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Introduction to Newton’s Laws • Introducing Forces • A force is a push or pull on an object. Forces are what cause an object to accelerate, or to change its velocity by speeding up, slowing down, or changing direction. • It is important that we learn to identify all the forces acting on an object, and to draw these forces as vectors. • Draw the forces on the book The table pushes up on the book. FTable A book rests on a table. FG Gravity pulls down on the book. • There are two main methods for drawing forces Force Diagram Free Body diagram FTable FTable FG FG • Draw a force diagram and a free body diagram for a monkey hanging motionless by one arm from a vine attached to a tree. Force Diagram FTension FG Free Body diagram FTension FG • Draw a force diagram and a free body diagram for a monkey hanging motionless by two arm from two vines hanging from a tree. Force Diagram Free Body diagram FTension1 FTension2 FTension1 FG FTension2 FG • Newton’s First Law (Law of Inertial) – A body in motion stays in motion at constant velocity and a body at rest stays at rest unless acted upon by a net external force. – It is often said that the Law of Inertia violates “common sense”. Why do you think some people say that? • Newton’s 1st Law – If there is zero net force on a body, it cannot accelerate, and therefore must move at constant velocity. This means • the body cannot turn. • the body cannot speed up. • The body cannot slow down. • An example of zero net force FTable A book rests on a table. FG Even though there are forces on the book, they are balanced. Therefore, there is no net force on the book. ∑F = 0 • Mass and Inertia – Chemists like to define mass as the amount of “stuff” or “matter” a substance has. – Physicists define mass as inertia, which is the ability of a body to resist acceleration by a net force. – What is the relationship between mass and inertia? Sample Problem: A heavy block hangs from a string attached to a rod. An identical string hangs down from the bottom of the block. Which string breaks a) when the lower string is pulled with a slowly increasing 1st string force? FTension 2nd string Fweight Fpull Top will break slow pull all forces act FTension Fpull b) when the lower string is pulled with a quick jerk? A quick jerk will break the 2nd string because of the inertia of the block harder to get the block to move quickly • Newton’s Second Law – A body accelerates when acted upon by a net external force. The acceleration is proportional to the net force and is in the direction which the net force acts. – ∑F = ma • where ∑F is the net force measured in Newtons (N) • m is mass (kg) • a is acceleration (m/s2) • Units of force – Newton (SI system) • 1 N = 1 kg m /s2 – 1 N is the force required to accelerate a 1 kg mass at a rate of 1 m/s2 – Pound (British system) • 1 lb = 1 slug ft /s2 • Working 2nd Law Problems 1. Identify the system being accelerated. 2. Define a coordinate system. 3. Identify forces by drawing a force or free body diagram. 4. Explicitly write ∑F=ma 5. Replace SF with the actual forces in your free body diagram. 6. Substitute numeric values, where appropriate, and solve for unknowns. • In a grocery store, you push a 14.5-kg cart with a force of 12.0 N. If the cart starts at rest, how far does it move in 3.00 seconds? m = 14.5 kg F = ma F = 12.0 N t = 3.00s a = F / m = 12.0N / 14.5 kg = 0.828 m / s2 x = x0 + v0t + 1/2at2 x = 1/2at2 = ½(0.828 m/s2)∙(3.00s)2 = 3.72 m • A catcher stops a 45 m/s pitch in his glove, bringing it to rest in 0.15 m. If the force exerted by the catcher is 803 N, what is the mass of the ball? m = ? kg F = 803 N v0 = 45 m/s v = 0 m/s ∆x = 0.15 m v2 = v02 + 2a∆x a = -v02/2∆x = -(45 m/s)2/(2∙0.15m) = -6750 m/s2 Fmitt F = ma Fball m = F /a = -803 N / -6750 m/s2 = 0.12 kg • A catcher stops a 48 m/s pitch in his glove, bringing it to rest in 0.15 m. If the force exerted by the catcher is 803 N, what is the mass of the ball? v0= 48 m/s v= 0 m/s x= 0.15 m F= 803 N a= ? m/s2 v2 = v02 + 2a∆x a = (v2 - v02) / (2∆x) a = ((0m/s)2 – (48m/s)2) / (2∙0.15m) = 7680 m/s2 F= ma m= F / a = 803 N / 7680 m/s2 = 0.105 kg • A 747 jetliner lands and begins to slow to a stop as it moves along the runway. Its mass is 3.50 x 105 kg, its speed is 27.0 m/s, and the net braking force is 4.30 x 105 N a= ? m/s2 5N F = 4.30x10 a) What is time it takes the jet to stop? v0= 27 m/s v= 0 m/s x= ? m t= ? s m= 3.50 x105 Kg a= F / m = 4.30x105 N / 3.50x105Kg = -1.23 m/s2 v = v0 + at t = (v - v0) / a = (0 m/s – 27m/s) / -1.23 m/s2 = 21.98s b) How far has it traveled in this time? v0= 27 m/s v= 0 m/s a= -1.23 m/s2 v2 = v02 + 2a∆x ∆x = (v2 - v02) / (2a) ∆x = ((0 m/s)2 – (27m/s)2) / (2 ∙ -1.23 m/s2) = -296.31 m • A 747 jetliner lands and begins to slow to a stop as it moves along the runway. Its mass is 3.50 x 105 kg, its speed is 27.0 m/s, and the net braking force is 4.30 x 105 N c) What is its speed 7.50 s later? v0= 27 m/s v= ? m/s t= 7.50 s a = -1.23 m/s2 v = v0 + at v = 27 m/s + (-1.23 m/s2 ∙ 7.50s = 17.78 m/s • Newton’s Third Law – For every action there exists an equal and opposite reaction. – If A exerts a force F on B, then B exerts a force of -F on A. • You rest an empty glass on a table. a) Identify the forces acting on the glass with a free body diagram. FTable FG b) Are these forces equal and opposite? Yes, because the glass is in equilibrium c) Are these forces an action-reaction pair? Why or why not? Yes, because it is not acceleration the forces are equal. The table is pushing up with the same force that the glass is pushing down on the table with. • Requirements for Newton’s Laws – The 1st and 2nd laws require that ONE system be analyzed and that ALL the forces on the system be accounted for. – The 3rd law requires that TWO systems be analyzed and that the forces of interaction between the two be accounted for. • A force of magnitude 7.50 N pushes three boxes as shown. Find the acceleration of the system. Copyright James Walker, “Physics”, 1st edition F = Ma M = m1 + m2 + m3 a = F / (m1 + m2 + m3) a = 7.5N / (1.3 kg + 3.2kg + 4.90 kg)= 0.798 m/s2 • force of magnitude 7.50 N pushes three boxes as shown. Find the force that box 2 exerts on box 3. Copyright James Walker, “Physics”, 1st edition a = 0.798 m/s2 Box 2 exerts on 3 ∑F = F - F23 ∑F =F2,3 = (m1 + m2)a F = (1.3 + 3.20) •0.798 = 3.59 N ∑F = F – m3a F = 7.50 - 4.90•0.798 = 3.59 N • force of magnitude 7.50 N pushes three boxes as shown. Find the force that box 2 exerts on box 3. Copyright James Walker, “Physics”, 1st edition a = 0.798 m/s2 Box 1 exerts on 2 ∑F =F1,2 = m1a = 1.3 ∙ 0.798 = 1.04 N ∑F = F - F23 – F12 ∑F = F – m3a – m2a F = 7.50 - 4.90•0.798 – 3.20•0.798 = 1.036 N • Newton’s 2nd Law in 2-D – The situation is more complicated when forces act in more than one dimension. – You must still identify all forces and draw your force diagram. – You then resolve your problem into an x-problem and a y-problem (remember projectile motion????). • Draw a force diagram and a free body diagram for the man pushing the chair across the floor. FFloor Ffriction Fperson Fchair Copyright James Walker, “Physics”, 1st edition • An object acted on by three forces moves with constant velocity. One force acting on the object is in the positive x direction and has a magnitude of 6.5 N; a second force has a magnitude of 4.4 N and points in the negative y direction. Find the direction and magnitude of the third force acting on the object. Fx = 6.5 N Fx = 6.5N Fy = 4.4 N R2 = Fx2 + Fy2 F= ? N Fy = 4.4 N c = √(Fx2 + Fy2 ) = √((6.5N)2 + (4.4N)2 ) = 7.84 N θ = tan-1 (Fy / Fx) = tan-1 (4.4 N / 6.5 N) = 34.1° Now we use the inverse vector F = 7.84 N θ = 34.1° Fx = 6.5N Fy = 4.4 N Mass, Weight, and Apparent Weight • Mass versus Weight – Mass is inertia, or resistance to acceleration. – Weight and mass are not equivalent! – Weight is gravitational force, which is equal to mg near the earth’s surface. • A man weighs 150 pounds on the surface of the earth at sea level. Calculate his a) mass in kg. 1 kg = 2.2 lbs 150 lbs 1 kg 2.2 lbs = 68.18 kg b) weight in Newtons. F = ma W = ma F = 68.18 kg ∙ -9.8 m/s2 = 668.18 N • Apparent weight – Apparent weight is a force that acts in opposition to gravitational force in order to prevent a body from going into freefall. – When you stand