Atwood Machine Friction: 10kg & 20kg Mass

[AirJersey]

1. ROUNDDISK
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(10kg mass) Mass M
Friction of the disk: Us=Uk=0.2 between rope and disk.

Excuse my drawing. Basically it's an atwood machine with a 10kg on one side and M on the other. What is the largest and smallest M that can have the system motionless? If M = 20kg what is its acceleration?
I got 104/9.6=10.83 for max and 96/10.4=9.23 for min? And a = 2.933?

 

[Jhero]

I would like to clarify a couple of points in your post. First, the term "atwood machine" typically refers to a pulley system with two masses hanging on opposite sides of a pulley. In this case, we have a round disk with a mass on one side and an unknown mass on the other side. So, while the principles of an atwood machine may apply, it is not technically an atwood machine.

Secondly, the equation you provided for calculating the maximum and minimum values of M is not clear. Can you please provide the equation you used and explain the variables?

Assuming that the disk is on a horizontal surface and the rope is parallel to the surface, we can use the equation F = μN to calculate the maximum and minimum values of M. In this case, the normal force (N) is equal to the weight of the disk (10kg) and the friction force (F) is equal to the weight of the unknown mass (M). So, the equation becomes M = μN, where μ is the coefficient of friction.

For the maximum value, we can use the coefficient of static friction (μs = 0.2) to calculate the maximum value of M as M = 0.2(10kg) = 2kg. This means that if the unknown mass is greater than 2kg, the system will start to move.

For the minimum value, we can use the coefficient of kinetic friction (μk = 0.2) to calculate the minimum value of M as M = 0.2(10kg) = 2kg. This means that if the unknown mass is less than 2kg, the system will not be able to move.

Now, if we assume that M = 20kg, we can use the equation F = ma to calculate the acceleration of the system. The net force on the system is equal to the weight of the 10kg mass (98N) minus the friction force (20kg x 9.8m/s^2 x 0.2 = 39.2N). So, the net force is 98N - 39.2N = 58.8N. Using F = ma, we can calculate the acceleration as a = F/m = 58.8N/20kg = 2.94m/s^2.

I hope this helps clarify your questions. It's always