Multiple Pulleys and Masses with Frictionby Wara Tags: acceleration, force of friction, tension, two mass, two pulleys 

#1
Mar1912, 03:12 PM

P: 9

1. The problem statement, all variables and given/known data
I have solved the question but I am still iffy as to whether my solution is correct or not. 2. Relevant equations Let m_{1} = 10kg. Let m_{2} = 20kg. (1) m_{1}a_{1} = T_{1}  μm_{1}g (2) m_{2}a_{a} = m_{2}g  T_{2} (3) 2T_{1} = T_{2} (4) 2a_{2} = a_{1} 3. The attempt at a solution T_{2} = m_{2}g T_{2} = (20kg)(9.8m/s^{2}) T_{2} = 196N T_{1} = [itex]\frac{T_{2} + μm_{1}g}{2}[/itex] T_{1} = [itex]\frac{196N + (0.32)(10kg)(9.8m/s^{2})}{2}[/itex] T_{1} = 113.68N Sub (3) and (4) into (2) m_{2}a_{2} = m_{2}g  T_{2} m_{2}([itex]\frac{a_{1}}{2}[/itex]) = m_{2}g  2T_{1} Called this new equation (5). (5) m_{2}([itex]\frac{a_{1}}{2}[/itex]) = m_{2}g  2T_{1} (1) m_{1}a_{1} = T_{1}  μm_{1}g (5) m_{2}([itex]\frac{a_{1}}{2}[/itex]) = m_{2}g  2T_{1} (1) + (5) m_{1}a_{1} + m_{2}([itex]\frac{a_{1}}{2}[/itex]) = T_{1}  μm_{1}g + m_{2}g  2T_{1} a_{1}(m_{1} + m_{2}([itex]\frac{1}{2}[/itex]) = μm_{1}g + m_{2}g  T_{1} a_{1}(10kg + 20kg([itex]\frac{1}{2}[/itex]) = (0.32)(10kg)(9.8m/s^{2})+(20kg)(9.8m/s^{2})113.68N a_{1} = 2.5448 m/s^{2} from (4) 2a_{2} = a_{1} a_{2} = [itex]\frac{a_{1}}{2}[/itex] a_{2} = [itex]\frac{2.548m/s^{2}}{2}[/itex] a_{2} = 1.274 m/s^{2} Therefore, T_{1} = 113.68N, T_{2} = 196N, a_{1} = 2.548 m/s^{2}, and a_{2} = 1.274 m/s^{2}. My question is if my solution is correct. Please check and reply back with suggestions or corrections. Thank you so much. 



#2
Mar1912, 07:53 PM

P: 9

Please help me with this question. have i done this question correctly?



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