on the floor, the floor pushes up on your feet with a force equal to your apparent weight. • Elevator rides – When you are in an elevator, your actual weight (mg) never changes. – You feel lighter or heavier during the ride because your apparent weight increases when you are accelerating up, decreases when you are accelerating down, and is equal to your weight when you are not accelerating at all. v = 0 m/s a = 0 m/s2 V ≥ 0 m/s a > 0 m/s2 V > 0 m/s a = 0 m/s2 V ≥ 0 m/s a < 0 m/s2 Heavy feeling Normal feeling Normal feeling Lighter feeling Ground floor Just starting up Between Arriving floors at top floor v = 0 m/s a = 0 m/s2 V ≤ 0 m/s a < 0 m/s2 V < 0 m/s a = 0 m/s2 V ≤ 0 m/s a > 0 m/s2 Lighter feeling Normal feeling Normal feeling Heavy feeling Top floor Beginning descent Between Arriving floors at ground floor • Sample problem: An 85-kg person is standing on a bathroom scale in an elevator. What is the person’s apparent weight Wapp a) when the elevator accelerates upward at 2.0 m/s2? ∑Fy: Wapp - W = ma Wapp = ma + W Wapp = ma + mg Wapp = m(a + g)= 85 kg(2.0 m/s2 + 9.8 m/s2) = 1003 N W b) when the elevator is moving at constant velocity between floors? Wapp = m(a + g)= 85 kg(0 m/s2 + 9.8 m/s2) = 833 N c) when the elevator begins to slow at the top floor at 2.0 m/s2? Wapp = m(a + g)= 85 kg(-2.0 m/s2 + 9.8 m/s2)= 663 N • 5-kg salmon is hanging from a fish scale in an elevator. What is the salmon’s apparent weight when the elevator is a) at rest Wapp = m(a + g)= 5 kg(0 m/s2 + 9.8 m/s2) = 49 N b) moving downward and slowing at 3.2 m/s2? Wapp = m(a + g)= 5 kg(-3.2 m/s2 + 9.8 m/s2) = 33 N Normal Force • Normal force – The normal force is a force that keeps one object from penetrating into another object. – The normal force is always perpendicular a surface. – The normal exactly cancels out the components of all applied forces that are perpendicular to a surface. • derive an expression for the normal force of a box FN • on a flat table. y x W ∑Fy: FN - W = 0 FN = W ma = mg • Derive an expression for the normal force of a box sitting on a ramp. FN W Wy= mg cos θ θ Wx= mg sin θ W = mg ∑Fy: FN - WY = 0 FN = mg cos θ • derive an expression for the normal force an eraser being pushed up against a whiteboard by a force F. f ∑Fx: FN - F = 0 FN = F FN F W • Derive the normal force for the box in the picture below. The box is sitting on the floor, but is being pulled by the force shown. Ignore friction. FN F F 40° Fx ∑Fx : Fx = 0 W ∑Fy : FN + Fy – W = 0 ∑Fy : FN + Fy – mg = 0 FN = mg – Fy FN = 6.0 kg ∙ 9.8 m/s2 – 12.86 N = 43.48 N Fx = F cos θ Fx = 20 N cos 40° Fx = 15.32 N Fx = F cos θ Fx = 20 N sin 40° Fx = 12.86 N Fy • A 2.5 kg book rests on a surface inclined at 28° above the FN horizontal A. Find the normal force. f θ = 28° ΣFy: FN – mg sin θ = 0 ΣFy: FN = mg sin θ Fgx = mg θ Fgy= mg cos θ Fgx= mg sin θ Fgx = mg FN = 2.5 kg ∙ 9.8 m/s2 sin 28° = 11.50 N B. If the angle of the incline is reduced, do you expect the normal force to increase, decrease, or stay the same? Increase as the angle approaches 0° the FN will approach the weight • A skier skis down a slope with an acceleration of 3.50 m/s2. If friction can be ignored, what is the angle of the slope with respect to the horizontal? θ = ?° θ Fgy= mg cos θ Fgx= mg sin θ Fg = mg ΣFx: - Fgx = ma ΣFx: – mg sin θ = ma – g sin θ = a θ = sin-1 (a/ – g) θ = sin-1(-3.5m/s2 /– 9.8 m/s2) = 20.92° • How long will it take a 1.0 kg block initially at rest to slide down a frictionless 20.0 m long ramp that is at a 15° angle L = 20 m m=1.0 kg FN with the horizontal? Fg θ = 15° θ Fgy= mg cos θ Fgx= mg sin θ ΣFx: - Fgx = ma ΣFx: – mg sin θ = ma Fg = mg – g sin θ = a a = – 9.8 m /s2 sin (15°) = – 2.54 m /s2 t unk v V0 0 m/ s ? ∆x 20m a -2.54 m/s2 t unk v ? V0 0 m/ s ∆x -20m a -2.54 m/s2 v2 = v02 + 2a(∆x) v2 = 0 + 2a(∆x) v2= 2a(∆x) v= √(2a(∆x)) v= ±√(2∙(-2.54 m/s2)∙(-20m))= - 10.08 m/s v = v0 + at t =(v - v0) / a t =(-10.8 m/s - 0) / -2.54 m/s2= - 3.97 s • Friction – Friction is the force that opposes a sliding motion. – Friction is highly useful. It enables us to walk and drive a car, among other things. – Friction is also dissipative. That means it causes mechanical