Ch 6 Using Newton s Laws. Applications to mass, weight, friction, air resistance, and periodic motion
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1 Ch 6 Using Newton s Laws Applications to mass, weight, friction, air resistance, and periodic motion
2 Newton s 2 nd Law Applied Galileo hypothesized that all objects gain speed at the same rate (have the same downward acceleration) regardless of mass. This theory was tested and found to be true. The weight force (F g ) exerted on a mass, m, was found to be g, (the ball s acceleration) so Newton s 2 nd Law becomes F g = mg.
3 Mass vs Weight Your MASS is the amount of MATTER that composes you. Your WEIGHT is the response of that mass to the pull of gravity. (No air resistance in this free fall). Though your WEIGHT will vary planet to planet due to a varying g, your mass will remain constant.
4 Weighing in an Elevator Look at the example problem on pg 128. In your coordinate system, positive direction is UP, so acceleration due to gravity is NEGATIVE. F net = sum of positive force of scale on you, F scale and negative weight force, F g. So F net = F scale F g. Solve for F scale and substitute ma for F net and mg for F g. F scale = ma + mg, F scale = m(a+g)
5 Practice Problem A 50-kg bucket is being lifted by a rope. The rope is guaranteed not to break if the tension is 500N or less. The bucket started at rest, and after being lifted 3.0m, it is moving 3.0m/s. Assuming that the acceleration is constant, is the rope in danger of breaking? Draw free body diagram and label all forces
6 Solution Known: Unknown: m = 50kg v = 3.0m/s F T =? v 0 =0m/s d= 3.0m Strategy: net force = vector sum F net = F T F g so, F T = F net + F g F T = ma + mg Because v 0 = 0m/s, then a = v 2 /2d F T = 570N, the rope is in danger of breaking.
7 Apparent Weight Typically a scale reads the support force of an object (the object s weight which is due to mass and gravity). If other forces are exerted on the object to increase or decrease the downward force, the scale reads the support which is the apparent weight. Example- stand on a scale and push up on the sink at the same time is your weight more or less?
8 Weightlessness If support (upward push) opposes gravity (downward pull), when the support becomes zero, as in an elevator falling at 9.8m/s2, a scale cannot push up on the object so the reading goes to zero. This is apparent weightlessness and does NOT mean the object has no weight. Normal Force- F N is the force pushing one surface against another. (The support of the object?) Perpendicular to the surface.
9 Proportionality Constants A constant that depends on the material used in the objects and the conditions of their movement (stationary or moving). For Friction, the constants are: μ s for the static coefficient of friction and μ k for the kinetic coefficient of friction.
10 How hard can you be pushed? A stationary object can only be pushed so hard, before it will move. Static friction force, F f, can grow, but has a limit. ( s F N ) This is the max friction before the object starts to move. Could you push harder on an object and it still NOT move? What does that do to the amount of friction to keep F net = 0?
11 Friction Due to surface irregularities between objects, a force of FRICTION opposes moving objects. Static Friction- a force existing between objects that don t move relative to each other. 0 F f static s F N Kinetic Friction- a force existing between objects that MOVE relative to each other. F f, kinetic k F N
12 Forces as Vectors It is important to note that the normal force, F N, and the force of friction, F f, are at right angles to each other. F push Use table p. 131 for friction coefficients. F N F f
13 Practice Problems A boy exerts a 36-N horizontal force as he pulls a 52-N sled across a cement sidewalk at a constant speed. What is the coefficient of kinetic friction between the sidewalk and the metal sled runners? Ignore air resistance.
14 Solution Draw a free body diagram There is not net horizontal or vertical force Since speed is constant, F f = force exerted by the boy, 36-N, but F f = μ k F N so, Ff 36N μ k = = = 0.69 FN 52N
15 Practice Problem #2 Suppose the sled runs on packed snow. The coefficient of friction is now only If a person weighing 650-N sits on the sled, what force is needed to pull the sled across the snow at constant speed?
16 Solution At constant speed, applied force = friction force, so F f = μ k F N F f = (0.12) (52 N N) F f = 84 N
17 Your turn to Practice Open your textbook to pg Do Ch 6 Rev. #s 26,27,29, 30-35*, 37, & 38
18 Air Drag and Terminal Velocity Fluids (liquids and gases) exert a friction force on moving objects. Amount of fluid friction depends on speed of the object, size and shape of the object, density, and the kind of fluid. Fluid force is called drag. When drag force = force of gravity, the object stops accelerating and reaches T.V. How can you change the terminal velocity an object reaches?
19 Periodic Motion As an object moves back and forth when disturbed, it has one position where it is in equilibrium and net force = zero. As disturbed by a net force, the object is pulled back toward equilibrium. This is Simple Harmonic Motion and the object is described in 2 areas: Period (T)- the time needed for one complete back and forth cycle. Amplitude- maximum displacement from equilibrium.
20 Pendulums The period of oscillation depends on the mass of the object and the strength of the spring, but NOT on the amplitude of the swing A pendulum bob swings back and forth with a period that depends on the length, l, of the string. T 2 l g Note: period does not depend on mass of the bob or amplitude of swing.
21 Resonance Applying a force to an object to cause it to vibrate is a Forced vibration. All objects have a natural frequency. Causing a vibrating object to vibrate at its natural frequency increases the AMPLITUDE of the vibration. This is called RESONANCE. Examples: coins in your car ashtray vibrate at a certain speed, a swing will only oscillate at it s natural freq. If you push 1x/ cycle it will go the highest.
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