calculate the shortest length of cable which can be used to support a load of 1000N if tension in the cable is not to exceed 710N?
Take AC=BC
There must be a figure that goes with this. Is this cable "horizontal" with a load in the center?
To calculate the shortest length of cable needed to support a load of 1000N with a maximum tension of 710N, we can use the concept of the right triangle.
Let's assume that point A represents the location where the cable is attached to the load, and point C represents the point where the cable is attached to a support point. We are given that AC = BC.
In this case, we can consider the load of 1000N as the hypotenuse of a right triangle, where the tension in the cable is one of the sides.
By using the Pythagorean theorem, we can calculate the tension (TB) by rearranging the equation to solve for TB:
TB^2 = AC^2 + BC^2
Since AC = BC, the equation simplifies to:
TB^2 = 2 * AC^2
TB = √(2 * AC^2)
Now, we know that the maximum tension in the cable should not exceed 710N. So we can set up the inequality:
TB ≤ 710N
√(2 * AC^2) ≤ 710N
Now, we can solve for AC:
AC^2 ≤ (710N)^2 / 2
AC ≤ √((710N)^2 / 2)
AC ≤ √(710N^2) / √2
AC ≤ 710N / √2
AC ≤ 710N * (√2 / 2)
Therefore, the shortest length of cable needed to support a load of 1000N, with a maximum tension of 710N, is 710N * (√2 / 2).