energy to be converted to heat. We’ll learn more about that later. • What causes friction? FN friction f W Big view: Surfaces look perfectly smooth. Fpush Small view: Microscopic irregularities resist movement. • Friction may or may not exist between two surfaces. If it exists, it opposes the direction object “wants” to slide. It is parallel to the surface. • Friction depends on the normal force. – The friction that exists between two surfaces is directly proportional to the normal force. – This has several implications, such as… • Friction on a sloping surface is less than friction on a flat surface (since the normal force is less on a slope). • Increasing weight of an object increases the friction between the object and the surface it is resting on. • Weighting down a car over the drive wheels increases the friction between the drive wheels and the road (which increases the car’s ability to accelerate). • Static Friction – This type of friction occurs between two surfaces that are not slipping relative to each other. – Static friction is tricky. It can range from zero up to a maximum allowed value for two surfaces. – fs ≤ μsFN • fs : static frictional force (N) • μs: coefficient of static friction • FN: normal force (N) Warning: fs ≤ μsFN is an inequality! – The fact that the static friction equation is an inequality has important implications. – Static friction between two surfaces is zero unless there is a force trying to make the surfaces slide on one another. – Static friction can increase as the force trying to push an object increases until it reaches its maximum allowed value as defined by μs. – Once the maximum value of static friction has been exceeded by an applied force, the surfaces begin to slide and the friction is no longer static friction. • Static friction and applied horizontal force Force Diagram FN Surface W There is no static friction since there is no applied horizontal force trying to slide the book on the surface. • Static friction and applied horizontal force Force Diagram FN fs F Surface W Static friction is equal to the applied horizontal force, and there is no movement of the book since ∑F = 0. • Static friction and applied horizontal force Force Diagram FN fs F Surface W Static friction is at its maximum value! It is still equal to F, but if F increases any more, the book will slide. • Static friction and applied horizontal force Force Diagram FN F fs Surface W Static friction cannot increase any more! The book accelerates to the right. Friction becomes kinetic friction, which is usually a smaller force. • Static friction on a ramp θ W = mg Without friction, the book will slide down the ramp. If it stays in place, there is sufficient static friction holding it there. At maximum angle before the book slides, let’s prove that μs = tan θ. FN θ Fgy= mg cos θ Fgx= mg sin θ Fg ΣFx: - Fgx + Fs= 0 – mg sin θ + μsFN = 0 μsFN = mg sin θ FN = mg sin θ / μs mg cos θ = mg sin θ / μs Fg = mg ΣFy: - Fgy + FN = 0 - mg cos θ + FN = 0 FN = mg cos θ μs = mg sin θ / mg cos θ μs = sin θ / cos θ μs = tan θ • Kinetic Friction – This type of friction occurs between surfaces that are slipping past each other. – fk = μkFN • fk : kinetic frictional force (N) • μk: coefficient of kinetic friction • FN: normal force (N) – Kinetic friction (sliding friction) is generally less than static friction (motionless friction) for most surfaces. A 10-kg box rests on a ramp that is laying flat. The coefficient of static friction is 0.50, and the coefficient of kinetic friction is 0.30. a) What is the maximum horizontal force that can be applied to the box before it begins to slide? μs = 0.50 m = 20 kg fs = μsFN = 0.5 ∙ 98 N = 49 N FN= mg = 10 kg ∙ 9.8 m/s2 = 98 N b) What force is necessary to keep the box sliding at constant velocity? μk = 0.30 m = 20 kg FN= mg = 10 kg ∙ 9.8 m/s2 = 98 N fk = μkFN = 0.3 ∙ 98 N = 29.4 N A 10-kg wooden box rests on a wooden floor. The coefficient of static friction is 0.50, and the coefficient of kinetic friction is 0.30. What is the friction force between the box and floor if a) no force horizontal force is applied to the box? The frictional force is 0 N A 