MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics. W02D3_0 Group Problem: Pulleys and Ropes Constraint Conditions

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1 MSSCHUSES INSIUE OF ECHNOLOGY Deprtment of hysics 8.0 W02D3_0 Group roblem: ulleys nd Ropes Constrint Conditions Consider the rrngement of pulleys nd blocks shown in the figure. he pulleys re ssumed mssless nd frictionless nd the connecting strings re mssless nd unstretchble. Denote the respective msses of the blocks s m, m 2 nd m 3. he upper pulley in the figure is free to rotte but its center of mss does not move. oth pulleys hve the sme rdius R. ) How re the ccelertions of the objects relted? b) Drw force digrms on ech moving object. c) Wht re Newton s equtions of motion for the system of pulleys nd objects? Solution: Choose n origin t the center of the upper pulley. Introduce coordinte functions for the three moving blocks, y, y 2 nd y 3. Introduce coordinte function y for the moving pulley (the pulley on the lower right in the figure). Choose downwrd for positive direction; the coordinte system is shown in the figure below then.

2 ) he length of string is given by l = y + y +! R () where! R is the rclength tht the rope is in contct with the pulley. his length is constnt, nd so the second derivtive with respect to time is zero, d l d y d y = = + = y, + y,. (2) dt dt dt hus block nd the moving pulley s components of ccelertion re equl in mgnitude but opposite in sign, =!. (3) y, y, he length of string is given by l = ( y " y ) + ( y " y ) +! R = y + y " 2y +! R (4) where! R is the rclength tht the rope is in contct with the pulley. his length is lso constnt so the second derivtive with respect to time is zero, d l d y2 d y3 d y 0 = = +! 2 = y,2 + y,3! 2y,. (5) dt dt dt dt We cn substitute Eqution (3) for the pulley ccelertion into Eqution (5) yielding the constrint reltion between the components of the ccelertion of the three blocks, b) Free ody Force digrms: 0 = (6) y,2 y,3 y, he forces cting on block re: the grvittionl force mg! nd the pulling force!,r of the string cting on the block. Since the string is ssumed to be mssless nd the pulley

3 is ssumed to be mssless nd frictionless, the tension in the string is uniform nd equl in mgnitude to the pulling force of the string on the block. he free body digrm is shown below. Newton s Second Lw pplied to block is then ˆ j : mg! = m y,. (7) he forces on the block 2 re the grvittionl force m2g! nd the string holding the block,! 2,r re shown below., with mgnitude. he free body digrm for the forces cting on block 2 Newton s second Lw pplied to block 2 is ˆ j : m2g! = m 2 y,2. (8) he forces on the block 3 re the grvittionl force m3g! nd the string holding the block,! 3,r re shown below., with mgnitude. he free body digrm for the forces cting on block 3 Newton s second Lw pplied to block 3 is

4 ˆ j : m3g! = m 3 y,3. (9)! he forces on the moving pulley re the grvittionl force m g = 0! (the pulley is ssumed mssless); string pulls down on the pulley on ech side with force,!,, which hs mgnitude. String holds the pulley up with force!, with the mgnitude equl to the tension in string. he free body digrm for the forces cting on the moving pulley re shown below. Newton s second Lw pplied to the pulley is ˆ j : 2! = m = 0. (0) y, Since the pulley is mssless we cn use this lst eqution to determine the condition tht the tension in the two strings must stisfy, 2 = () We re now in position to determine the ccelertions of the blocks nd the tension in the two strings. We record the relevnt equtions s summry. 0 = (2) y,2 y,3 y, m g! = m (3) y, m g! = m (4) 2 2 y,2 m g! = m (5) 3 3 y,3 2 =. (6) Optionl d) Solve for the ccelertions of the objects nd the tensions in the ropes. here re five equtions with five unknowns, so we cn solve this system. We shll first use Eqution (6) to eliminte the tension in Eqution (3), yielding

5 m g! = m. (7) 2 y, We now solve Equtions (4), (5) nd (7) for the ccelertions, y,2 y,3 y, m = g! (8) 2 = g! m3 (9) 2 = g!. m (20) We now substitute these results for the ccelertions into the constrint eqution, Eqution (2), 4! 4 " 0 = g # + g # + 2g # = 4g # $ + + %. (2) m2 m3 m & m2 m3 m ' We cn now solve this lst eqution for the tension in string, 4g 4g m m2 m3 = =! 4 " m m + m m + 4m m # + + $ m m m % 2 3 & (22) From Eqution (6), the tension in string is 8g m m2 m3 = 2 =. (23) m m + m m + 4m m We find the ccelertion of block from Eqution (20), using Eqution (22) for the tension in string, 2 8g m m m m + m m! 4m m y, = g! = g! = g m m m + m m + 4m m m m + m m + 4m m (24) We find the ccelertion of block 2 from Eqution (8), using Eqution (22) for the tension in string, 4g m m3! 3m m3 + m m2 + 4m2 m3 y,2 = g! = g! = g. (25) m m m + m m + 4m m m m + m m + 4m m

6 Similrly, we find the ccelertion of block 3 from Eqution (9), using Eqution (22) for the tension in string, 4 g m m m m! 3m m + 4m m y,3 = g! = g! = g m3 m m3 + m m2 + 4m2 m3 m m3 + m m2 + 4m2 m3. (26) s check on our lgebr we note tht =, y 2, y 3, y m m + m m! 4m m! 3m m + m m + 4m m m m! 3m m + 4m m 2g + g + g m m + m m + 4m m m m + m m + 4m m m m + m m + 4m m =

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