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# Gen Physics I PHYS 2010

ETSU

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This 33 page Class Notes was uploaded by Iva Cormier on Sunday October 11, 2015. The Class Notes belongs to PHYS 2010 at East Tennessee State University taught by Brian Espino in Fall. Since its upload, it has received 16 views. For similar materials see /class/221405/phys-2010-east-tennessee-state-university in Physics 2 at East Tennessee State University.

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Date Created: 10/11/15

Day 8 Ch 4 Forces Normal Forces Looking at Newton s 2nOI Law 2F ma A 10 kg book is sitting on a 20 kg desk What normal force does the ground exert on the desk Free body diagrams Since the book sits still on the desk a O WbFnb Also the desk sit still on the ground a O ZFmaOwbwdFndON Wb Wd 39 Fnd remember that forces are vectors F b 39 mbg 10 k898msz 98 N upward Fnd 39 mbg 39 mag 10 kg 20 kg 98msz 294 N upward Notice that to find the normal force the ground exerts on the desk you use the weight ofthe book and the desk Weight of Object on Inclined Plane F n Opp Fn is the adjacent component of the weight What about the opposite component FP is the opposite parallel component ofthe weight vector It is parallel to the inclined plane This is the component of the weight that causes the object to slide down the plane FPWsin9mgsin9 What happens when we increase 9 from O to 90 degrees Let mass 50 kg P 50 kg98msz sin 0 0 P 50 kg98msz sin 10 85 N P 50 kg98msz sin 30 245 N P 50 kg98msz sin 45 346 N P 50 kg98msz sin 60 424 N P 50 kg98msz sin 90 490 N 39I39I39I39I39I39I 39I39I39I39I39I39I Normal Forces from Tuesday What happens when we increase 9 from O to 90 degrees FN 50 kg98msz cos 9 N 50 kg98msz cos 0 490 N N 50 kg98msz cos 10 483 N N 50 kg98msz cos 30 424 N N 50 kg98msz cos 45 346 N N 50 kg98msz cos 60 245 N F F F F F FN 50 kg98msz cos 90 O N As the angle of inclination increases FN decreases FP increases As the slope increases gravity has a IN greater effect on the object When incline is vertical Object resembles falling body FP weight mg What acceleration will the block have Let m 50 kg 9 300 The block is free to accelerate along the incline So FP is the accelerating force Free body diagram IN Free body diagram Fnet is the total or resultant force It is directed along the plane illepplying the 2nd Law Fnet ma 50kga a Fnetm a 50kgg sin 3050 kg a 49 ms2 Another way of looking at incline plane problems rotating the coordinate system Rotate coordinates b anle of incline 9 Now the Free Body diagram The normal component of the weight is again balanced out by the normal force Fnet is the parallel component of the weight Vector addition Again solving for acceleration a Fnetm a mg sin 9m g sin 9 98mszsin 30 49ms2 Vector addition V FN This tells us that the acceleration down an incline plane is equal to gravity times the inclination angle ag n9 Example A 2 kg box is slides down a frictionless incline plane 9 200 The block starts from rest After the block moves 2 meters what will its velocity be First we use the force work to find Fnet ma a k a g sin 9 m m a g sin 20 34ms2 Now use the acceleration and solve for the velocity of the block after sliding 2 meters 12 2 123 2an O ms2 234ms22m Vf 37 ms direction is down the incline What force PULLis needed to pull a sled up a frictionless hill at constant velocity mass of sled 10 kg 0 Angle of hill is 45 degrees What force PULLis needed to pull a sled up a frictionless hill at constant velocity mass of sled 10 kg Angle of hill is 45 degrees F PULL Free body diagram of sled 450 450 W Since the velocity is constant the sum of all the forces needs to be zero So the PULL force needs to balance out the parallel component of the weight vector PULL mg sin 45 10kgg sin 45 69 N Friction Friction is due to surfaces not being perfectly smooth Friction force ALWAYS opposes the direction of motion The result of friction forces is energy heat Two types of friction Static friction force keeps the object at rest and is needed to be overcome so the object can begin motion Kinetic friction occurs when one object is sliding against another surface Friction When is friction used is ordinary life car brakes driving around a turn walking rubbing your hands together to make warmth Friction How to calculate frictional force Frictional force Ff depends on the material of the two surfaces involves Given by coefficient of friction See table on page 101 for examples There are two coefficients static p5 and kinetic pk Ff also depends on the normal force Static friction force Ff g psFN Kinetic friction force Ff kaN Static friction If S HsFN The static friction force can vary from O to usFN usFN is known as the maximum static friction force This is the force needed to be overcome to start sliding an object across the surface See figure 419 on page 101 Example What coefficient of friction is needed to keep a block from sliding down an incline FN The static friction force balances out the parallel component of The weight Fs balances out FP From earlier we know FP mg sin 9 When the box is about to slide dqNvn Ff psFN psmg cos 9 Therefore nsmg cos 9 mg sin 6 Solve for us gives as tan 9 This tells us the static coefficient 5 needed to keep an object from sliding down an incline depends only on the incline In this example you are holding a book still up against the wall How hard do you have to push to keep the book from sliding Free body diagram for the book f IPush FN W lfthe book is not accelerating the sum ofthe forces is zero FN and Fpush counteract each other in the horizontal direction Ff IPush FN w Weight and friction force balance each other out in the vertical direction The friction is what is holding up the book Let the book have mass 5 kg and us 4 W 5kgg 49 N For W Ff Ff uSFN 04FN 04 F 49 N 04 F push F 1225 N push push Kinetic Friction Problem You are pushing a block up an incline The angle of inclination is 30 degrees The mass of the block is 20 kg The coefficients of friction are us 03 and uk 02 What force must be applied to the box to keep the speed constant FN Free body diagram F appHed W Since block is sliding across surface use uk 02 Since velocity is constant a O Fnet ma 0 Free body diagram Iapplied We are only concerned with motion along the incline Fapplied Ff and W have components along the incline Writing 2nd Law along incline we get applied39 Ff 39 mg sin 30 ma O Fapplied Ff mg sin 30 Hkmgkos 30mgsin 3O Fapplied 0220kggcos 30 20kggsin 30 131 N Pulling with force F on box at an angle Free Body Diagram 2nd LaW equations give US FN F Xdirection F cos 9 m ax Y direction F sin 9 FN W m ay 0 FN W F sin 9 w Normal force is reduced You hang a mass from a scale When you jerk the scale upwards what does the scale read When you accelerate the scale downward what does it read When you jerk the scale upwards the reading will increase When you jerk the scale down the reading decreases Jerking the scale upwards The force exerted by the scale T on the mass is in the gositive y direction W is in the negative y direction Acceleration is in the gositive y direction 2nOI Law 2F ma T Wma TWma T mg ma mga Jerking the scale downwards The force exerted by the scale T on the mass is in the gositive y direction W is in the negative y direction Acceleration is in the negative y direction 2nOI Law 2F ma T W ma TW ma T mg mamg a Example m1 20 kg Mass 1 is on frictionless surface Mass 2 is allowed m2 10 kg to fall What accelerations do the objects have They are tied to together Free Body Diagrams Mass 1 Mass 2 Do Newton s 2nOI Law for each mass Mass 1 moves horizontally picking right for positive T mla1 Mass 2 moves vertically picking up for positive T m2g m2a2 Important a1 a2 in magnitude since they are tied together T mla and T m2g m2a mla ng mza 77128 m1 1712 a

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