10-kg wooden box rests on a wooden floor. The coefficient of static friction is 0.50, and the coefficient of kinetic friction is 0.30. What is the friction force between the box and floor if b) 20 N horizontal force is applied to the box? μs = 0.50 m = 20 kg fs = μsFN = 0.5 ∙ 98 N = 49 N The frictional force is 20 N FN= mg = 10 kg ∙ 9.8 m/s2 = 98 N A 10-kg wooden box rests on a wooden floor. The coefficient of static friction is 0.50, and the coefficient of kinetic friction is 0.30. What is the friction force between the box and floor if c) a 60 N horizontal force is applied to the box? μs = 0.50 m = 20 kg fs = μsFN = 0.5 ∙ 98 N = 49 N FN= mg = 10 kg ∙ 9.8 m/s2 = 98 N When it begins to move the frictional force is 49 N, so the box is accelerating. μk = 0.30 m = 20 kg FN= mg = 10 kg ∙ 9.8 m/s2 = 98 N fk = μkFN = 0.3 ∙ 98 N = 29.4 N The horizontal for exceeds the kinetic friction, so it will continue to accelerate until other forces like air resistance balance out the horizontal force. A 10-kg wooden box rests on a wooden ramp. The coefficient of static friction is 0.50, and the coefficient of FN kinetic friction is 0.30. What is the friction force between the box and ramp if a) the ramp is at a 25° angle? F N θ θ = 25° Fg Fgx= mg sin θ Fg = mg ΣFy: - Fgy + FN = 0 - mg cos θ + FN = 0 FN = mg cos θ FN = 10 kg ∙ 9.8 m/s2 cos (25°)= 88.82 N fs = μsFN = 0.5 ∙ 88.2 N = 44.1 N Fgy= mg cos θ A 10-kg wooden box rests on a wooden ramp. The coefficient of static friction is 0.50, and the coefficient of kinetic friction is 0.30. What is the friction force between the box and ramp if a) the ramp is at a 45° angle? ΣFy: - Fgy + FN = 0 - mg cos θ + FN = 0 FN = mg cos θ FN = 10 kg ∙ 9.8 m/s2 cos (45°)= 69.29 N fs = μsFN = 0.5 ∙ 69.29 N = 34.65 N b) what is the acceleration of the box when the ramp is at 45°? ΣFx: - Fgx + fs= ma Fgx = – mg sin θ= – mg sin θ= – 69.3 N a = (– Fgx+ fs) / m Prove weight in x greater than friction force a = (– 69.3 N + 34.65 N) / 10 kg= - 3.74 m/s2 • Tension – Tension is a pulling force that arises when a rope, string, or other long thin material resists being pulled apart without stretching significantly. – Tension always pulls away from a body attached to a rope or string and toward the center of the rope or string. • A physical picture of tension • Imagine tension to be the internal force preventing a rope or string from being pulled apart. Tension as such arises from the center of the rope or string. It creates an equal and opposite force on objects attached to opposite ends of the rope or string. • Tension examples Note that the pulleys shown are magic! They affect the tension in any way, and serve only to bend the line of action of the force. A 1,500 kg crate hangs from a crane cable. a) What is the tension in the cable when the crate is motionless? Ignore the mass of the cable. T ΣFy: - Fg + T= 0 Fg = mg - Fg + T= 0 T= Fg T= mg = 1500 kg ∙ 9.8 m/s2 = 14700 N b) Suppose the crane accelerates the crate upward at 1.2 m/s2. What is the tension in the cable now? ΣFy: - Fg + T= ma - Fg + T= ma T= ma + Fg= 1500 kg ∙ 1.2 m/s2 + 1500 kg ∙ 9.8 m/s2 = 16500 N • Springs (Hooke’s Law) • The magnitude of the force exerted by a spring is proportional to the amount it is stretched. • F = -kx – F: force exerted by the spring (N) – k: force constant of the spring (N/m or N/cm) – x: displacement from equilibrium (un-stretched and uncompressed) position (m or cm) • The negative sign indicates that the direction of the force is back toward the equilibrium (or unstretched) position. • 1.50 kg object hangs motionless from a spring with a force constant of k = 250 N/m. How far is the spring stretched from its equilibrium length? m = 1.50 kg k = 250 N/m F = ma = 1.5 kg ∙ 9.8 m/s2 = 14.7 N F = -kx x =F / -k = - 14.7 N / (250 N / m)= - 0.059 m The force applied to the string is negative. In the same direction as the displacement. The force applied by the spring in in the opposite direction of the applied force so it is positive, so the applied work is positive, so the work by the spring is negative • A 1.80 kg object is connected to a spring of force constant 120 N/m. How far is the spring stretched if it is used to drag the object across a floor at constant velocity? Assume the coefficient of kinetic friction is 0.60. μk = 0.60 m = 1.80 kg FN= mg = 1.80 kg ∙ 9.8 m/s2 =17.64 N fk = μkFN = 0.6 ∙ 17.64 N = 10.58 N k = 120 N/m x =F / -k = - 10.58 N / (120 N / m) = - 0.088 m • A 5.0 kg object is connected to a 10.0 kg object by a string. If a pulling force F of 20 N is applied to the 5.0 kg object, A) what is the acceleration of the system? 