Question

1) Figure 5.2-1 shows the free body diagrams and the resulting equations of motion that are found by applying Newton's second law of motion to the Atwood's machine. From the equations of motion shown in the figure, derive equations (5.2) and (5.3). (Hint: Think two equations and two unknowns.) 2) Consider an Atwood's machine with m, =(110. 00.1) g and m, = (175. 00.1) g. Determine the acceleration of the masses.

Answer 1

Solution Atwood's Machine llllll Consider two masses m, and me connected by a string Passing over a pulky. (which is massless and i friction less) let the acceleration of masses lee 'al as shown - Free Body Diagram YON Los тя е tension in the Stering, Free body diagr -ame of the bodies are 21 ma ma Now, for mass mit Fret = mia = T-mig - 0 for mass m2 Fret = maa = mag-T-@ adding equation ☺ and (mitme) a = mag-mig

= mz- Putting in een ☺ we get m. la toma de - T-mig T = my (onts) - mig T= mig (m. in the town) T= mig x 2m2 9 X 2 m2 m, tm2 - T = 2mimag mitm © m. = (10.0 Fog. me = (175.0 500g Now absoluted error in ma-m, = 0.2 absolute beror in mitma= 0.2 Tome value of acceleration sa = (175 -110)x9.88 (175+110)

a= 637 2 - 2.24 me? Teme value = 2.24m/8² sulatus error in oteta acceleration = 0.2 + others 65 49 = 0.0031 + 0.0007 -0.0038 o absolute error in a = 0.0038 X 2.24 = 0.0085 & acceleration = (2-24 + 0.38%) m/s2 or =(12.24 + 0.0085) w/8?