Determine the resultant couple moment of the two couples that act on the pipe assembly. The distance from A to B is d = 400 mm. Express the result as a Cartesian vector.

#### Solution:

Show me the final answerโ

To solve this problem, we will first express a position vector from A to B. Note that d = 400 mm as stated in the question.

Locations of points A and B are:

A:(0.35i+0j+0k)\,\text{m}

B:(0.35i-0.4\cos30^0+0.4\sin30^0)\,\text{m}

B:(0.35i-0.4\cos30^0+0.4\sin30^0)\,\text{m}

(Simplify the trigonometric values)

B:(0.35i-0.346j+0.2k)\,\text{m}

We can now write a position vector from A to B as follows:

r_{AB}=\left\{(0.35-0.35)i+(-0.346-0)j+(0.2-0)k\right\}\,\text{m}

r_{AB}=\left\{0i-0.346j+0.2k\right\}\,\text{m}

r_{AB}=\left\{0i-0.346j+0.2k\right\}\,\text{m}

Let us now calculate the coupling moments created by the forces applied to the pipe assembly. Looking at the diagram again, take note of the forces already expressed in Cartesian vector form.

Force 1:

M_1=r_{AB}\times F_1

M_1=\begin{bmatrix}\bold i&\bold j&\bold k\\0&-0.346&0.2\\0&0&35\end{bmatrix}

M_1=\left\{-12.11i+0j+0k\right\}\,\text{N}\cdot\text{m}

M_1=\begin{bmatrix}\bold i&\bold j&\bold k\\0&-0.346&0.2\\0&0&35\end{bmatrix}

M_1=\left\{-12.11i+0j+0k\right\}\,\text{N}\cdot\text{m}

Force 2:

M_2=r_{AB}\times F_2

M_2=\begin{bmatrix}\bold i&\bold j&\bold k\\0&-0.346&0.2\\-50&0&0\end{bmatrix}

M_2=\left\{0i-10j-17.3k\right\}\,\text{N}\cdot\text{m}

M_2=\begin{bmatrix}\bold i&\bold j&\bold k\\0&-0.346&0.2\\-50&0&0\end{bmatrix}

M_2=\left\{0i-10j-17.3k\right\}\,\text{N}\cdot\text{m}

The resultant coupling moment is the addition of both of these moments.

M_c=M_1+M_2

M_c=\left\{-12.11i+0j+0k\right\}+\left\{0i-10j-17.3k\right\}

M_c=\left\{-12.11i-10j-17.3k\right\}\,\text{N}\cdot\text{m}

M_c=\left\{-12.11i+0j+0k\right\}+\left\{0i-10j-17.3k\right\}

M_c=\left\{-12.11i-10j-17.3k\right\}\,\text{N}\cdot\text{m}

#### Final Answer:

M_c=\left\{-12.11i-10j-17.3k\right\}\,\text{N}\cdot\text{m}

Hello dear

Why are you used the both sine and cosine at point B? And which is opposite / adjust? When you’re solving B?

Hi! We must use cosine and sine to figure out the location of point B with respect to the origin. Using sine give us the height or the z-value of point B and using cosine gives us the length, or in other words, the y-value of point B. Without using trigonometry, we have no way to figure out those values. The only one we can figure out is the x-position of point B because that’s simply 350 mm. When you say “which is opposite/adjust” I assume you are referring to trig functions? In this case, sine gives us opposite over hypotenuse, which means it’s 0.4sin30, and cosine is simply adjacent over hypotenuse. I hope that helps!

One last thought, it might be very helpful for you to review the following: http://www.questionsolutions.com/forces-cartesian-vector-form/

Good luck with your studies! ๐

Hello, may I know why you use M2 = rab x F2, instead of M2 = rac x F2?

The couple moment is applied at both A,B and C. But the 50N force goes through the line of action at B from C, so we can just consider A and B.

But would we get the wrong answer if we use rac?

Try it out and let me know ๐

For Mc, the resultant vector, should not be rAC instead of rAB?

The line of action of C goes through B, so you can use either.