The pipe assembly is subjected to the 80-N force. Determine the moment of this force about point B.

#### Solution:

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This question is the second part of the same question. Please refer to the previous question as basic steps will be skipped.

The force applied at C must be expressed in Cartesian vector form (full steps shown in previous question).

F=\left\{44.5i+53.1j-40k\right\}

We now need a position vector from B to C. B is the location where we are calculating the moment, and C is where the force is applied.

To express the position vector, we need to write down the locations of points B and C.

B:(0i+0.4j+0k)\text{m}

C:(0.55i+0.4j-0.2k)\text{m}

C:(0.55i+0.4j-0.2k)\text{m}

The position vector is then:

r_{BC}=\left\{(0.55-0)i+(0.4-0.4)j+(-0.2-0)k\right\}

r_{BC}=\left\{0.55i+0j-0.2k\right\}

r_{BC}=\left\{0.55i+0j-0.2k\right\}

A position vector, denoted \mathbf{r} is a vector beginning from one point and extending to another point. It is calculated by subtracting the corresponding vector coordinates of one point from the other. If the coordinates of point A was (x_A,y_A,z_A) and the coordinates of point B was(x_B,y_B,z_B), then r_{AB}\,=\,(x_B-x_A)i+(y_B-y_A)j+(z_B-z_A)k

We can now take the cross product between the position vector and the force, which will give us the moment created at B.

M_B=r_{BC}\times F

M_B=\begin{bmatrix}\bold i&\bold j&\bold k\\0.55&0&-0.2\\44.5&53.1&-40\end{bmatrix}

M_B=\left\{10.6i+13.1j+29.2k\right\}\,\text{N}\cdot\text{m}

M_B=\begin{bmatrix}\bold i&\bold j&\bold k\\0.55&0&-0.2\\44.5&53.1&-40\end{bmatrix}

M_B=\left\{10.6i+13.1j+29.2k\right\}\,\text{N}\cdot\text{m}

#### Final Answer:

M_B=\left\{10.6i+13.1j+29.2k\right\}\,\text{N}\cdot\text{m}