How Does Friction Affect the Motion of Three Blocks?

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The discussion focuses on calculating the mass m3 required for a system of three blocks to move at constant speed, given a coefficient of kinetic friction of 0.560 and specific masses for m1 and m2. The initial equations presented include tensions and friction but contain errors in the application of Newton's second law. It is emphasized that the tensions acting on m1 and m2 are different and should be labeled as T1 and T2, respectively. Participants are encouraged to derive separate equations for each mass and combine them to find the correct values for acceleration and tensions. The clarification on using distinct tension variables is crucial for solving the problem accurately.
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You are given that the coefficient of kinetic friction between the block and the table in the figure is 0.560, and m1 = 0.150 kg and m2 = 0.250 kg.
(a) What should m3 be if the system is to move with a constant speed?

this is what i have:
T2 - T1 - f = m3a

m2g - m1g - ukg = m3
2.45 - 9.65 - 5.488 = 4.508kg

the answer is wrong, what is it that i am doing wrong?
 

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complexc25 said:
this is what i have:
T2 - T1 - f = m3a
This is good.

m2g - m1g - ukg = m3
2.45 - 9.65 - 5.488 = 4.508kg
This is not good. For one thing, the tensions are not equal to the weights.

Hint: Apply Newton's 2nd law to each mass separately. You took care of M3, now come up with equations for M1 & M2.

You'll combine all three equations to solve for the acceleration and the tensions.
 
i don't understand, i have for:
m1
T - m1g = m1a
m2
m2g - T = m2a
 
complexc25 said:
i don't understand, i have for:
m1
T - m1g = m1a
m2
m2g - T = m2a
Looks good, except that the tensions are different. Use T1 for m1 and T2 for m2.
 

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