of the ramp is 30° N mg cosq (1 kg) g cos 30 (0.87 kg) g Ffr mg sin q (1 kg) g sin 30 (0.5 kg) g 6/28/04 36 Tension A 16 lb. bowling ball is hung from a rope attached to a scale which is attached to a wall. The scale reads 16 lbs. It is now attached to another bowling ball. What does it read? A) 0 B) 16 lbs C) 32 lbs 6/28/04 37 Tension in Two Dimensions T3 30o 60o T2 30o 60o T1 Consider the mass: T1 m Summing forces: Fnet , y ma y 0 m mg T1 mg 0 6/28/04 T1 mg 38 Tension in 2-D (Continued) Consider the knot: T3 T3 60o 30o T2 30o 60o m 30o 60o T1 y-direction: 6/28/04 T2 T1 x-direction: Fnet , y may 0 Fnet , x max 0 T2 sin 60 T3 sin 30 T1 0 T2 cos 60 T3 cos 30 0 T2 sin 60 T1 T3 sin 30 cos 30 T2 T3 cos 60 39 Tension in 2-D (Continued) What we know: T1 mg cos 30 T2 T3 cos 60 T2 sin 60 T1 T3 sin 30 Jumping through the algebraic hoops (left as an exercise for the ambitious student): mg T3 0.5 mg cos 30 tan 60 sin 30 T2 0.866 mg T1 mg 6/28/04 40 Example: Fuzzy Dice With no acceleration Frope Fnet,y = may = 0 Fg Frope- Fg = 0 Frope= -Fg = mg What happens when we accelerate? 6/28/04 41 Fuzzy Dice… Find q: q Frope x direction: Frope sin q ma y direction: ma Frope cosq mg 6/28/04 Fnet , y ma y 0 Frope cos q mg 0 Fg Frope sin q Fnet , x max Frope cos q mg a tan q g e.g. a=g θ=45° 42 Another Example: Find the acceleration of this system: Start with free body diagrams of both blocks a1 N T T m2 q 6/28/04 a2 x m2 FG Recall: a1 = a2 = a y FG 43 Solution: Summing forces in the x direction: a1 N T Fnet , x ma1, x T m1 g sin q m1a T m1 (a g sin q ) Summing forces in the y direction: FG 6/28/04 N m1 g cosq 0 N m1 g cosq 44 Solution y T a2 x Summing forces in the y direction: Fnet , y m2 a2, y T m2 g m2 a m2 T m2 (a g ) FG 6/28/04 45 Solution m2 What we have: N m1 g cosq T m1 (a g sin q ) T m2 (a g ) q The tensions must be equal to each other m2 (a g ) m1 (a g sin q ) m2 a m2 g m1a m1 g sin q a (m1 m2 ) m2 g m1 g sin q 6/28/04 m2 g m1 g sin q a m1 m2 46 Fan Cart With a Sail: Will It Move? A) Yes, in the same direction as without the sail B) Yes, in the opposite direction C) No D) No Clue 6/28/04 47