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Solve friction problems

Nuffield Free-Standing Mathematics Activity. Solve friction problems. What forces are acting on the sledge?. What force is making the suitcases accelerate?. The friction model. Before sliding occurs …. Friction is just sufficient to maintain equilibrium and prevent motion. F < F MAX.

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Solve friction problems

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  1. Nuffield Free-Standing Mathematics Activity Solve friction problems

  2. What forces are acting on the sledge? What force is making the suitcases accelerate?

  3. The friction model Before sliding occurs … Frictionis just sufficient to maintain equilibrium and prevent motion F< FMAX On the point of sliding and when sliding occurs … F = mR where F is the friction, R is the normal contact force and m is a constant called the coefficient of friction

  4. 5 kg R 15N F 5g 15N Friction problems Example If m = 0.4, will the box move? Think about What is the smallest force that will make the box slide along the table? Solution Vertical forces: R= 5g Maximum possible friction FMAX = m R = 19.6 N = 0.4  5g The pushing force is less than 19.6 N where g = 9.8 ms–2 The box will not move

  5. smooth R T F 0.4g 0.2g 1 = 2 0.4g Friction problems Example If the package is on the point of moving, find m. 400 grams Think about What forces are acting on the package? Vertical forces: R= 0.4g 200 grams Solution On the point of moving F= m R = m 0.4g F= T As the pulley is smooth T= 0.2g m 0.4g= 0.2g m = where g = 9.8 m s–2

  6. (u + v)t s = ut + at2 s = 2 1 2 More difficult friction problems may require the use of … Think about Why does the friction model allow the use of these equations? Newton’s Second Law Resultant force = mass  acceleration where the force is in newtons, mass in kg, and acceleration in m s–2 Equations of motion in a straight line with constant acceleration v = u + at v2 = u2 + 2as where u is the initial velocity, v is the final velocity, a is the acceleration, t is the time and sis the displacement

  7. 20 m s–1 R F 1200g More difficult friction problems Example The car brakes sharply then skids. a the deceleration If m = 0.8, find 1.2 tonnes b the distance travelled in coming to rest Solution Think about What is the friction when the car is skidding? a Vertical forces: R = 1200g = 0.8  1200g = 9408 N F= m R Newton’s Second Law gives: –9408 = 1200a a = –7.84 m s–2 Think about Which equation can be used to find the distance the car travels as it comes to a halt? b 02 = 202 - 2  7.84s v2 = u2 + 2as where g = 9.8 m s–2 15.68s = 400 s = 25.5 metres

  8. Solve friction problems • When can you use F= mR? • How does the friction model allow you to use F = ma and the constant acceleration equations to solve problems? • Can you think of other situations when friction prevents an object from moving? • Can you think of other situations when friction causes an object to accelerate? Reflect on your work

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