Tension problem and concept

[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.

[e(ho0n3]

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.

[speeding electron]

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.

[Doc Al]

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.