* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Survey
Document related concepts
Lagrangian mechanics wikipedia , lookup
Renormalization group wikipedia , lookup
Fictitious force wikipedia , lookup
Modified Newtonian dynamics wikipedia , lookup
Belt (mechanical) wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Routhian mechanics wikipedia , lookup
Rigid body dynamics wikipedia , lookup
Continuously variable transmission wikipedia , lookup
Centripetal force wikipedia , lookup
Relativistic quantum mechanics wikipedia , lookup
Work (physics) wikipedia , lookup
Equations of motion wikipedia , lookup
Transcript
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.01 W02D3_0 Group Problem: Pulleys and Ropes Constraint Conditions Consider the arrangement of pulleys and blocks shown in the figure. The pulleys are assumed massless and frictionless and the connecting strings are massless and unstretchable. Denote the respective masses of the blocks as m1 , m2 and m3 . The upper pulley in the figure is free to rotate but its center of mass does not move. Both pulleys have the same radius R . a) How are the accelerations of the objects related? b) Draw force diagrams on each moving object. c) What are Newton’s equations of motion for the system of pulleys and objects? Solution: Choose an origin at the center of the upper pulley. Introduce coordinate functions for the three moving blocks, y1 , y2 and y3 . Introduce a coordinate function yP for the moving pulley (the pulley on the lower right in the figure). Choose downward for positive direction; the coordinate system is shown in the figure below then. a) The length of string A is given by l A y1 yP R (1) where R is the arclength that the rope is in contact with the pulley. This length is constant, and so the second derivative with respect to time is zero, 0 d 2l A d 2 y1 d 2 yP 2 2 a y ,1 a y , P . dt 2 dt dt (2) Thus block 1 and the moving pulley’s components of acceleration are equal in magnitude but opposite in sign, a y , P a y ,1 . (3) lB ( y3 yP ) ( y2 yP ) R y3 y2 2 yP R (4) The length of string B is given by where R is the arclength that the rope is in contact with the pulley. This length is also constant so the second derivative with respect to time is zero, 0 d 2lB d 2 y2 d 2 y3 d 2 yP 2 a y ,2 a y ,3 2a y , P . dt 2 dt 2 dt 2 dt 2 (5) We can substitute Equation (3) for the pulley acceleration into Equation (5) yielding the constraint relation between the components of the acceleration of the three blocks, 0 a y ,2 a y ,3 2a y ,1 . (6) b) Free Body Force diagrams: The forces acting on block 1 are: the gravitational force m1g and the pulling force T1,r of the string acting on the block 1. Since the string is assumed to be massless and the pulley is assumed to be massless and frictionless, the tension TA in the string is uniform and equal in magnitude to the pulling force of the string on the block. The free body diagram is shown below. Newton’s Second Law applied to block 1 is then ˆj : m g T m a . 1 A 1 y ,1 (7) The forces on the block 2 are the gravitational force m2g and the string holding the block, T2, r , with magnitude TB . The free body diagram for the forces acting on block 2 are shown below. Newton’s second Law applied to block 2 is ˆj : m g T m a . 2 B 2 y ,2 (8) The forces on the block 3 are the gravitational force m3g and the string holding the block, T3, r , with magnitude TB . The free body diagram for the forces acting on block 3 are shown below. Newton’s second Law applied to block 3 is ˆj : m g T m a . 3 B 3 y ,3 (9) The forces on the moving pulley are the gravitational force mP g 0 (the pulley is assumed massless); string B pulls down on the pulley on each side with a force, TP ,B , which has magnitude TB . String A holds the pulley up with a force TP , A with the magnitude TA equal to the tension in string A . The free body diagram for the forces acting on the moving pulley are shown below. Newton’s second Law applied to the pulley is ˆj : 2T T m a 0 . B A P y, P (10) Since the pulley is massless we can use this last equation to determine the condition that the tension in the two strings must satisfy, 2TB TA (11) We are now in position to determine the accelerations of the blocks and the tension in the two strings. We record the relevant equations as a summary. 0 a y ,2 a y ,3 2a y ,1 (12) m1 g TA m1 a y ,1 (13) m2 g TB m2 a y ,2 (14) m3 g TB m3 a y ,3 (15) 2TB TA . (16) Optional d) Solve for the accelerations of the objects and the tensions in the ropes. There are five equations with five unknowns, so we can solve this system. We shall first use Equation (16) to eliminate the tension TA in Equation (13), yielding m1 g 2TB m1 a y ,1 . (17) We now solve Equations (14), (15) and (17) for the accelerations, a y ,2 g TB m2 TB m3 2T a y ,1 g B . m1 a y ,3 g (18) (19) (20) We now substitute these results for the accelerations into the constraint equation, Equation (12), 0g 1 TB T 4T 1 4 g B 2 g B 4 g TB . m2 m3 m1 m2 m3 m1 (21) We can now solve this last equation for the tension in string B , TB 4g 4 g m1 m2 m3 . 1 1 4 m1 m3 m1 m2 4 m2 m3 m2 m3 m1 (22) From Equation (16), the tension in string A is TA 2TB 8 g m1 m2 m3 . m1 m3 m1 m2 4 m2 m3 (23) We find the acceleration of block 1 from Equation (20), using Equation (22) for the tension in string B, a y ,1 g 2TB 8 g m2 m3 m m m1 m2 4 m2 m3 . g g 1 3 m1 m1 m3 m1 m2 4 m2 m3 m1 m3 m1 m2 4 m2 m3 (24) We find the acceleration of block 2 from Equation (18), using Equation (22) for the tension in string B, a y ,2 g TB 4 g m1 m3 3 m1 m3 m1 m2 4 m2 m3 g g . m2 m1 m3 m1 m2 4 m2 m3 m1 m3 m1 m2 4 m2 m3 (25) Similarly, we find the acceleration of block 3 from Equation (19), using Equation (22) for the tension in string B, a y ,3 g m m 3 m1 m2 4 m2 m3 TB 4 g m1 m2 . g g 1 3 m3 m1 m3 m1 m2 4 m2 m3 m1 m3 m1 m2 4 m2 m3 As a check on our algebra we note that 2 a1, y a2, y a3, y 2g m1 m3 m1 m2 4 m2 m3 3 m1 m3 m1 m2 4 m2 m3 m m 3 m1 m2 4 m2 m3 g g 1 3 m1 m3 m1 m2 4 m2 m3 m1 m3 m1 m2 4 m2 m3 m1 m3 m1 m2 4 m2 m3 0. (26)