The curved rod lies in the x–y plane


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 O.

The curved rod lies in the x–y plane

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

Solution:

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Let us first express the force being applied at the end of the rod in Cartesian vector form. To do so, we first need a position vector starting from the location of the force to the location where it is headed. The locations of points A and C in Cartesian form is:

A:(3i+3j+0k)

C:(4i+0j-2k)

 

Now, we can write our position vector.

r_{AC}=\left\{(4-3)i+(0-3)j+(-2-0)k\right\}

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

 

The magnitude of this position vector is:

magnitude of r_{AC}=\sqrt{(1)^2+(-3)^2+(-2)^2}=3.74

The magnitude is equal to the square root of the sum of the squares of the vector. If the position vector was r\,=\,ai+bj+ck, then the magnitude would be, r_{magnitude}\,=\,\sqrt{(a^2)+(b^2)+(c^2)}. In the simplest sense, you take each term of a vector, square it, add it together, and then take the square root of that value.

 

The unit vector of this position vector is:

u_{AB}\,=\,\left(\dfrac{1}{3.74}i-\dfrac{3}{3.74}j-\dfrac{2}{3.74}k\right)

The unit vector is each corresponding unit of the position vector divided by the magnitude of the position vector. If the position vector was r\,=\,ai+bj+ck, then unit vector, u\,=\,\dfrac{a}{\sqrt{(a^2)+(b^2)+(c^2)}}+\dfrac{b}{\sqrt{(a^2)+(b^2)+(c^2)}}+\dfrac{c}{\sqrt{(a^2)+(b^2)+(c^2)}}

 

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

F=80\left(\dfrac{1}{3.74}i-\dfrac{3}{3.74}j-\dfrac{2}{3.74}k\right)

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

 

The next step is to express another position vector from O to A. This is the location of where the moment is calculated to the location where the force is being applied.

The curved rod lies in the x–y plane

From the image, we see that points O and A are located at:

O:(0i+0j+0k)

A:(3i+3j+0k)

 

The position vector from O to A is:

r_{OA}=\left\{(3-0)i+(3-0)j+(0-0)k\right\}

r_{OA}=\left\{3i+3k+0k\right\}

 

We can now calculate the moment by taking the cross product between the position vector and the force.

M_O=r_{OA}\times F

M_A=\begin{bmatrix}\bold i&\bold j&\bold k\\3&3&0\\21.4&-64.2&-42.8\end{bmatrix}

M_A=\left\{-128i+128.4j-257k\right\}N\cdot m

 

Final Answer:

M_A=\left\{-128i+128.4j-257k\right\}N\cdot m

 

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 4, question 4-36.

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