Physics 2514 Lecture 12 P. Gutierrez Department of Physics & Astronomy University of Oklahoma P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 1 / 13
Goal Goals for today s lecture: Application of Newton s second law. Equilibrium Static and Dynamic. Define weight. Introduce friction Static and Dynamic P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 2 / 13
Clicker An elevator suspended by a cable is moving upward and slowing to a stop. Which free-body diagram is correct? P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 3 / 13
Equilibrium Newton s Second Law. An object of mass m subjected to forces F 1, F 2,... will undergo an acceleration a given by a = F net m where F net = n i=1 F i is the vector sum of all forces acting on the object. The acceleration vector a points in the same direction as the force vector F net. Condition for equilibrium. F i = 0 a = 0 i P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 4 / 13
Equilibrium Newton s Second Law. An object of mass m subjected to forces F 1, F 2,... will undergo an acceleration a given by a = F net m where F net = n i=1 F i is the vector sum of all forces acting on the object. The acceleration vector a points in the same direction as the force vector F net. Condition for equilibrium. F i = 0 a = 0 i P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 4 / 13
Equilibrium Newton s Second Law. An object of mass m subjected to forces F 1, F 2,... will undergo an acceleration a given by a = F net m where F net = n i=1 F i is the vector sum of all forces acting on the object. The acceleration vector a points in the same direction as the force vector F net. Condition for equilibrium. F i = 0 a = 0 i P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 4 / 13
Equilibrium Equilibrium i F i = 0. Static equilibrium system is at rest. Dynamic equilibrium system moves with a constant velocity; a = 0. P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 5 / 13
Example An accident victim with a broken leg is placed in traction. The patient wears a special boot with a pulley attached to the sole. The foot and boot together have a mass of 4.0 kg and the doctor decides to hang a 6.0 kg mass from the rope. The boot is held suspended by the ropes and does not touch the bed. P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 6 / 13
Example What is the tension in the rope? P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 7 / 13
Example What is the tension in the rope? T F G P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 7 / 13
Example What is the tension in the rope? T F G The force due to gravity F G = ma F G = mg P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 7 / 13
Example What is the tension in the rope? T F G The force due to gravity F G = ma F G = mg F i = T m 6 g = 0 T = m 6 g = 6.0 kg 9.8 m/s 2 = 58.8 N i P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 7 / 13
Example The net traction force needs to pull straight out on the leg. What is the proper angle θ for the upper rope? P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 8 / 13
Example The net traction force needs to pull straight out on the leg. What is the proper angle θ for the upper rope? y F m 4 g T 1 T 2 x P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 8 / 13
Example The net traction force needs to pull straight out on the leg. What is the proper angle θ for the upper rope? y F m 4 g T 1 T 2 x T = T 1 = T 2 = 58.8 N Fy = T sinθ m 4 g T sin(15 ) = 0 sinθ = (m 4 g + T sin(15 ))/T θ = 67.5 Fx = T cos(15 ) + T cos θ F = 0 F = 79.0 N P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 8 / 13
Example The net traction force needs to pull straight out on the leg. What is the proper angle θ for the upper rope? y F m 4 g T 1 T 2 x T = T 1 = T 2 = 58.8 N Fy = T sinθ m 4 g T sin(15 ) = 0 sinθ = (m 4 g + T sin(15 ))/T θ = 67.5 Fx = T cos(15 ) + T cos θ F = 0 F = 79.0 N P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 8 / 13
Example The net traction force needs to pull straight out on the leg. What is the proper angle θ for the upper rope? y F m 4 g T 1 T 2 x T = T 1 = T 2 = 58.8 N Fy = T sinθ m 4 g T sin(15 ) = 0 sinθ = (m 4 g + T sin(15 ))/T θ = 67.5 Fx = T cos(15 ) + T cos θ F = 0 F = 79.0 N P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 8 / 13
Friction Next to gravity, friction is the most common force that we interact with. It allows us to walk, slows us down, and prevents us from moving. There are 3 types of friction (all act to oppose motion) Static fraction f s µ s N when no motion occurs; Kinetic friction f k = µ k N, when object is moving; Rolling friction f r = µ r N, when objects rolls. P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 9 / 13
Clicker Joe and Bill are playing tug-o-war. Joe is pulling with a force of 200 N. Bill is simply hanging on, but skidding towards Joe at a constant velocity. What is the magnitude of the force of friction between Bill s feet and the ground. A) 200 N B) 400 N C) 0 N D) 300 N D) None of the above P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 10 / 13
Example A 50 kg steel box is in the back of a dump truck. The truck s bed, also made of steel, is slowly tilted. At what angle will the file cabinet begin to slide? Brief description: A 50 kg steel box is on a steel incline plane. What is the maximum angle the incline can have for the box to remain in static equilibrium? P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 11 / 13
Solution Known µ s = 0.80, m = 50 kg Unknown angle θ of incline Normal force n Equations of motion: } mg sin θ µ s n = 0 mg cos θ + n = 0 µ s = tanθ θ = tan 1 µ s = 38.7 P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 12 / 13
Announcements Read chapter 6. Reading quiz on chapter 5 and 6; due by 11:59 PM today. New homework available. P. Gutierrez (University of Oklahoma) Physics 2514 February 21, 2011 13 / 13