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Nuffield Free-Standing Mathematics Activity Solve friction problems © Nuffield Foundation 2011 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 < 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 Friction problems Example If m = 0.4, will the box move? 5 kg 15N Think about What is the smallest force that will make the box slide along the table? Solution R Vertical forces: R = 5g 15N F Maximum possible friction FMAX = m R = 0.4 5g = 19.6 N 5g where g = 9.8 ms–2 The pushing force is less than 19.6 N The box will not move Friction problems Example If the package is on the point of moving, find m. smooth 400 grams Think about What forces are acting on the package? Vertical forces: Solution R = 0.4g On the point of moving R F = m R= m 0.4g F=T T F 200 grams As the pulley is smooth T = 0.2g m 0.4g = 0.2g 0.4g where g = 9.8 m s–2 m = 0.2g = 1 0.4g 2 More difficult friction problems may require the use of … Newton’s Second Law Think about Why does the friction model allow the use of these equations? 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 s = ut + 1 at2 2 s= (u + v)t 2 v2 = u2 + 2as where u is the initial velocity, v is the final velocity, a is the acceleration, t is the time and s is the displacement More difficult friction problems Example The car brakes sharply then skids. 20 m s–1 1.2 tonnes If m = 0.8, find a the deceleration b the distance travelled in coming to rest Solution Think about What is the friction when R a Vertical R = 1200g the carforces: is skidding? F = m R = 0.8 1200g = 9408 N Newton’s Second Law gives: –9408 = 1200a a = –7.84 m s–2 F 1200g where g = 9.8 m s–2 Think about Which equation can be 2 = distance find0the car b v2used 202 - 2 the 7.84s = u2 +to2as travels as it comes to a halt? 15.68s = 400 s = 25.5 metres Solve friction problems Reflect on your work • 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?