Solving Homework Problem on \alpha = 37, m1 = 10 kg

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Homework Help Overview

The discussion revolves around a physics problem involving a pulley system with two masses, m1 and m2, where m1 is given as 10 kg and the angle α is 37 degrees. The participants are tasked with finding the relationship between the two masses and their accelerations under different conditions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to derive equations related to the forces acting on the masses and questions the relationship between tensions in the ropes. Participants discuss the implications of the system being at rest and the conditions for acceleration.

Discussion Status

Participants are actively engaging with the problem, raising questions about the relationships between the variables involved. Some guidance has been offered regarding the relationship between the accelerations of m1 and m2, but no consensus has been reached on the specifics of these relationships.

Contextual Notes

There is a mention of the system being at rest for the first part of the problem, which influences the equations being considered. Additionally, participants are exploring the implications of the constant length of the rope in the pulley system.

asi123
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Homework Statement



[tex]\alpha[/tex] = 37, m1 = 10 kg
Also attachment.

Homework Equations





The Attempt at a Solution



ok, the first question is to find m2 when the system is in rest and the second one is to find the acceleration when m1 = m2.
I found 3 equations:
n1 = m1 * g * cos([tex]\alpha[/tex] )
m1 * g * sin([tex]\alpha[/tex] ) = T2
T1 + T2 = m2 * g
I need another one I think...

Could it be that T1 = T2?
And as for the second one, how can I find the equations between the accelerations? (there is always one...)

10x in advance.
 

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Yes, the tensions of lengths of rope connected by frictionless pulleys have the same tension. But in your Newton's law F=ma equations, I only see F parts, I don't see any ma parts. You don't even have an acceleration to solve for...
 
Dick said:
Yes, the tensions of lengths of rope connected by frictionless pulleys have the same tension. But in your Newton's law F=ma equations, I only see F parts, I don't see any ma parts. You don't even have an acceleration to solve for...

Yeah, because the first set of questions is for acceleration = 0 (the system is in rest).
Here are the equations for the second question:

T - m1 * g * sin([tex]\alpha[/tex]) = m1 * a1
m2 * g - 2 * T = m2 * a2

I need the accelerations equation and I don't know how to come up with that...

10x.
 
Good point, a=0 for the first part. Sorry, I missed that. But for the second part there is another relation between a1 and a2. Consider the length of rope going through each part of the pulley system. Total length of rope is constant. One acceleration is twice the other. Which is which?
 
Dick said:
Good point, a=0 for the first part. Sorry, I missed that. But for the second part there is another relation between a1 and a2. Consider the length of rope going through each part of the pulley system. Total length of rope is constant. One acceleration is twice the other. Which is which?

I have no idea...
I would like to think m2 is faster cause the g force is stronger with him :smile: But I'm not sure.
I didn't get you, how can I see it through the length of the rope?

10x.
 
If m1 descends 1 meter down the ramp, how much does m2 rise? You can tell just by looking at the length of the rope. The total length of the rope on m2's side of the pulley decreases by 1 meter.
 

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