Tension
A common form of tension is mental or emotional strain. But that's not what this site is about so if your looking for a way to deal with your tension, go away.
The tension I'm refering to deals with physics. The specific problem I'm refering is when you have an object suspended by 2 lines. Here is an example:
Let's say you have a problem as shown above. The object being suspended is 50N. The line connected to it is T3. This in turn is head up by both T1 and T2. Since the force pulling the object downis 50N, to hold it up, the force of T3 must also be 50N. You also have the force of T2, which is 80N. To find the force of T1, take T2 and make a right triangle out of it as shown. The angle between side b and T2 is, will say 40 degrees. Side a is vertical like T3 and has the same value, which is 50N. T1 will be the same as side b. The formula to solve this is simple. It is:
a^2 + b^2 = c^2
Put in the numbers from the problem and you have:
50^2 + b^2 = 80^2
2500 + b^2 = 6400
Now subtract 2500 from both sides to get:
b^2 = 3900
b = 62.44997998
Since T1 is the same as b, T1 is about 62.45N.
The other type of tension problem looks very similar to the first, but there is enough of a difference that it must be solved almost entireley differently. It also has T3 attached to two other lines. However, in the first problem, one line (T1) was horizontal and the other was at an angle. In this problem, both T1 and T2 are at an angle. Here's an example:
In this problem, the object is 200N. In order to hold that, T3 must once again have an upward force equal to the force pulling the object down. T3 is attached to both T1 and T2, but this time both are at an angle and you are only given the value of T3. This problem takes more work than the other one, but you have all the information you need. The first formula you need is:
(T1 * cos a) + (T2 * cos b) = 0
Plug the numbers into the problem and you have:
(T1 * cos 15) + (T2 * cos 20) = 0
Figure out the cosine of each angle and you have:
(T1 * 0.97) - (T2 * 0.94) = 0
Now add (T2 * 0.0.94) to both sides and you should have:
(T1 * 0.97) = (T2 * 0.94)
Next, divide both sides by 0.94 to get:
T1 * 1.03 = T2
The next formula you need to finish this problem is:
(T1 * sin a) + (T2 * sin b) - Fw = 0
Plug the numbers into the problem and you have:
(T1 * sin 15) + (T2 * sin 20) - 200 = 0
Figure out the sine of each angle and you get:
(T1 * 0.26) + (T2 * 0.34) - 200 = 0
You also have the answer from the other last formula, which brings us to:
(T1 * 0.26) + ((T1 * 1.03) * 0.34) - 200 = 0
Solve everything in the parentheses to get:
(T1 * 0.26) + (T1 * 0.35) - 200 = 0
Add both T1's to get:
T1 * 0.61 - 200 = 0
Add 200 hundred to both sides to get:
T1 * 0.61 = 200
Finally, divide both sides by 0.61 to get T1
T1 = 328N
At this point, solving T2 is easy. Remember that the answer from using the first formula was:
T1 * 1.03 = T2
You now know what T1 is, so that leaves you with:
328 * 1.03 = T2
That information boils down to:
338N = T2
Once solved, the entire problem should look like:
Have a nice day.
