Frictions

March 21, 2018 | Author: jhon | Category: Friction, Force, Mass, Gravity, Orbit


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Friction and weight are common experiences and taken for granted, but were not always understood. Aristotle believed that objects that contained earth fell because in Earth was at the center of the universe, and that was the natural order of things. We know know that weight, which is the force of gravity between objects, not only makes apples fall from trees, but keeps the Moon circling the Earth, and the Earth circling the sun. Aristotle also believed (and some students still believe) that the natural state of an object is to be at rest. That is, without a force pushing it, an object will slow down to a stop. This view overlooks friction, which is the force that causes most objects we see to come to rest. Now we know that there is nothing pushing planets along their orbit, and that without gravity to hold them in orbit, they would continue to move in a straight line until interfered with. A. Weight, Vertical Forces, and Elevator Problems Examples 1. What is the weight of a 200 kg object? 2. A woman exerts a force of 500 N straight up to lift a 35 kg basket. What is the acceleration of the basket? Answers to Examples 1 and 2 1.Near the Earth's surface, objects fall at 9.8 m/s/s. If air resistance is negligible then gravity is the only force acting on the falling object. Applying Newton's Second Law of Motion, a 200 kg object will have a weight of F = ma = (200 kg)(9.8 m/s/s) = 1960 N 2. Free body diagram: Sum the forces: Fnet = FA + mg Fnet = 500 N + (35 kg)(-9.8 m/s2) Fnet = 157 N [up] Second Law: Fnet = ma Fnet = (35.0 kg)a Substitute: 157 N [up] = (35.0 kg)a a = 4.49 m/s2 [up] Problems - Part A (See below for answers) 1. What is the acceleration of a falling crate (mass including parachute 132 kg) when the upward force or air resistance is one-fourth its weight? 2. The payload lifted by a balloon has a mass of 6.9 kg. The balloon exerts an upward force of 92.8 N on the payload. 0 N of tension. What is the greatest combined mass of the pail and water if they are lifted (a) with constant speed? (b) with an acceleration of 1.0 m/s. How much tension must a rope withstand when used to accelerate a 1300 kg car vertically upward at 0. The balloon accelerates from rest for 10. at which point the payload is released. What is the velocity of the payload at the moment of its release? 3.35 m/s/s? 9. What is his velocity of impact? 8. A 20. Horizontal and Friction Problems Example 3.Part B (See below for answers) . 4.0 N. What is the acceleration of the mass? Answer to Example 3: Free body diagram (whenever there are two or more forces): Sum of forces (vertical forces cancel as evidenced by lack of acceleration in the vertical dimension) Fnet = T + Ff Fnet = (100 N) + (-20.0 kg mass is pulled by along a surface by a horizontal force of 100 N.7 s. 5.0 m/s2. How long will it take for the crate to rise 8. Friction is 20. A constant force of 600 N is applied straight up on a 50 kg crate.a. What is the tension in the cable supporting the cable? 7. What is the force exerted on the elevator floor by a 75 kg person. What is the acceleration of the balloon and the payload? b. You are standing on a bathroom scale in an elevator and wish to calculate what the scale would read while experiencing the following scenarios: (just derive a working equation) a)You are standing in the elevator that just began accelerating downwards as it approaches and comes to a stop at the 11th floor b) You are standing in the elevator that just began accelerating upwards as it approaches and comes to a stop at the 10th floor.90 m/s/s? B. An elevator of mass 3000 kg is ascending at a steady speed of 2. An 80-kg stuntman steps out of a window 30 meters above layers of mattresses. An elevator is moving downward with an acceleration of 2. Air resistance exerts 100-N on the stuntman.0 m from rest? 6.0 N) = 80 N Problems . The string used to lift a pail of water will break if there is more than 51. 