Two Masses One on Table, One Hanging

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AI Thread Summary
The discussion revolves around a physics problem involving two masses, one on a table and one hanging, connected by a string. The key equations used include F=ma and the frictional force equation, with a coefficient of friction of 3.00 between the block and the table. Participants clarify that the tension in the string is influenced by both masses, and both blocks will accelerate at the same rate due to the inextensibility of the string. Newton's third law is applied to explain the interaction between the two blocks, emphasizing that the hanging block exerts a pull on the block on the table and vice versa. Overall, the focus is on understanding the forces acting on each mass and their resulting acceleration.
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Homework Statement



A block of mass1 5.0 kg is on a table that bonds a coefficient of friction of 3.00. Mass1 is attached to Mass2 by a string, and Mass2 is hanging over the table. The system is released from rest for it to accelerate.

Homework Equations



F=ma
F fric = μFnormal

The Attempt at a Solution



\sum Fx of Mass1 = F tension - F friction

\sum Fy of Mass1 = 0

\sum Fx of Mass2 = 0

\sum Fy of Mass2 = FW - F tension

Is tension created by the mass1 or mass2?
im guessing both? but that doesn't make any sense...
 
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1. Where is m3?
2. Is coefficient of friction of 3.00 is between the block and table?
 
azizlwl said:
1. Where is m3?
2. Is coefficient of friction of 3.00 is between the block and table?

no m3...

2. yes
 
ok wrong interpretation.

Find net force on each block.
Both will accelerate at equal rate for the string is massless and inextensible.
 
Last edited:
oops one on table, one hanging
 
Use Newton's 3rd law.
The hanging block will pull the block on the table.
The block on the table will pull the hanging block with equal force.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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