Friction Forces. Review important points by referring the following section taken from your textbook.

Transcription

1 Friction Forces Review important points by referring the following section taken from your textbook.

2 Physics McGray-Hill Ryerson 2002 pp

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6 In grade 11, you did problems where the object would be moved along a horizontal surface. When Applied Force is Parallel to Motion F N motion F f Fa Fg When examining this, we would have grouped horizontal force vectors in the horizontal net force calculations. Fnet x = F a + F f and Fnet x = m a x equating the above gives F a + F f = m a x And, then the vertical force vectors are grouped in the vertical net force calculations. F nety = F N + F g and F nety = m a y F N + Fg = m a y

7 When Forces Are Not Parallel to Motion In grade 12, we do problems where forces are exerted at angles. Pulling up at an angle: F N motion Fa F ay F f θ F ax Fg θ positive...ccw F AX is positive. Fay is positive. When examining this, we would have grouped horizontal force vectors in the horizontal net force calculations. F ax = F cos θ Fnet x = F ax + F f and Fnet x = m a x equating the above gives F ax + F f = m a x F cos θ + F f = m a x And, then the vertical force vectors are grouped in the vertical net force calculations. F ay = F sin θ F nety = F N + F g + F ay and F nety = m a y F N + Fg + F ay = m a y 0 The vertical acceleration, a y, would equal zero. So, the vertical net force would equal zero. This means that F N + Fg + Fay = 0 N [up] So, F N = 0N Fg Fay F N = Fg Fay When calculating friction here... Ff = - μ F N = - μ ( - Fg - Fay)

8 Pushing down at an angle. motion F N F f θ Fax Fay Fa Fg θ is in quad IV here so...can use the CW measurement if desired. Fax is positive. Fay is negative. When examining this, we would have grouped horizontal force vectors in the horizontal net force calculations. F ax = F cos θ Fnet x = F ax + F f and Fnet x = m a x equating the above gives F ax + F f = m a x F cos θ + F f = m a x And, then the vertical force vectors are grouped in the vertical net force calculations. F ay = F sin θ F nety = F N + F g + F ay and F nety = m a y F N + Fg + F ay = m a y 0 The vertical acceleration, a y, would equal zero. So, the vertical net force would equal zero. This means that F N + Fg + Fay = 0 N [up] So, F N = 0N Fg Fay F N = Fg Fay When calculating friction here... F f = - μ F N = - μ ( - Fg - Fay)

9 Let's reflect for a moment on what we see in the last three examples: F f F N Fg motion Fa Here, the normal force is equal and opposite to the weight. 0 Fnet y = may Fnet y = Fn + Fg 0 N = Fn + Fg Fn = - Fg and Fn = Fg F N motion Here, the normal force is opposite in direction to the weight but it has smaller magnitude than the weight. Fa F ay Fn < Fg F f θ F ax Here, the Fay supports some of the weight of the object. The normal force exerted on the object by the supporting surface will be less. Fg 0 Fnet y = may Fnet y = Fn + Fg + Fay 0 N = Fn + Fg + Fay Fn = - Fg - Fay...where Fay itself is pos. and the Fg is neg. motion F N Here, the normal force is opposite in direction to the weight but it a larger magnitude than the weight. F f Fg θ Fa Fax Fay Fn > Fg The Fay pushes the object into the surface. The normal force exerted on the object by the supporting surface must be larger because it is now opposing weight and the Fay. 0 Fnet y = may Fnet y = Fn + Fg + Fay 0 N = Fn + Fg + Fay Fn = - Fg - Fay...where Fay itself is neg. and the Fg is neg.

10 In PHY 621, you will be expected to show the reader why the normal force equals whatever it equals. Will need net force statements. Need two net force statements. One is the definition of vertical net force which is the sum of the vertical forces acting on the object. Fnet = Σ F One is use of Newton's second law. Fnet y = ma y

11 Try these examples. Answers provided on following slides. Draw a free body diagram! Showing the work is part of the solution. So, show givens and all equations you need to use. Combine equations then as required and at your comfort level.

