Transcript F - Physics
Forces in One Dimension Chapter 4 Physics Principles and Problems Zitzewitz, Elliot, Haase, Harper, Herzog, Nelson, Nelson, Schuler and Zorn McGraw Hill, 2005 Remember! When using F = ma it is the net force! • Kamaria is learning how to ice skate. She wants here mother to pull her along so that she has an acceleration of 0.80m/s2. If Kamaria’s mass is 27.2kg, with what force does her mother have to pull her along (neglect ice resistance)? 1. Fnet = Fmother on Kamaria + ( - Fkanaria on mother) 2. a = Fnet / m 3. a = Fmother on Kamaria + ( - Fkanaria on mother) / m 4. (a • m) - ( - Fkanaria on mother) = Fmother on Kamaria 5. Fmother on Kamaria = (0.80m/s2 • 27.2kg) + 0 N Applying Newton’s Laws • The mass of an object is a magnitude of that object’s amount of matter. It never changes (i.e. 12g of iron on earth is still 12g of iron on the moon). • The weight of a object is a force (hence, weight’s units are Newtons) and is directly related to the mass (kg) and acceleration (m/s2) of the object. What is your weight on earth? What is your weight on the moon? The Elevator Problem http://plus.maths.org/issue38/features/livio/figure8.jpg Does your weight change in an elevator? Understanding Weight • Apparent Weight - the force an object experiences as a result of ALL the forces acting upon it, resulting in acceleration. • Weightlessness - when an object’s apparent weight is zero as a result of no contact forces being exerted on the object (actual weight is not zero however). Real and Apparent Weight • Your mass is 75.0kg and you are standing on a scale in an elevator. Starting from rest the elevator accelerates upward at 2.0m/s2 for 2.0s and then continues at a constant speed. What is the reading of the scale during rest and during acceleration? http://sol.sci.uop.edu/~jfalward/physics17/chapter4/elevator.jpg Solution • F = ma • Fnet = Fscale + ( -Fg) • Fscale = Fnet + Fg At Rest Accelerating Fscale = 0 + Fg Fscale = Fnet + Fg Fscale = mg Fscale = ma + mg Fscale = (75kg)(10m/s2) Fscale = (2.0m/s2)(75kg) + (75kg)(10m/s2) Fscale = 750 N Fscale = 900 N Drag Force • The force exerted by a fluid on the object moving through that fluid. This force is dependent upon the properties of both the object (shape, mass) and fluid. http://www.fluent.com/about/news/newsletters/02v11i1/img/a9i5_lg.gif http://www.swe.org/iac/images/prafoil.jpg http://newsimg.bbc.co.uk/media/images/42381000/jpg/_42381388_swimmers416. Terminal Velocity - the constant velocity that an object obtains when the drag force equals the force of gravity (acceleration = 0). http://www.iop.org/activity/education/Teaching_Resources/Teaching%20Advanced%20Physics/Mechanics/Images%20200/img_mid_4140.gif Newton’s Third Law • Forces come in an interaction pair (two forces that are in opposite direction and have equal magnitudes). • FA on B = - FB on A • The force of A on B is equal in magnitude and opposite direction of the force of B on A. • Action and Reaction. http://www.primidi.com/images/newton_action_reaction_law.jpg Tension - the force exerted by a string or rope. • A 50kg bucket is being lifted by a rope. The rope will not break if the tension is 525 N or less. The bucket started at rest, and after being lifted 3m, it is moving at 3m/s. If the acceleration is constant, is the rope in danger of breaking? http://upload.wikimedia.org/wikipedia/commons/8/8f/Helicopter_at_Yellowstone_198 Solution • Fnet = Ftension + ( -Fg) • Ftension = Fnet + Fg • Ftension = ma + mg Use vf2 = vi2 + 2ad to solve for a. since vi2 is 0 then a = vf2 / 2d Therefore, 1. FT = (m)(vf2 / 2d) + (m)(g) 2. FT = (50kg)((3m/s)2/2(3m) + (50kg)(10m/s2) 3. FT = 570 N, Yes it will break!