Newton's Newton's Second Law of Motion Newton's First Law of Motion
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Newton's First Law of Motion Newton's Three Laws of Motion Any object at rest or constant velocity (in a straight line), will remain at rest or constant velocity (in a straight line), unless acted upon by an outside force. This is called INERTIA. Inertia is the tendency of an object to keep doing what it is already doing. Inertia has units of kilograms "kg". The greater the mass of an object, the greater the force is needed to change its motion. Newton's Second Law of Motion The net force on an object is equal to the product of its mass and its acceleration. ƩF = ma Net Force = mass x acceleration The metric system unit for force is called the Newton (N). (Can you guess why?) 1 Newton = 1 kg.m / s2 The English unit for force is called the Pound! one pound = 4.448 N Nov 10­10:30 PM Nov 10­10:35 PM Nov 10­10:45 PM 1 Newton's Third Law of Motion For every action, there is an equal and opposite reaction. Common Forces FA = Applied Force (can be broken down into componets) Action = Force Reaction = Force Force = push or pull It is impossible to apply a force on an object, such that it does not apply an equal and opposite force on the object pushing or pulling on it. Nov 10­11:06 PM Free Body Diagram FA = FAX + FAY FN = Normal Force GENERAL PROCESS FOR NEWTON’S LAWS PROBLEMS Step 1: Create a free­body diagram A. Remove the object from its surroundings. B. Replace all interactions with force vectors, taking care to indicate directions correctly. C. Include mg (down), normal force, friction force, applied forces, contact forces, etc. D. Any force that acts at an angle, indicate its X and Y components on the FBD FG or mg = Force of Gravity FA FF = Force from Friction Nov 10­11:15 PM Nov 10­9:35 PM 2 FBD Step 2: Decisions to make Step 3: Add X components and Y components of all forces. FA A. Which way is the object moving? Make that the positive direction. FA Sums are X and Y components of net force ΣF. (ΣFX and ΣFY) The positive direction! ΣFX column ΣFY column B. Is the object accelerating in the X direction? If yes, then ΣFX = maX If no, then ΣFX = 0 C. Is the object accelerating in the Y direction? mg FAY FN FF If yes, then ΣFY = maY If no, then ΣFY = 0 Place the right side Place the right side of the Y force of the X force equation from step equation from step 2B here 2B here FAX Step 4: Use the columns to write force equations and solve for whatever is needed. Nov 10­9:35 PM Nov 11­8:07 AM Nov 11­8:14 AM 3 FBD FAY FA The Coefficient of Friction A 20.0 kg box is being pulled by a rope with an applied force of 100.0 newtons. Draw the FBD for this situation. μ = 0.22 FAX mg FF = coefficient of friction FA Coefficient of Static Friction FN "mu" Ff = FN Steel on Steel 0.74 Aluminum on Steel 0.61 Copper on Steel 0.53 Rubber on Concrete 1.0 Wood on Wood 0.25 ­ 0.5 Glass on Glass 0.94 Waxed wood on Wet snow 0.14 Waxed wood on Dry snow 0.04 Metal on Metal ﴾lubricated﴿ 0.15 Ice on Ice 0.1 Teflon on Teflon 0.04 Synovial joints in humans 0.01 FAX FAY FN mg FF Nov 10­9:35 PM Nov 12­8:35 AM Nov 10­9:35 PM 4 A 150 kg box is being pulled by a rope at an angle theta (20 degrees) with an applied force of 600 newtons. Draw the FBD for this situation. μ = 0.11 A 12.5 kg box is being pushed with an applied force of 50.0 newtons. Draw the FBD for this situation. μ = 0.11 A 2.5 kg box is being pushed at an angle theta (380) with an applied force of 16.00 newtons. Draw the FBD for this situation. μ = 0.22 FA FA FA mg FAY FN FAX FF mg FAY mg FN FF FAY FN FF FAX FAX Nov 10­9:35 PM Nov 10­9:35 PM Nov 10­9:35 PM 5 mg FAY FN FF FA FAX FF FParallel mg FN FA FAY mg FF FAX FN FAY FAX FA FParallel Nov 10­9:35 PM Nov 10­9:35 PM Nov 10­9:35 PM 6 FA FAY mg FF FAX FN FParallel FA FAY mg FF FAX FN FParallel Nov 19­2:57 PM Nov 19­2:57 PM 7