12 Forces Applied At Angles Ex. Vivien pushes a trolley by applying a force of 250 N [E 20 down]. The trolley moves against a frictional force of N. The trolley has a mass of 28.7 kg. a) What is the horizontal acceleration of the trolley? Free body diagram showing forces (labelled), angle, and direction of motion if moving. Forces drawn as exerted throu centre of mass of object. Givens listed b) What is the coefficient of static friction between the wheels and the floor?

13 Forces Applied At Angles Ex. Vivien pushes a trolley by applying a force of 250 N [E 20 down]. The trolley moves against a frictional force of N. The trolley has a mass of 28.7 kg. motion Ff FN Fg Fa Fax Fay a) What is the horizontal acceleration of the trolley? Fa = 250 N [E 20 o down] θ = - 20 o I will use the CW angle. Ff = N [E] m = 28.7 kg θ F ax = Fa cos θ Free body diagram showing forces (labelled), angle, and direction of motion if moving. Forces drawn as exerted throug centre of mass of object. Fnet x = F ax + F f and Fnet x = m a x equating the above gives F ax + F f = m a x Givens listed Fa cos θ + F f = a x = Fa cos θ + F f m m a x a x = 250 N cos (-20 o ) + (-132.6N) 28.7 kg ax = N N 28.7 kg = N 28.7 kg = m/s2 = 3.57 m/s 2 [E] b) What is the coefficient of static friction between the wheels and the floor? Fg = mg F ay = F sin θ Fa = 250 N [E 20 o down] θ = - 20 o Ff = N [E] m = 28.7 kg F nety = F N + F g + F ay and F nety = m a y = 0N [up] g = m/s 2 [up] F N + Fg + F ay = 0N [up] ay = 0 m/s 2 [up] F N = - Fg - Fay Givens listed. Note that here I am using negative up direction for g...which I do most of the time. If you chose positive down, then your signs will be different then mine. F N = - mg - Fa sinθ =-( 28.7 kg)(-9.81 m/s 2 ) - 250N sin (-20 o ) = N N = N [up] Ff = μ Fn μ = Ff = ( N) = Fn N μ = 0.361

14 Ex. Nicolas pulls a wagon by applying a force of 150 N [ N 40 o up]. The wagon is moving with a constant speed. The coefficient of sliding friction between the runners and the grass is a) What is the force of sliding friction opposing this motion? b) What is the mass of the wagon? Ans: N [N] Ans: kg

15 Ex. Nicolas pulls a wagon by applying a force of 150 N [ N 40 o up]. The wagon is moving with a constant speed. The coefficient of sliding friction between the runners and the grass is motion FN Fa Fay θ Ff Fax Fg a) What is the force of sliding friction opposing this motion? Fa = 150 N [N 40 o up] θ = 40 o I will use the CCW angle. ax = 0 m/s 2 [E] Moving at a constant velocity, no acceleration. F ax = Fa cos θ 0 Fnet x = F ax + F f and Fnet x = m a x = 0 N [E] equating the above gives F ax + F f = Fa cos θ + F f = 0 N 0N F f = - Fa cos θ F f = N cos (40 o ) = N = N [E] F f b) What is the mass of the wagon? Fa = 150 N [E 40 o up] θ = 40 o Ff = N [E] m =? μ = g = m/s 2 [up] F f = μ Fn FN = F f = ( N) = μ FN = N [up] ay = 0 m/s 2 [up] Fg = mg F ay = F sin θ F nety = F N + F g + F ay and F nety = m a y = 0N [up] 0 F N + Fg + F ay = 0N [up] Fg = - FN - F ay mg = - FN - Fay m = - Fn - Fay g = - Fn - Fa sin θ g = N - 150N sin 40 o m/s 2 = N N m/s 2 = N m/s 2 = kg m = 37.6 kg

16 Try this worksheet! Start easy. Most challenging is at the end.

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