Pulley and block system tension & acceleration

[heyrefusuck]

For the following: 2kg. hanging mass and 8 kg. tabled mass are connected by a massless string passing over a frictionless pulley with a mass of 6kg. The 2kg. mass moves vertically and the 8kg. mass moves on a horizontal frictionless surface. The moment of I of the pulley is 1/2MR^2. The radius of the pulley is 0.1m

Q1. What is the accel. of the 8 kg. mass on the table. My answer was that since the pulley system is frictionless then the tension on both masses will be equal. Therefore there will be no resultant force causing an acceleration on the 8kg. mass on the table. Is that right?

Q2. What is the tension on the string connected to the 2kg. mass. Answer: The formula that I presume I use is T = (2m1m2 /m1 + m2) 9.8
Am I on the right track? What would be the final unit in the amount of tension?

Thanks!

 

[Doc Al]

Q1. What is the accel. of the 8 kg. mass on the table. My answer was that since the pulley system is frictionless then the tension on both masses will be equal. Therefore there will be no resultant force causing an acceleration on the 8kg. mass on the table. Is that right?

No.
(1) The tension is not the same throughout the string, since the pulley has mass.
(2) Even if the tension was equal on each mass, that would not imply no net force on the 8 kg mass. (What other horizontal forces act on that mass?)

Q2. What is the tension on the string connected to the 2kg. mass. Answer: The formula that I presume I use is T = (2m1m2 /m1 + m2) 9.8

Rather than use some canned formula, derive your own starting with basic principles. Analyze the forces on the two masses and the pulley and apply Newton's 2nd law to each. You'll end up with three equations and three unknowns.

Am I on the right track? What would be the final unit in the amount of tension?

Tension, like any other force, will have units of Newtons.