- #1
Mcmenhweilleisi
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- 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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SUMMARY
The discussion focuses on solving a physics problem involving equilibrium with a string positioned horizontally at the center. Participants emphasize the application of equilibrium conditions, specifically the equations for force and moment: $$\sum F_x=0$$, $$\sum F_y=0$$, and $$\sum M=0$$. The tension forces, denoted as ##T_1, T_2, T_3##, and the angle ##\theta## are critical variables, with the suggestion that the moment equation can be treated similarly to a force balance equation. The conversation highlights the importance of free body diagrams (FBD) for analyzing forces at knots in the system.
PREREQUISITES- Understanding of equilibrium conditions in physics
- Familiarity with free body diagrams (FBD)
- Knowledge of tension forces and their components
- Basic grasp of moment equations in mechanics
- Study the application of equilibrium conditions in static systems
- Learn how to construct and analyze free body diagrams (FBD)
- Explore the relationship between tension forces and angles in equilibrium
- Investigate the concept of moments and their role in mechanical systems
Students in physics, particularly those studying mechanics, as well as educators and tutors seeking to enhance their understanding of equilibrium problems involving strings and forces.
- #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.