Hanging Weight

Problem: Three cords attached to a ceiling support an 80 kg mass as shown. The points of support form a right triangle. Point P is 80 cm below the ceiling. Determine the tensions in the cords.

Solution: The free body diagram is shown as part of the picture, and is taken about point P. It is not difficult to write a vector for the weight:

But a vector for each tension requires this procedure:

1. Write a position vector for the direction of the force (e.g., from P to A)

2. Find the magnitude of the position vector

3. Divide the position vector by its magnitude to get a unit vector in the direction

4. Multiply the unknown magnitude of the tension by the unit vector

Step 1. To write a position vector from P to A, think of using your fingers to "walk there". You'd have to travel 0.2 m in the x-direction, then 0.15 m in the negative y-direction, and then 0.8 m in the z-direction:

Step 2. To get its magnitude, I use the 3-d Pythagorean Theorem:

Step 3. Dividing to get a unit vector, you can see that the units divide out:

Step 4. Therefore, the force in cord #1 is:

You can do the work for the other two forces; the results are:

Since the sum of the vector forces must be the null vector, we write

At this point, I could substitute each of the forces into this equation to get a rather long expression. What comes next? This expression is the null vector if the independent portions - the x, y, and z components - each sum to be a null vector. So I simply look through each vector and sum the coefficients (the scalar parts):

x: T₁(0.239) - T₂ (0.122) - T₃ (0.116) = 0

y: -T₁(0.179) - T₂ (0.183) + T₃ (0.349) = 0

z: T₁(0.954) + T₂ (0.976) + T₃ (0.930) = (80)(9.8)

Now I use Excel and its matrix functions to solve this:

1. Find the inverse matrix of the coefficients

2. Multiply the inverse matrix by the column matrix of constants to get the solution

Here are the results as they appear on the spreadsheet:

matrix of coefficients:	0.23862	-0.12194	-0.11625	column of constants:	0.0
	-0.17897	-0.18291	0.348743		0.0
	0.95448	0.975537	0.929981		784

inverse matrix:	2.793843	2.64E-06	0.349231	solution matrix: T₁	273.7968
	-2.73354	-1.82236	0.341692	T₂	267.8866
	9.8E-07	1.911628	0.35843	T₃	281.0092

A review on how to use Excel...

The second condition of equilibrium - the sum of the torques - was not needed, because the free body was about a point and a point cannot support torques. (Only extended bodies can have torques, since a torque requires a moment arm.)