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Transcript
Unit 3.8: Net Forces with Inclines For most motion problems, the object in question is moving horizontally or vertically but not in both directions at the same time. Objects that are sliding up or down hills present difficulty due to the number of forces that are acting on the object as it slides because of the number of forces that are at angles to the axis of origin (as shown to the right-where the dotted line is the hills surface). As shown in example 3.7 we would have to find each of the components of all the forces and then add and subtract the forces that are acting in the same dimension (horizontally & vertically). This means doing a lot of trigonometry to find all these components. However, there is an easier fix to this problem. Since friction acts along the surface of the hill and the normal force acts perpendicular to it, we can tilt the x-y axis to match the tilt of the hill as shown to the left. When tilting the axis, rotate the x & y axes until they match the angle of the hill’s tilt. This reduces the number of components that must be calculated. In this set up, the force of friction and the x component of gravity are working against each other. However, since the object is not sinking through the hill nor is it shooting off of it, the normal force and the y component of gravity cancel each other out. (Remember that Fg -weight- is found by multiplying mass-kg-by acceleration 2 due to gravity-9.8 m / s ) Examples: 1. A 300 N box slides down a 20 incline at constant velocity. a) How much are the horizontal and vertical components of the box’s weight? b) How much is the net force acting on the box? c) How much is the normal force acting on the box? d) How much is the friction acting on the box to make its velocity constant? Since this box is moving at constant velocity, the net force acting on the box is zero and thus all the forces in the same directions must cancel out. This means that x and F f must cancel each other out and thus must be equal to each other. Furthermore, y and FN must also cancel each other out and thus must be equal. a) x component: y component: x x 300 sin 20 102.6 300 y cos 20 y 300 cos 20 281.9 300 sin 20 b) Net Force is zero due to the constant velocity c) Since Net Force is zero, y and FN must be equal so that they can cancel with each other. Thus FN =281.9 N d) Again, since the Net Force is zero, x and F f must be equal so that they can cancel with each other. Thus F f =102.6 N 2. A force of 200 N is used to push a 27 kg suitcase up a 38 inclined ramp at constant velocity. a) How much does the suitcase weigh? b) How much is the normal force? c) How much is the friction? a) Since weight is mass times by acceleration due to gravity, Fg ma 27(9.8) 264.6 N b) Since the suitcase is moving at constant velocity, the net force is zero which means that all the forces cancel out. Thus, y FN must be equal so that they can cancel. Thus and y y 264.6 cos 38 208.5 264.6 y FN 208.5 N cos 38 c) Again since the net force is zero, x, F p , and F f cancel with each other. Because there are 3 forces this time, the two directed down the incline (x & F f ) must add up to and equal the one directed up the incline ( F p ). Thus x x 264.6 sin 38 162.9 264.6 x F f FP 162.9 F f 200 F f 37.1N sin 38 3. A 100 kg rock remains stationary on First calculate the weight of the rock a 50 inclined hill. How much Fg mg 100(9.8) 980 N friction must be acting on the rock to keep is stationary? Again Now find the horizontal ( x) component of the weight sin 50 x x 980 sin 50 750.7 N 980 Since the rock is at rest it is therefore in equilibrium and thus has a net force of zero. This means that x and F f must be equal in order to cancel with each other. F f x 750.7 N 4. A man pushes a 200 N box down a 15 inclined hill by pushing with a force of 35 N. If friction produces 40 N on the box, calculate the amount of acceleration the box experiences as it slides down the hill. First find the x-component of the box’s weight: sin 15 x x 200 sin 15 51.76 N 200 Now find the net horizontal force acting on the box: Fnet x Fp F f Fnet 51.76 35 40 46.76 N Now find the mass of the box: Fg mg 200 m(9.8) 200 20.41kg 9.8 m Now using the net force and mass (calculated above) find the acceleration that the box experiences: Fnet ma 46.76 20.41(a) a 5. An 18 kg wagon at rest on a 30 inclined hill rolls to the bottom of the hill in 4.5 s and is going 12 m/s by the time it reaches the bottom. a) What is the weight of the wagon? b) Find the horizontal (x) component of the wagon’s weight. c) What is the acceleration that the wagon experiences as it rolls to the bottom of the hill? d) How much is the net horizontal force acting on the wagon? e) How much friction is acting on the wagon? 46.76 2.29m / s 2 20.41 a) Fg ma 18(9.8) 176.4 N b) sin 30 x x 176.4 sin 30 88.2 N 176.4 c) Use v v0 at thus 12 0 a(4.5) a 12 4.5 2.67m / s 2 d) Since Fnet ma 18(2.67) 48.06 N e) Since Fnet 48.06 N this means that one of the two horizontal forces is this much larger than the other and since, by definition, net force is the sum of all forces then Fnet x F f 48.06 88.2 F f F f 40.14 N