- #1
Mcmenhweilleisi
- 1
- 0
- Homework Statement
- Find
Tension T1
Tension T2
Tension T3
- Relevant Equations
- Obtain on the left weigs 40N and on the right 50N
Equilibrium with the string in the center exactly horizontal
- Thread starter Mcmenhweilleisi
- Start date
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Homework Help Overview
The discussion revolves around a physics problem involving equilibrium conditions for a system with a string positioned horizontally in the center. Participants are exploring the application of force and moment equations to analyze the tensions in the strings and the angles involved.
Discussion Character
- Exploratory, Assumption checking, Mathematical reasoning
Approaches and Questions Raised
- Participants discuss applying equilibrium conditions, including summing forces and moments. There is mention of using free body diagrams (FBD) to analyze the forces acting on knots in the system. Some participants express uncertainty about the concept of moments in relation to strings and question the number of equations and unknowns involved.
Discussion Status
The discussion is active, with participants offering different perspectives on how to approach the problem. Some guidance has been provided regarding the use of equilibrium equations, and there is acknowledgment of the equivalence of different methods being discussed. However, there is no explicit consensus on the best approach yet.
Contextual Notes
Participants are reminded to show their work for further assistance, indicating a focus on individual attempts at solving the problem. There is an emphasis on adhering to homework guidelines, which may influence the nature of the responses.
- #2
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For homework help you have to show us your best attempt at solving the problem yourself. See the guidelines:
https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/
- #3
Delta2
Homework Helper
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A few things to get you started:
Consider the piece of string in the middle and apply the equilibrium conditions for it. So it will be
$$\sum F_x=0$$$$\sum F_y=0$$ $$\sum M=0$$
So you will have 3 equations and 3 unknowns (##T_1,T_{3x} , T_{3y}##. You can find angle by ##\tan\theta=\frac{T_{3x}}{T_{3y}}##.
I think that ##T_2## will be equal to ##T_{3x}## or (##T_{1x}## ) not very sure about this.
Consider the piece of string in the middle and apply the equilibrium conditions for it. So it will be
$$\sum F_x=0$$$$\sum F_y=0$$ $$\sum M=0$$
So you will have 3 equations and 3 unknowns (##T_1,T_{3x} , T_{3y}##. You can find angle by ##\tan\theta=\frac{T_{3x}}{T_{3y}}##.
I think that ##T_2## will be equal to ##T_{3x}## or (##T_{1x}## ) not very sure about this.
Last edited:
- #4
Frabjous
Gold Member
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Shouldn’t you just sum the forces on each side of the second string. 4 eqs and 4 unknowns (T1, T2, T3, θ). I have no idea what a moment means for a string
- #5
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Actually, the 4 equations are decoupled. One can find T1 and T2 from the FBD of the knot on the left and then find the rest from the FBD of the knot on the right.caz said:
@Mcmenhweilleisi : Please show us your work to get more help.
- #6
Delta2
Homework Helper
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Well you apply the moment equation as the string was some sort of rigid body (since it is taut).caz said:
I think the two ways are equivalent, because the way I see the moment equation it is essentially a force balance equation e.g $$T_{3y}l-50l=0$$ where l the length of the second string.
- #7
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Quite so.Delta2 said:
We have one set of forces acting at one knot, and another set at the other knot.
Writing that the resultant of one set has no moment about the other knot is the same as writing that that resultant has no component normal to the string.