Two Masses, a Pully, and an Inclinde Plane

AI Thread Summary
The discussion revolves around a physics problem involving two blocks connected by a pulley, with one block on an inclined plane experiencing friction. The user is attempting to solve for the mass of block 2 (m2) using the equation m2a = T - f - m2gsin(30), where T is the tension, f is the frictional force, and g is the acceleration due to gravity. The user is confused about how to isolate m2 since it appears on both sides of the equation and is unsure about the correct values to substitute. Clarifications about the tension value and the application of Newton's second law for both masses are suggested to aid in solving the problem. The discussion emphasizes the need for careful algebraic manipulation to find the correct mass of block 2.
Robdog
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Homework Statement



Block 1, of mass m1 = 0.700kg , is connected over an ideal (massless and frictionless) pulley to block 2, of mass m2. Assume that the blocks accelerate downword with an acceleration of magnitude = 0.250m/s^2 and that the coefficient of kinetic friction between block 2 and the plane is u = 0.200.

Tried to attach a good picture of the angle at which the block is 30

Homework Equations



m2a=T-f-m2gsin(30)

With T=6.69
F=un Which n=8.5 so then it would be 1.7

The Attempt at a Solution



well to set it up I am not sure if I am right but i tried to sub all know values into the above eq

M2(2.50m/s^2)=6.69-1.7-M2(9.8m/s^2)Sin(30)

i keep geting wrong answer + i don't konw how to solve when M2 is on both sides? What do i do. I know the answer has to be smaller than .700kg because if it were larger they would not move.
 

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  • MLD_2l_2_v2_2_a.jpg
    MLD_2l_2_v2_2_a.jpg
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any idea?
 
Robdog said:
m2a=T-f-m2gsin(30)
Good. That's Newton's 2nd law applied to mass 2. Do the same for mass 1.

With T=6.69
How do you know the tension?
 
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