- 42,788
- 10,490
If 2T1 - T2=ma and m=0 then...?Ithilrandir said:3. Resultant force = 2T1 - T2
4. If the pulley has negligible mass, F = ma shouldn't be calculable?
The discussion centers on the analysis of an Atwood machine involving two masses, M1 and M2, with M1 weighing 400 grams and M2 weighing 200 grams. The initial calculations for tension and acceleration were incorrect due to the misapplication of forces and failure to account for the geometry of the system, specifically the angle θ and the effects of the pulley. The correct acceleration is determined to be g/9 (1.09 m/s²) and the tension in the system is recalculated to be 2.17782 N after resolving the relationships between the tensions and applying Newton's laws correctly.
PREREQUISITESPhysics students, mechanical engineers, and anyone interested in understanding dynamics and forces in pulley systems.
If 2T1 - T2=ma and m=0 then...?Ithilrandir said:3. Resultant force = 2T1 - T2
4. If the pulley has negligible mass, F = ma shouldn't be calculable?
Yes!Ithilrandir said:2T1 = T2?
Thanks, I solved the answer. I was under the impression that the pulley accelerates with force = ma and trying to solve it abstractly that way.haruspex said:Yes!