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Aim: More friction

Aim: More friction. Do Now: Why do objects at rest have more friction than objects in motion?. A piece of steel whose mass is 100 kg is placed on a steel beam as a force is applied to it causing it to move with a constant velocity. Draw a free-body diagram labeling all the forces. F N. F F.

miroslav-solovyov
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Aim: More friction

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  1. Aim: More friction Do Now: Why do objects at rest have more friction than objects in motion?

  2. A piece of steel whose mass is 100 kg is placed on a steel beam as a force is applied to it causing it to move with a constant velocity. Draw a free-body diagram labeling all the forces FN FF F Fg

  3. What is the weight? Fg = mg Fg = (100 kg)(9.8 m/s2) Fg = 980 N What is the normal force? FN = Fg FN = 980 N What is the net force? 0 N (velocity is constant) FN = 980 N FF F Fg = 980 N

  4. A new force of 700 N is applied. Does steel move? Solve for the maximum static force of friction FF = µsFN FF = (0.74)(980N) FF = 725.2 N The steel does not move. To get it to move, you must have a force greater than 725.2 N

  5. A car whose mass is 1,000 kg is accelerating at a rate of 5 m/s2 with a force of 10,684 N on a surface containing friction. Draw and label a free-body diagram FN FF F = 10,684 N Fg

  6. What is the weight? Fg = mg Fg = (1,000 kg)(9.8 m/s2) Fg = 9,800 N What is the normal force? FN = Fg FN = 9,800 N What is the net force? FNet = ma FNet = (1,000 kg)(5 m/s2) FNet = 5,000 N = 9,800 N FN F = 10,684 N FF Fg = 9,800 N

  7. What is the force of friction? FN = 9,800 N FF = 5,684 N F = 10,684 N Fg = 9,800 N FNet = F – FF 5,000 N = 10,684 N – FF FF = 5,684 N

  8. What is the coefficient of friction? FF = µkFN 5,684 N = µk(9,800 N) µk = 0.58 What two objects are in contact with one another? Rubber on wet concrete

  9. An 80-kilogram skier slides on waxed skis along a horizontal surface of snow at constant velocity while pushing with his poles. What is the horizontal component of the force pushing him forward? F = Ff (constant velocity) F = μkFN F = μkFg (FN = Fg) F = μkmg F = (0.05)(80 kg)(9.8 m/s2) F = 39.2 N

  10. A car’s performance is tested on various • horizontal road surfaces. The brakes are • applied, causing the rubber tires of the car to slide along the road without rolling. The tires encounter the greatest force of friction to stop the car on • dry concrete • wet concrete • dry asphalt • wet asphalt

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