Newtons 3rd Law (Ropes, Pulley)

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PhysicsNoob76
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Homework Statement


A 2.5 kg mass is attached to a 3.5 kg mass by a light string that passes over a friction-less pulley. The masses are released.

a) What is the magnitude of the acceleration of the masses?
b) What is the tension in the string?

Homework Equations



(I think) Ft - mg = ma

The Attempt at a Solution



I drew 2 free body diagrams for both the weights and they both have Fg and Ft on the y.

Handwritten derivation on lined paper: F - mg = ma, then F - (2.5)(9.8) = (2.5)(a), combined with (3.5)(9.8) - F = (3.5)(a), leading to -15a = ... and a = -6.54.


On this 6th line, starting with (2.5)(a)... I made it so that Ft2 = Ft2
My answer was suppose to be 6.54 m/s in the work I showed because the acceleration of the bigger weight will accelerate negatively. So ignore the minus sign for my answer to a =
I think I've done the best I could, but my answer is still wrong, not sure why.
 
Last edited:
The magnitude of the acceleration will be the same, but in opposite directions.
 
lewando said:
The magnitude of the acceleration will be the same, but in opposite directions.

Yeah, I knew that, I should have included that too. I need actual with with the question though.
 
You have 2 equations, 2 unknowns (one needs fixing-- one of the masses should be accelerating upwards-- use "a", the other, downwards-- use "-a"). Solve for a.
 
lewando said:
You have 2 equations, 2 unknowns (one needs fixing-- one of the masses should be accelerating upwards-- use "a", the other, downwards-- use "-a"). Solve for a.

I've done some changes, but I am still getting it wrong...
 
Since the heavier one is accelerating downward, replace "a" in line 2 with "-a". Try again.
 
lewando said:
Since the heavier one is accelerating downward, replace "a" in line 2 with "-a". Try again.

Ohhhhhhh, thank youuuuuuuuuuuuu. I re-did my whole thing and now it works!