F = ma. F = mg. Forces. Forces. Free Body Diagrams. Find the unknown forces!! Ex. 1 Ex N. Newton s First Law. Newton s Second Law
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1 Forces Free Body Diagrams Push or pull on an object Causes acceleration Measured in Newtons N = Kg m s Shows all forces as vectors acting on an object Vectors always point away from object Used to help find net force Contact Forces pplied Force Frictional Force Tensional Force Normal Force Spring Force Field Forces Gravitational Force Electrical Force Magnetic Force F f F N F g F pull Find the unknown forces!! Ex. Ex. 00 N F Newton s First Law Law of Inertia Resistance to change motion 75 N 50 N Objects in motion stay in motion Objects at rest stay at rest et =? et = 00N 75 N et = 5 N Upward et = 0 N Downward et = F 50 N = 0 N F = 40 N Equilibrium balanced forces, net force = 0 Net force sum of all forces 3 4 Newton s Second Law net force will cause acceleration friction force from a fluid (gases and liquids) force mass F = ma acceleration Terminal Velocity constant velocity of falling when F drag = F g Gravity force F = mg Mass and weight are not the same!!! 5 6
2 Newton s Third Law Each action has an opposite and equal reaction F on B = F B on Interaction Pair action / reaction forces Solving Tips. Draw the problem and choose coordinates. Determine known and unknown forces. 3. Create a free body diagram showing the net force. 4. Use Newton s laws to link acceleration and net force. 5. Solve equations for the unknowns 7 8 Combining Forces Normal Force +X F N = mg F N = mg + F hand F N = mg F string 9 0 Friction Factor lways against motion Two branches of friction (3 Types) Kinetic (Moving) Sliding Rolling Static (Stationary) Friction of fluids is called viscosity Kinetic Friction Friction Force F f = k = friction factor = Normal Force F f,static s
3 The force of static friction is not constant! Static friction is equal to pulling force until the object begins to move The maximum static friction is equal to s F f,static,max F applied stays still F f,static,max = F applied constant speed, a=0 F f,s F f F f,s,max F f = s, F f,k F f = k, Kinetic Friction F f,static,max F applied accelerates F pplied 3 4 Thrust from Friction and Motion What is the maximum acceleration a car can achieve if the tires/road friction coefficient is equal to 0.7? (ignore drag) et,x = F thrust F drag = ma F Drag The maximum thrust cannot exceed road friction F thrust = F f,s,max = s = s mg F g F thrust From et,x ma = s mg 5 a = s mg = (0.7)(9.8m/s ) = 6.9m/s 6 The most important rule: This means that: Static Equilibrium F = 0 F x = 0 and F y = 0 Static Equilibrium LWYS FOLLOW THESE STEPS:. Draw a labeled free body diagram. Break angled forces into components 3. Write net equations ( et,x = ) Only use the components of angled forces!! et,x = 0 and et,y = 0 4. Solve for unknowns one at a time 7 8 3
4 Vector Direction ( Common Ways) Labeled degrees north or south of x axis Degrees from east direction (0 ). 80 N of East Or S of East Or N of West 30 Or 50 Working with Forces at an ngle When a force is at an angle: break into x and y components Do not use the original force again!! dd x and y components separately Find the new resultant force and its angle F = F x +F y = Tan O Using the angle from the x axis: X Component y Component F x = F cos F y = F sin 9 0 Right Triangle Help You will typically want to work with the angle from the x axis. = Tan O = angle = H Cos θ djacent 90 O = H Sin θ 90 Opposi ite SOH Sin θ = O CH TO Cos θ = H Tan θ = H = Sin O = Cos = Tan H H O Opposite djacent Hypotenuse O = H Sin θ = H Cos θ H = O Sin θ O = Tan θ = O H = Tan θ Cos θ O = H = H O H = +O O Forces at an ngle Breaking into components Forces on a Ramp Breaking into components m Find F P Components F Px = F P Cos F Py = F P Sin Use components for net equations etx = F Px ety = F N + F Py F g F Py F N F P F Px F g = mg F N should not equal F g!! F N = F g F py if a y = 0 3 Find F g Components F gx = F g Cos (90 ) F gy = F g Sin (90 ) Or just use SOH CH TO F gx = F g Sin F gy = F g Cos Use components for net equations etx = F gx ety = F N F gy F g = mg F N should not equal F g!! Many cases, F N will equal F gy 4 4
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