# Solve friction problems

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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. - PowerPoint PPT Presentation

### Text of Solve friction problems

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

• The friction modelBefore sliding occurs Friction is just sufficient to maintain equilibrium and prevent motion F < FMAXOn the point of sliding and when sliding occurs F = mRwhere F is the friction, R is the normal contact forceand m is a constant called the coefficient of friction

• Friction problemsIf m = 0.4, will the box move?Solution Vertical forces: R = 5g where g = 9.8 ms2Maximum possible frictionFMAX = m RThe pushing force is less than 19.6 NExample= 0.4 5g= 19.6 N The box will not moveThink about What is the smallest force that will make the box slide along the table?

• If the package is on the point of moving, find m.400 gramsFriction problemsExample200 gramsSolution Vertical forces: R = 0.4g where g = 9.8 m s2As the pulley is smoothF = m Rm 0.4g = 0.2g m =On the point of moving T = 0.2g F = T = m 0.4gThink about What forces are acting on the package?

• More difficult friction problemsResultant forceNewtons Second Law= mass accelerationwhere the force is in newtons, mass in kg, and acceleration in m s2Equations of motion in a straight line with constant accelerationv = u + atv2 = u2 + 2aswhere u is the initial velocity, v is the final velocity, a is the acceleration, t is the time and s is the displacementmay require the use of Think about Why does the friction model allow the use of these equations?

• More difficult friction problemsThe car brakes sharply then skids. Solution Vertical forces: R = 1200g where g = 9.8 m s2Newtons Second Law gives:Example= 9408 N9408 = 1200a1.2 tonnesIf m = 0.8, finda the decelerationb the distance travelled in coming to restF = m R= 0.8 1200ga = 7.84 m s2

v2 = u2 + 2as02 = 202 - 2 7.84s15.68s = 400 s = 25.5 metres a bThink about What is the friction when the car is skidding?Think about Which equation can be used to find the distance the car travels as it comes to a halt?

• 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 Solve friction problems

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