TwoBody System: Two Hanging Masses


 Felix Jenkins
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1 Specific Outcome: i. I can apply Newton s laws of motion to solve, algebraically, linear motion problems in horizontal, vertical and inclined planes near the surface of Earth, ignoring air resistance. Newton s Laws and Pulley Systems TwoBody System: Two Hanging Masses TwoBody System: One Hanging Mass ThreeBody System: Two Hanging Masses Newton s Laws and Pulley Systems We will deal with three different pulley systems Each pulley system: 1) is frictionless, and 2) has a net force supplied only by the hanging mass or masses Mass of the pulley itself is insignificant This first system has 2 competing hanging masses The system s net force is the magnitude of the difference in both of their F g values The pulley turns in the direction of the heavier mass; both masses accelerate together as a system 1
2 Steps to solve for this system: 1) Calculate F net from the difference in F g between the two hanging masses 2) Calculate magnitude of acceleration from F net and the system s total mass 3) Directions: the larger mass is moving down, the smaller mass is moving up 75 kg. Determine the acceleration of each mass. F ga = m A g = (55 kg)(9.81 m/s 2 ) = N F gb = m B g = (75 kg)(9.81 m/s 2 ) = N F net = F ga  F gb = N N = N 75 kg. Determine the acceleration of each mass. m system = m A + m B = 55 kg + 75 kg = 130 kg a system = F net m system a A = 1.5 m/s 2 [up] = N 130 kg = m/s2 a B = 1.5 m/s 2 [down] This second pulley system has one hanging mass that supplies all of the net force (F g = F net ) The magnitude of acceleration is the same for the surface mass and the hanging mass The pulley turns in the direction of the hanging mass, which moves down m B m A 2
3 The surface mass moves toward the pulley, ignoring friction Steps: 1) Find F net from F g of the hanging mass 2) Find acceleration of the system from the combined mass of both objects ex. A 5.00 kg lab cart is accelerated on a frictionless table by a 2.00 kg mass 5.00 kg accelerating straight down, as shown. What is the acceleration of the lab 2.00 kg cart? F net = F g = mg = (2.00 kg)(9.81 m/s 2 ) = N m system = 5.00 kg kg = 7.00 kg F N a system = net = = 2.80 m/s 2 m system 7.00 kg a = 2.80 m/s 2 [right] ex. Find the acceleration of the two masses shown in the diagram. Assume there is no friction. m system = 1.0 kg kg = 3.0 kg 2.0 kg F net = F g = mg = (1.0 kg)(9.81 m/s 2 ) = 9.81 N F net 9.81 N a system = = = 3.3 m/s 2 m system 3.0 kg a cart = 3.3 m/s 2 [right] 1.0 kg a hanging = 3.3 m/s 2 [down] ex. Find the acceleration of the two masses shown in the diagram. The 10 kg cart experiences 5.0 N of friction. 2.0 kg F g = mg = (2.0 kg)(9.81 m/s 2 ) = 19.62N F net = F g + F f = 19.62N + (5.0 N) = N m system = 2.0 kg + 10 kg = 12 kg F N a system = net = = 1.2 m/s 2 m system 12 kg a cart = 1.2 m/s 2 [right] a hanging = 1.2 m/s 2 [down] 3
4 One last pulley system involves: two competing hanging masses a surface mass that is pulled in the direction of the heavier hanging mass all 3 masses accelerate together (system) m C Steps to solve a two pulley system: 1) Calculate F net from the difference in F g between the two hanging masses 2) Calculate acceleration from F net and the combined mass of the system 3) Assign directions based on which hanging mass is heavier m A m B ex. A 15 kg mass lies on a surface. The mass is connected to a 10 kg mass and a 13 kg mass, each hanging by pulleys on each end. Calculate the magnitude of acceleration of the pulley system. F ga = m A g = (10 kg)(9.81 m/s 2 ) = 98.1 N F gb = m B g = (13 kg)(9.81 m/s 2 ) = N F net = F ga  F gb = 98.1 N N = N ex. A 15 kg mass lies on a surface. The mass is connected to a 10 kg mass and a 13 kg mass, each hanging by pulleys on each end. Calculate the magnitude of acceleration of the pulley system. m system = 15 kg + 10 kg + 13 kg = 38 kg F net N a = 0.77 m/s 2 system = = 38 kg m system 4
5 Tension in a pulley string is applied force After finding acceleration, determine tension by doing the following: 1) Choose one of the hanging masses of the system (and use its value of mass) 75 kg. Determine the tension in the string. [Refer to the first example in the whole lesson] F net = F A + F g becomes ma = F A + mg if you chose the 75 kg mass, moving down: 2) Use F net = F A + F g (or ma = F A + mg) to F A = ma mg find tension, F A = (75 kg)( m/s 2 ) (75 kg)(9.81 m/s 2 ) = 6.2 x 10 2 N [up] 75 kg. Determine the tension in the string. [Refer to the first example in the whole lesson] if you chose the 55 kg mass, moving up: F A = ma mg = (55 kg)(1.509 m/s 2 ) (55 kg)(9.81 m/s 2 ) = 6.2 x 10 2 N [up] IDENTICAL! 5
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