Tension Problem

The weight of a crate is 566 newtons. Find the tension in each of the supporting cables shown in the figure.

The Newton is the unit used for Force and is equivalent to the force required to give a mass of 1 kg an acceleration of 1 ms^-2. We are going to calculate the tension force in newtons that exists in each supporting cable. We can begin this problem by imagining the weight of the box as a force exerted downward and coinciding with the z-axis. This can be written in component form as W=<0,0,-115>. The end point where all of the cables connect occurs at A=(0,0,-115). If we take the point, (0,0,0), at the origin we can imagine it to exist on the horizontal plane which represents the surface that the cables are suspended from. We can then subtract the coordinates of A from this point to yield a weight vector in component form. Next we can create coordinate points which represent where each cable is connected to the horizontal plane. This is done using their distances from the x and y axis, and yields the three points: B-(0,70,0), C-(-60, 0, 0), and D-(45, -65, 0). Next, we can subtract the x, y, and z coordinates of A from each coordinate of each connecting point to obtain the line segments AB=<0,70,115>, AC=<-60, 0, 115>, and AD=<45, -65, 115>. Each are in component form so that they are dissociated from a particular place in space as is the weight force vector.

Now we can divide AB by the magnitude of ||AB|| which will yield a unit vector in the direction of B. Then all that is missing from the unit vector is the magnitude which gives us the length of the supporting cable. After repeating this procedure for vectors C and D  we obtain the following, where × signifies the multiplicative operator and not the cross product: Vector B = ||B|| ×(<0,70,115>)/(25√(29)), Vector C= ||C|| × (<-60,0,115>/(5√673), and Vector=||D||×(<45, -65, 115>)/(5√779). 

Now in order for the system to be in Mechanical Equilibrium the net sum of Force vectors must be equivalent to 0 which means B+C+D+W= 0, or B+C+D=-W. With this parameter in mind we can map the vectors into an augmented matrix and solve for each magnitude of each cable to obtain the amount of tension in each cable, which of course depends on the length of the cable in this context. The matrix looks will look like this:


After using a calculator to go through the steps of Gaussian Elimination and putting the matrix into Row Reduced Echelon Form, we find that the tension in AB is ≈ 229.704 Newtons, AC≈178.753 Newtons, and the tension in AD ≈256.421 Newtons


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