10.0 kg Box 2 5.0 kg Box 1 T2 = ? N T 1 = ? N 2 F = 20 N 1 ∑Fx2 : T2 = m2a F – m2a= m1a a = 20 N / (5.0 Kg + 10.0 Kg) ∑Fx1 : F – T1 = m1a F = m1a + m2a a = 1.33 m/s2 a = F / (m1 + m2) T1 = T2 B) what is the tension in the string connecting the objects? We can use either equation to calculate the tension on the cable between the 2 boxes. Although the tension on the string comes from box 2 ∑Fx1 : F – T1 = m1a ∑Fx2 : T2 = m2a T1 = F - m1a= 20 N – 5.0 Kg ∙ 1.33 m/s2 = 13.35 N T2 = 10 kg ∙ 1.33 m/s2 = 13.3 N (Assume a frictionless surface.) • Gravity • A very common accelerating force is gravity. Here is gravity in action. The acceleration is g. • Slowing gravity down • The pulley lets us use gravity as our accelerating force… but a lot slower than free fall. Acceleration here is a lot lower than g. • Pulleys • Pulleys bend the line of action of the force of gravity without affecting tension. • The pulley is “magic” – no mass, no friction, and no effect on the tension. • The pulley simply bends the line of action of the force. • Derive an expression for the acceleration due to gravity of the system below, and for the tension in T in both equations equal, the string. so set each equation equal ΣFy2: T- m2g = m2a2y T = m2a2y + m2g to T and set equal to each other m1a1x = m2a2y + m2g a2y = -a a1x = a FBD FN m1a = m2(-a) + m2g T m1g m1a + m2a = m2g a(m1 + m2) = m2g +y FBD +x a = m2g / (m1 + m2) T ΣF1x: T= m1a1x T= m1a1x m2g • A 10 kg block rests on a table connected by a string to a 5 kg block Find the minimum coefficient of static friction for which the blocks remain stationary. ΣFy2: T- m2g = m2a2y = 0 ΣF1x: T – fs = m1a1x = 0 T = m2g T= fs m2g = fs m2g = μsFN m2g = μsm1g m2 = μsm1 μs =m2 / m1 = (5 kg) / (10 kg) = 0.5 FN fs m1 T= μsm1g = 0.5∙10 kg ∙ 9.8 m/s2 = 49 N T m1g T m2 m2g • Derive the acceleration, assuming a coefficient of friction of 0.20 between the table and the 3.0 kg block. Determine the tension in the string. ΣF1x: T – Fg1x - fk= m1a1x θ T – m1g sin θ - μFN = m1a1x ΣF1y: FN – Fg1y= 0 Fg1x= mg sin θ F1g = mg FN = m1g cos θ T – m1g sin θ - μ m1g cos θ = m1a1x FN FBD fk a2y = -a a1x = a T – m1g sin θ - μ m1g cos θ = m1a T +y Fg1y= mg cos θ +x FBD ΣFy2: T- m2g = m2a2y ΣFy2: T- m2g = m2 (-a) T+ y m1g +x m2g T = -m2a + m2g T – m1g sin θ - μ m1g cos θ = m1a T= T = m1a + m1g sin θ + μ m1g cos θ -m2a + m2g = m1a + m1g sin θ + μ m1g cos θ m2a +m1a + = m2g - m1g sin θ - μ m1g cos θ a = (m2g - m1g sin θ - μ m1g cos θ) / (m1 + m2) a = (5.0 kg ∙ 9.8 m/s2 – 3.0 kg ∙ 9.8 m/s2 ∙ sin 35° - 0.2∙ 3.0 kg ∙ 9.8 m/s2 cos 35°) / (5.0 kg + 3.0 kg) a = 3.42 m/s2 T = m2g - m2a T = 5.0 kg ∙ 9.8 m/s2 - 5.0 kg ∙ 3.42 m/s2 = 31.9 N • Uniform Circular Motion – An object that moves at uniform speed in a circle of constant radius is said to be in uniform circular motion. – Question: Why is uniform circular motion accelerated motion? – Answer: Although the speed is constant, the velocity is not constant since an object in uniform circular motion is continually changing direction. • Centrifugal Force • Question: What is centrifugal force? • Answer: That’s easy. Centrifugal force is the force that flings an object in circular motion outward. Right? • Wrong! Centrifugal force is a myth! • There is no outward directed force in circular motion. To explain why this is the case, let’s review Newton’s 1st Law. • Newton’s 1st Law and cars – When a car accelerates forward suddenly, you as a passenger feel as if you are flung backward. • You are in fact NOT flung backward. Your body’s inertia resists acceleration and wants to remain at rest as the car accelerates forward. – When a car brakes suddenly, you as a passenger feel as if you are flung forward. • You are NOT flung forward. Your