What is the equation for equilibrium of a picture suspended by two strings?

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Homework Help Overview

The problem involves a picture suspended by two strings, with given tensions and angles. Participants are exploring the equilibrium conditions of the system, particularly focusing on the relationship between the tensions in the strings and their angles with respect to the horizontal.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between the tensions in the strings and their angles, questioning whether the equation T1cos(65) = T2cos(32) accurately represents the equilibrium condition. There is uncertainty about equating the tensions directly versus their components.

Discussion Status

The discussion is ongoing, with participants providing insights into the vector nature of forces and the conditions for equilibrium. Some guidance has been offered regarding the necessity of considering both x and y components of the forces, but no consensus has been reached on the specific calculations or interpretations.

Contextual Notes

Participants are working under the assumption that the picture is in equilibrium, which implies that the net forces acting on it must sum to zero. There is also mention of gravitational forces acting on the picture, but details on their magnitude or effects are not provided.

grahamfw
Okay here's the problem:

A picture hangs on a wall suspended by two strings. The tension in string 1 is 1.7 N and is at an angle of 65 degrees above horizontal (from left). String 2 is at an angle of 32 degrees above horizontal (from right). The strings connect to the same point that holds the picture.

a.) Is the tension in string 2 greather than, less than, or equal to 1.7 N? Explain.
b.) Verifiy your answer to part a by calculating the tension in string 2.

If I remember correctly, the tensions times the cosine of the angles are equal so:

T1cos(65) = T2cos(32). But I do not understand why. Isn't that just equating the x components of the tension and not the tensions themselves? Maybe I have been misled...

Thanks in advance.
 
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Notice that in the expression T1cos(65) = T2cos(32), you are using the magnitudes of the tensions, i.e. T1 and T2. If you write out what is happening using vectors, you will see that the equation you gave is a direct result of that.
 
You cannot equate the tension forces directly since they are acting in different directions. What you can say is that the picture is in equilibrium, so the algebriac sum of the components of the forces acting along any given axis will equal zero.
 
Last edited:
Equilibrium

grahamfw said:
If I remember correctly, the tensions times the cosine of the angles are equal so:

T1cos(65) = T2cos(32). But I do not understand why. Isn't that just equating the x components of the tension and not the tensions themselves? Maybe I have been misled...
Everything is derived from the fact that the picture is in equilibrium. Which means that the net force (both x and y components) must equal zero. That equation you wrote is just a statement that the x-components of the forces on the picture must add to zero. (To complete this problem you'll also need the equation that comes from setting the sum of the y-components equal to zero.)

Note: the only forces on the picture are the two string tensions and gravity.
 

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