99 kg)(6) = 30. A jet engine generates 160 kN of force as it propels a 20.81 = 4. A 49-N block is pulled by a horizontal force of 50.1.99 kg In the horizontal axis. What is the coefficient of friction? Answer to example 4: Fg = 49-N T = 50. Coefficient of Friction Problems Example 4.000kg plane down a runway. If 40 kN of friction opposes the plane. What is the coefficient of friction? Free-body diagram: Sum the forces: In the vertical axes forces are equal and opposite as there is no vertical acceleration. FN = -Fg FN = mg = 49 N Also.0 N a = 6 m/s/s. Fnet = T + Ff Fnet = (50 N) + Ff Newton's Second Law: Fnet = ma Fnet = (4. m = 49/9.0 N .0 N along a rough horizontal surface at a constant acceleration of 6 m/s/s. how much time it will take the plane to reach a speed of 33m/s from rest? C. 8 m/s2. each with mass = M. The acceleration of gravity is 9.51. A 2500kg car traveling 14.056. a physics major.95 kg slides with an initial speed of 4.5. What is the coefficient of friction between the tires and the surface? . Calculate the tension in the horizontal rope connecting the cars.00 Neglect any friction force acting on car B..0m. A 4. Two large blocks.10g.260. Car A is accelerating at 1m/s/s. What is the maximum force. whereas kinetic friction coefficient is . Determine (a) the coefficient of static friction and (b) the coefficient of kinetic friction between the couch and the carpet. the man continues to push with 700 N and his daughter.0 kg toboggan rests on a frictionless icy surface and a 2.Part C (See below for answers) 1. Both cars have the same mass ma = mb = 1000 kg. horizontal surface.0 N along a rough horizontal surface at constant velocity. If the coefficient of static friction between the crate and the plane is 0. If the coefficient of friction between road and tires on a rainy day is 0. are placed on a frictionless surface. Car A is towing car B. The block now encounters a rough patch with a coefficient of kinetic friction given by µk = 0. A car is traveling at 60. 12.0 m/s on an icy. a) what is the kinetic friction force.10 m/s2. In attempting to move a heavy couch (mass 200 kg) across a carpet. 6.30. The coefficient of static friction between the box and toboggan is .6 mph on a horizontal highway. What minimum coefficient of static friction must exist between your feet and the floor if you are not to slide? 11. A 15.65.0 kg box rests on top of the toboggan. What is the coefficient of friction? 2. Once the couch starts moving. The box is pulled by a 30 N force along the horizontal parallel to the icy floor. what is the minimum distance in which the car will stop? Answer in units of m. A packing crate is placed on a plane inclined at an angle of 35° from horizontal.0 kg block is pulled by a horizontal force of 30. What are the magnitudes and directions of the resulting accelerations of the block and toboggan? 7. What is the acceleration of the block when it is in the rough patch? 9. a man finds he must exert a horizontal force of 700 N to get the couch to barely move.33 m/s on a smooth. estimates that it then accelerates at 1. that can be applied to the first large block to make all four move with the same acceleration? 8. will the crate slide down the plane? 10. The coefficient of friction between the top and bottom blocks is µ. Suppose that you are standing on a train accelerating at 0. what horizontal force is required to move the crate at a steady speed across the floor? What horizontal force is required if µk is zero? 4. with mass = m. is placed on each of the two larger. A 300 lb golf cart is traveling at 60 f/sec on a surface with a coefficient of kinetic friction equal to 0. level road approaches an intersection. and the static coefficient of friction between its tires and the road is µs = 1.60. If the coefficient of kinetic friction between a 35 kg crate and the floor is .Problems . with a cord connecting the two top (small) blocks. Car A has four-wheel drive. 3. Another cord is used to pull the first large block. A smaller block. if the brakes are locked? b) what distance will it take to stop if the brakes are locked? 