The curved rod lies in the x–y plane and has a radius of 3 m. If a force F = 80 N of acts at its end as shown, determine the moment of this force about point B.

#### Solution:

Show me the final answer↓

**Please refer to the previous question if a step is unclear. This is the second part of the same question.**

The 80 N force must be first expressed in Cartesian vector form. To do so, we need a position vector from A to C.

r_{AC}=\left\{1i-3j-2k\right\}

The magnitude of this position vector is:

The unit vector of this position vector is:

We can now express the force in Cartesian form by multiplying the magnitude of the force by the unit vector.

F= \left\{21.4i-64.2j-42.8k\right\}

The next step is to write a position vector from B to A. B is the location where we are calculating the moment and A is the location where the force is applied.

r_{BA}=\left\{2.12i+0.88j+0k\right\}

The moment can be found by taking the cross product between the position vector, r_{BA} and the force, F.

M_B=\begin{bmatrix}\bold i&\bold j&\bold k\\2.12&0.88&0\\21.4&-64.2&-42.8\end{bmatrix}

M_B=\left\{-37.6i+90.7j-155k\right\}

#### Final Answer:

Why 3-3cos45

I am not sure which part you are referring to. Please let us know and we’d love to help.

Could you please explain why j for vector rBA is 3-3sin45?

A position vector, is a vector beginning from one point and extending to another point. We calculate it by subtracting the corresponding vector coordinates of one point from the other. In this case, for j, it’s 3j-3sin45j. We need to subtract 3sin45 because that’s the location of point B with respect to the y-axis from the origin. Hope that helps! 🙂