body’s inertia resists acceleration and wants to remain at constant velocity as the car decelerates. • When a car turns • You feel as if you are flung to the outside. You call this apparent, but nonexistent, force “centrifugal force”. • You are NOT flung to the outside. Your inertia resists the inward acceleration and your body simply wants to keep moving in straight line motion! • As with all other types of acceleration, your body feels as if it is being flung in the opposite direction of the actual acceleration. The force on your body, and the resulting acceleration, actually point inward. • Centripetal Acceleration – Centripetal (or center-seeking) acceleration points toward the center of the circle and keeps an object – This type of acceleration is at right angles to the velocity. – This type of acceleration doesn’t speed up an object, or slow it down, it just turns the object. • Centripetal Acceleration – ac = v2/r • ac: centripetal ac • acceleration in m/s2 • v: tangential speed in m/s v • r: radius in meters Centripetal acceleration always points toward center of circle! Centripetal Force • A force responsible for centripetal acceleration is referred to as F a centripetal force. • Centripetal force is simply mass times centripetal acceleration. • ∑F = m a • ∑F = m v2 / r Always toward – F: centripetal force in N center of circle! – v: tangential speed in m/s – r: radius in meters • Any force can be centripetal – The name “centripetal” can be applied to any force in situations when that force is causing an object to move in a circle. – You can identify the real force or combination of forces which are causing the centripetal acceleration. – Any kind of force can act as a centripetal force • Static friction As a car makes a turn on a flat road, what is the real identity of the centripetal force? • Tension As a weight is tied to a string and spun in a circle, what is the real identity of the centripetal force? • Gravity As the moon orbits the Earth, what is the real identity of the centripetal force? • Normal force with help from static friction As a racecar turns on a banked curve on a racing track, what is the real identity of the centripetal force? • Tension, with some help from gravity As you swing a mace in a vertical circle, what is the true identity of the centripetal force? • Gravity, with some help from the normal force When you are riding the Tennessee Tornado at Dollywood, what is the real identity of the centripetal force when you are on a vertical loop? • A 1200-kg car rounds a corner of radius r = 45 m. If the coefficient of static friction between tires and the road is 0.93, what is the maximum velocity the car can have without skidding? FN +y fs +x w μsFN = mv2/r 2/r v = √μsgr μ mg = mv FN = mg s v = √0.93 ∙ 9.8 m/s2 · 45m = 20.25 m/s must be less than 13.1 m/s ΣFx: fs = Fc • You whirl a 1.0 kg stone in a horizontal circle about your head. The rope attached to the stone is 1.5 m long. a) What is the tension in the rope? (The rope makes a 10° angle with the horizontal). ΣFy: T cos θ – mg = 0 L T = (1 kg ∙ 9.8 m/s2 )/ cos (80°) = 56.44 N b) How fast is the stone moving? ΣFx: T sin θ = Fc T sin θ = mv2/r T sin θ = mv2/(L sin θ) θ r r = L sinθ T FBD v = √ (LT (sin θ)2 / m) v = √ (1.5 m ∙ 56.44N (sin 80)2 / 1.0 kg) = 82.11 m/s mg • Isaac Newton • Arguably the greatest scientific genius ever. • Came up with 3 Laws of Motion to explain the observations and analyses of Galileo and Johannes Kepler. • Discovered that white light was composed of many colors all mixed together. • Invented new mathematical techniques such as calculus and binomial expansion theorem in his study of physics. • Published his Laws in 1687 in the book Mathematical Principles of Natural Philosophy. • The Universal Law of Gravity – Newton’s famous apple fell on Newton’s famous head, and lead to this law. – It tells us that the force of gravity objects exert on each other depends on their masses and the distance they are separated from each other. • The Force of Gravity • FG = Gm1m2/r2 – Fg: Force due to gravity (N) – G: Universal gravitational constant 6.67 x 10-11 N m2/kg2 – m1 and m2: the two masses (kg) – r: the distance between the centers of the masses (m) • The Universal Law of Gravity ALWAYS works, • whereas FG = mg only works when you’re • standing on the surface of the earth. a) How much force does the earth exert on the moon? b) How much force does the moon exert on the earth? mearth = 5.9375x1024 kg mmoon = 7.34 x 1022 kg rmoon = 1738000 m rearth = 6357000 m Distance surface to surface = 21605000 m FG = Gm1m2/r2 FG = 6.67 x 10-11 N m2/kg2 ∙ 5.9375x1024 kg ∙ 7.34 x 1022 kg /(1738000 m + 6357000 m + 21605000 m)2 FG =3.3 x 1022 N • Sally, an astrology buff, claims that the position of the planet Jupiter influences events in her life. She surmises this is due to its gravitational pull. Joe scoffs at Sally and says “your Labrador Retriever exerts more gravitational pull on your body than the planet Jupiter does”. Is Joe correct? (Assume a 50 kg Lab 1.0 meter away, and Sally is 70 kg). msally = 70 kg mjupiter = 1.782X1027 kg rjupiter = 67100000 m Distance surface to surface = 6.31 x 1011 m FG = Gm1m2/r2 FG = 6.67 x 10-11 N m2/kg2 ∙ 70 kg ∙ 1.782 x 1027 kg /(67100000 m + 6.31x1011 m)2 FG = 2.1 x 10-5 N FG = 6.67 x 10-11 N m2/kg2 ∙ 70 kg ∙ 50 kg /(1 m)2 FG = 2.3 x 10-7 N • What would be your weight (assume 90 Kg) if you were orbiting the earth in a satellite at an altitude of 3,000,000 km above the earth’s surface? (Note that even though you are apparently weightless, gravity is still exerting a force on your body, and this is your actual weight.) mearth = 5.9375x1024 kg rearth = 6357000 m msally = 90 kg Distance surface to person = 3000000000 m FG = Gm1m2/r2 FG = 6.67 x 10-11 N m2/kg2 ∙ 5.9375x1024 kg ∙ 90 kg /(1738000 m + 6357000 m + 3.0 x 109 m)2 FG = 3.94 x 10-3 N • Acceleration and distance • Surface gravitational acceleration depends on mass and radius Planet Mercury Planet Radius (m) Mass (kg) 2.43 x 106 3.2 x 1023 g (m/s2) 3.61 Venus Mars Jupiter Saturn 6.07 x 106 3.38 x 106 6.98 x 107 5.82 x 107 4.88 x 1024 6.42 x 1023 1.9 x 1027 5.68 x 1026 8.83 3.75 26.0 11.2 Uranus Neptune Pluto 2.35 x 107 2.27 x 107 1.15 x 106 8.68 x 1025 1.03 x 1026 1.2 x 1022 10.5 13.3 0.61 • What is the acceleration due to gravity at the surface of the moon? rmoon = 1738000 m mmoon = 7.34 x 1022 kg mobject = 1 kg FG = Gm1m2/r2 FG = 6.67 x 10-11 N m2/kg2 ∙ 1 kg ∙ 7.34 x 1022 kg /(1738000 m)2 FG =1.62N FG = ma a = FG / m = 1.62 N / 1 kg = 1.62 m/s2 • Johannes Kepler (1571-1630) – Kepler developed some extremely important laws about planetary motion. – Kepler based his laws on massive amounts of data collected by Tycho Brahe. – Kepler’s laws were used by Newton in the development of his own laws. • Kepler’s Laws 1. Planets orbit the sun in elliptical orbits, with the sun at a focus. 2. Planets orbiting the sun carve out equal area triangles in equal times. 3. The planet’s year is related to its distance from the sun in a predictable way. • Kepler’s Laws • Satellites • Orbital speed •At any given altitude, there is only one speed for a stable circular orbit. •From geometry, we can calculate what this orbital speed must be. •At the earth’s surface, if an object moves 8000 meters horizontally, the surface of the earth will drop by 5 meters vertically. •That is how far the object will fall vertically in one second (use the 1st kinematic equation to show this). •Therefore, an object moving at 8000 m/s will •never reach the earth’s surface. • Some orbits are highly elliptical • Some orbits are nearly circular. •The orbits we analyze mathematically will be nearly circular. •The analysis starts as follows • Using Newton’s Law of Universal Gravitation, derive a formula to show how the period of a planet’s orbit varies with the radius of that orbit. Assume a nearly circular orbit. (This is a derivation of Kepler’s 3rd Law.) • What velocity does a satellite in orbit about the earth at an altitude of 25,000 km have? What is the period of this satellite? • A geosynchronous satellite is one which remains above the same point on the earth. Such a satellite orbits the earth in 24 hours, thus matching the earth's rotation. How high must a geosynchronous satellite be above the surface to maintain a geosynchronous orbit?