5. A block of mass 1. F. The brakes lock and the car skids 25. Fnet = 0.000-kg load sits unsecured on the flat bed of a 20.4226)} = 29.5 .5 . D. Fnet = 0.8 N Problems .25){(118 N) .5 T = 36. A 10. and FN = mg . T(. Calculate the minimum stopping distance for which the load will not slide forward relative to the truck.T(0.4226) In the horizontal Fnet = Th .Part D .Tv Since there is no acceleration in the vertical axis.9063) = 29.9063) Ff = µFN = (0.0 m/sec.Ff Since there is no acceleration in the horizontal axis.Tsin25º FN = (118 N) .T(0.1057) Since Th = Ff . A block is pulled by a string that makes an angle of 25º to the horizontal.FN .25.1057) T(.0 kg and the coefficient of friction is 0. what force would keep the block moving at a constant velocity? Answer to Example 5: Free-body diagram: Sum the forces: in the vertical Fnet = mg . The load has a coefficient of static friction of 0. If the mass of the block is 12. Problems Involving 2-Dimensional Forces Example 5.500 with the truck bed.T(.T(.8) .13.000-kg truck moving at 12.8007) = 29.Tv = mg . and Th = Ff Th = Tcos25º = T(.Tsinø = (12 kg)(9. Calculate a. the horizontal friction force on the mower. FN = Fy and Ff = Fx . components along the y-axis cancel. An 18-kg suitcase is pulled at constant speed by a 43-N force applied to an attached strap. FN = mgcosø and Ff = mgsinø But Ff = µFN .0 m/s/s? 5. We find the maximum angle the plane can be inclined without the crate sliding. Find the acceleration of the box. while the vertical component of the force is 300N up. Since acceleration is zero.1. and the coefficient of kinetic friction between the box and floor is . If the coefficient of kinetic friction between the box and the floor is . normal force and coefficient of friction. 7. A 4. normal force and coefficient of friction. A 40-kg package and a 10-kg package slide vertically up a wall.0 N directed along the handle. then b. A 100kg block is pulled at a constant speed along a horizontal surface..1 kg box is pushed along a horizontal surface by a force F of magnitude 21 N at an angle of 35º with the horizontal. so .0 kg lawn mower at constant speed with a force of F = 88. The horizontal component of the force is 400N. 3. Find the kinetic friction.0 degrees to the horizontal.386. 6. 2. which is at an angle of 45. A force of 185N is applied to the rope. What angle does the strap make with the horizontal? b.0 degrees with the horizontal. A strap attached to a 16. a. What is the normal force exerted on the suitcase? C9. 4. Therefore. The angle between the cable and the wall is 15º. while the vertical component of the force is 300N down. Find the kinetic friction. Friction is 27 N. Calculate the tension in the strap and the normal force on the chest when the chest moves with constant speed. 8. A person pushes a 14.450.0 kg box makes an angle of 25.9 kg chest makes an angle of 46. A rope attached to a 35. The friction between the wall and the 40-kg package is 60 N. The coefficient of kinetic friction between the chest and the floor is 0.8° above the horizontal. In other words.20. A 100kg block is pushed at a constant speed along a horizontal surface. The horizontal component of the force is 400N. Calculate acceleration of the box. and components along the x-axis cancel. a. What is the tension in the upper cable if the packages move at a constant speed? b) What is the tension in the cable connecting the packages if they are lowed at 1. the normal force exerted vertically upward on the mower by the ground. 65. ø = 33. Since acceleration is zero.0° The crate will slide. net force is zero. (a) If the coefficient of static friction is 0.mgsinø = µmgcosø . Therefore.312 A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 24° above the horizontal. what minimum force magnitude is required from the rope to start the crate moving? (b) If the coefficient of kinetic friction = 0. Since µ = 0. what is the magnitude of the initial acceleration (m/s^2) of the crate? . µ =Ff / N = 400/1281 = 0. mg + Fy = N (100)(9.47. since net force is zero. Ff = Fx and kinetic friction is 400N. This yields µ = tanø .81) + 300 = N N = 1281N Finally. Further.29. D7.
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