Calculating Tension in Wires Supporting a Lithograph

[mikefitz]

Homework Statement

2. A 45-N lithograph is supported by two wires. First wire makes a 15 degrees angle with the vertical and the second one makes a 25 degrees angle with the vertical. Find tension in the second wire.

I already have the answer to this problem, what I would like is some clarification on how tension is calculated. My book tells me I need to split tension into x and y; for example- x:-T1x+T2x=0 , y:T1y+T2y-mg=0


I understand that since the net Force on the system is zero that T added to mg is set to zero. Part of the solution goes something like this: (T2 ( sin25/sin15)) * cos15 + T2 cos25=mg

I am confused as to 1. why you split tension into x and y - I thought tension only acted in one direction, and 2. how does using sin and cos help in finding the answer? I guess if you could explain this to me I would understand how tension works better; thanks.

 

[Ertosthnes]

Nope, tension acts in both directions. The wires are pulling the lithograph upward (since they are supporting its weight), and since the wires are at angles to the vertical, they are also pulling the lithograph in slightly different directions on the x-axis. If you cut one of the wires, for instance, the lithograph would not only swing downward, but also across.

As for using sine and cosine; basic trigonometry. Say you want to find the horizontal components of the tension. Well, if you draw a right angle triangle:

sine(theta) = opposite/hypothenuse, and

hypothenuse*sine(theta) = opposite

In this case, theta and the tension of the hypothenuse are both given, so you can simply plug those values into obtain the tension of the opposite side (the pull of the wire along the x-axis).

 

[PhanthomJay]

Homework Statement

2. A 45-N lithograph is supported by two wires. First wire makes a 15 degrees angle with the vertical and the second one makes a 25 degrees angle with the vertical. Find tension in the second wire.

I already have the answer to this problem, what I would like is some clarification on how tension is calculated. My book tells me I need to split tension into x and y; for example- x:-T1x+T2x=0 , y:T1y+T2y-mg=0


I understand that since the net resultant[/color] Force on the system is zero that the sum of the vertical components of the two tension forces is equal to mg. What must be the sum of the horizontal components of the two tension forces?[/color] . Part of the solution goes something like this: (T2 ( sin25/sin15)) * cos15 + T2 cos25=mg

I am confused as to 1. why you split tension into x and y - I thought tension only acted in one direction,tension always acts in one direction, pulling away from the object in the direction of the wire...this gives it both a vertical and horizontal component, T1x, T1y, etc.[/color] and 2. how does using sin and cos help in finding the answer? I guess if you could explain this to me I would understand how tension works better; thanks.

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