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These problems are modified versions of the Atwood machine.
They are best approached looking at free body diagrams (FBDs) of each component and then applying F=m*a
When the pulley is massless and frictionless, there are some simplifications to the problem.
First, the tension in the cord is the same everywhere
(there is a special case where that is not true. I won't go into it here since it may confuse. If you want to know more about the case of "slack cord", send me an email).
Second, the acceleration of the masses are equal.
Okay, I like to start with a FBD of the hanging mass. Since it is much more massive than the one on the slope, I will make the assumption that the mass will descend when released from rest if the tension is sufficient to break static friction on the slope. So positive will be in that direction of motion
Now let's look at the mass on the slope.
The forces parallel to the slope are
T, the tension in the cord
10*9.81*sin(30), the force of gravity parallel to the slope
10*9.81*cos(30)*µ, this is the force of friction
now let's set up F=m*a and test to see if the system will accelerate.
solve for a
let's test to see if we get a>0 when µ=0.6
a=6.51 so it will move
Once moving, µ=0.5, so the actual
Just for fun, let's find T from the FBD of the hanging mass
Check by plugging T and a into the FBD equation for
the mass on the incline
You will get the two sides equal if all of the math and algebra are done right. I get 66.5=66.5