The magnitude of the tension in each supporting cable is approximately 70.71 N.
Let T1 be the tension in the left cable and T2 be the tension in the right cable. The weight of the body is 100.0 N.
Horizontal forces:
The horizontal components of the tensions T1 and T2 must balance each other out. Since the angles of the cables are 45 degrees, the horizontal components of the tensions are equal to T1cos(45) and T2cos(45), respectively. Therefore:
T1cos(45) = T2cos(45)
Vertical forces:
The vertical components of the tensions T1 and T2 must add up to the weight of the body. Since the angles of the cables are 45 degrees, the vertical components of the tensions are equal to T1sin(45) and T2sin(45), respectively. Therefore:
T1sin(45) + T2sin(45) = 100.0 N
Solving the equations:
We can now solve the two equations for T1 and T2. One way to do this is to eliminate one of the variables. For example, we can solve the first equation for T2:
T2 = T1
Then we can substitute this expression for T2 in the second equation:
T1sin(45) + T1sin(45) = 100.0 N
2T1sin(45) = 100.0 N
T1*sin(45) = 50.0 N
Finally, we can solve for T1:
T1 = 50.0 N / sin(45)
T1 ≈ 70.71 N
It is important to note that this is an approximate solution because we used rounded values for sin(45) and cos(45). The exact values of the tensions would be slightly different.
Complete Question:
Find the magnitude of the tension in each supporting cable shown below: In each case, the weight of the suspended body is 100.0 N and the masses of the cables are negligible.