#### Transcript Force - springsphysics

Chapter 4: Newton’s Laws and Forces I. II. III. Newton’s Laws Forces Free-Body Diagrams Acceleration… Chapters 2 and 3 described “how much” acceleration was and what it meant for motion. Chapter 4 describes “why” acceleration occurs in the first place. Newton’s First Law An object at rest remains at rest; an object in motion continues in motion with constant velocity unless acted upon by a net external force. sometimes known as the Law of Inertia Inertia: the tendancy of an object to resist change in motion not accelerate (change in speed and/or direction) inertia is a measurement of mass more massive objects have more inertia more resistance to change in motion Newton’s Second Law Acceleration is directly proportional to the net force and inversely proportional to mass. a F m F ma Newton’s Third Law For every action, there is an equal and opposite reaction. sometimes known as the Law of Action-Reaction Walking… You are able to travel because you push the earth backwards. You experience an equal and opposite force, pushing you forwards. Push on a desk with your hand. your hand is distorted, showing a force is acting on it you can feel the desk pushing back on your hand Push harder! the harder you push on the desk the harder the desk pushes back on your hand You only feel the forces exerted on you. The forces must be equal, however the acceleration need not be equal. Force: interaction between two objects may cause an acceleration if it does, the velocity changes speed or direction unit: Newton (N) 1 N = 1 kg∙ m/s2 Kinds of Forces: contact force physical contact Ff , FN , FP field force no contact Fg , FE Field Theory: the presence of an object effects the space around it and any object in that space. Weight (Fg) force due to gravity from the center of mass towards the center of the earth Fg = mg Weight (Fg) not the same as mass! mass is always the same scalar, kg weight can change! different on moon, etc gearth ≠gmoon force vector, N 1 pound = 4.448 N Normal Force (FN): from the contact surface perpendicular to the surface amount varies FN FN Normal Force (FN): For flat surfaces… FN = weight = mg …and pulled up: FN = mg - FP (b/c it feels lighter to the table) …and pushed down: FN = mg + FP (b/c it feels heavier to the table) For inclines… Only one component applies. FN = mgcos FN FN Friction (Ff) from the contact surface opposite the direction of motion two types static kinetic Ff = kFN Ff = sFN (max) Friction (Ff) an intermolecular force between surface atoms independent of the surface area depends upon the types of surfaces described by a coefficient of friction, μ μ has no units μs >μk Push or Pull (FP): acts in the direction of the push or pull amount varies Tension (FT) acts along the string tension is the same at all points along the string string is taut (not slack) Free-Body Diagrams: very important ! force is a vector magnitude and direction arrows objects can be modeled as boxes points Drawing Free Body Diagrams: draw one body/object in position define coordinate axes show and label all forces: gravity pulling down, the ground pushing back up, friction, push/pull tension PRACTICE PROBLEMS: 1. Your car stalls on the way up an incline. Show a free body diagram for the car as you push it parallel to the incline. PRACTICE PROBLEMS: 2. You drag a block along a rough, flat surface using a rope. The rope is at an angle of 25 upward from the horizontal. Show a free-body diagram for this. PRACTICE PROBLEMS: 3. You are lifting a bucket vertically. Show a free-body diagram for the bucket. Solving Force Problems… draw a free-body diagram use the picture to decide: x- or y- direction direction of forces (+ or –) use N2L to “do the math” forces are vectors components F m a F x max F y may Practice Problems… A 12 kg block is being pulled by a string to the right at a constant velocity. The surface is rough, where .20 is the coefficient of friction between the block and the surface. Calculate the tension in the string. Practice Problems… 1. Your new college roommate has put a box in front of your closet. The box is quite massive (60kg), so you slide it out of the way. The floor is dirty and rough, with a coefficient of friction .31. a) b) 2. If you push the box to the right with a force of 900N, what is the acceleration of the box? 12m/s2 If you push for 2s, how far does the box travel? 24m A 40N pulling force is needed to keep a block traveling at a constant velocity to the left. a) b) If the coefficient of friction between the block and the surface is .25, what is the mass of the block? 16kg If you double the mass of the block, what force is now required to keep it traveling at a constant velocity? 79N Practice Problems… While mowing the lawn, you pull the 25kg mower to the right with a 75N force at an angle of 60o above the horizontal. Calculate the normal force acting on the mower. Calculate the coefficient of kinetic friction that must be present to keep the lawnmower traveling at a constant velocity. Practice Problems: 1. A young girl is walking her Chihuahua (m=4kg) down the street. To get the dog to move, the girl pulls on the leash at an angle of 45o above the horizontal with a force of 10N. a) b) 2. If the dog can exert a resistive force of 5N, what acceleration does the dog experience? How much resistive force must the dog exert in order to not move at all? .5m/s2 7.1N A boat is traveling through the bay at a constant speed to the right while someone parasails behind them. The rope attached to the parasail has a tension of 450N and is at an angle of 60o above the horizontal. a) If the resistive force of the water is 200N, what is the “pulling” force of the motor? 425N b) If the “pulling” force of the motor were increased by 300N, what would be the resulting acceleration? The boat’s mass is 1200kg. .25m/s2 Practice Problems: A skier travels down a slope of 20o. The surface is relatively smooth, with a coefficient of friction of .05. Calculate the normal force. (variables only ) What acceleration does the skier experience? Practice Problems: 1. A 20kg box is sliding up a 15o hill. If the coefficient of friction between the box and the hill is .23, a) b) 2. what acceleration does the box experience? If the box stopped in 2.8s, how fast was it traveling? -4.72m/s2 13.2m/s Two young children are sledding down a 12o slope of fresh snow. The coefficient of friction between the sled and the hill is .13 and their combined mass is 90kg. a) b) c) Calculate the force due to friction. 112N What acceleration do the children experience? How fast will they be going after sliding 30m? .79m/s2 6.88m/s Practice Problems: A 3.0kg block is connected by a light string over a pulley to a 5.0kg block which hangs over the edge of the table. The coefficient of friction between the block and the table is .28. Calculate the tension in the string. (variables only ) Calculate the acceleration of the system. Calculate the tension in the string. Practice Problems: 1. A 12.0kg block is connected by a light string over a pulley to a 18.0kg block which hangs over the edge of the table. The coefficient of friction between the block and the table is .31. a) b) Calculate the tension in the string. Calculate the acceleration of the system. 177N 11.7m/s2