If the couple moment acting on the pipe 2


If the couple moment acting on the pipe has a magnitude of 400 N•m, determine the magnitude F of the vertical force applied to each wrench.

If the couple moment acting on the pipe

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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The couple moment can be found by taking the cross product between a position vector from A to B and force F.

M_c=r_{AB}\times F

 

Let us first express a position vector from A to B. The coordinates of A and B are:

A:(0.65i+0.4j+0k)\,\text{m}

B:(0.3i+0.2j+0k)\,\text{m}

 

The position vector is:

r_{AB}=\left\{(0.3-0.65)i+(0.2-0.4)j+(0-0)k\right\}

r_{AB}=\left\{-0.35i-0.2j+0k\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 now need to express from F in Cartesian vector form. Force F is only applied in the z-axis, thus it will only have a k component.

F=\left\{0i+0j+Fk\right\}

 

Let us now take the cross product:

M_c=\begin{bmatrix}\bold i&\bold j&\bold k\\-0.35&-0.2&0\\0&0&F\end{bmatrix}

M_c=\left\{-0.2F\bold i+0.35F\bold j\right\}

 

The magnitude of this couple moment is:

magnitude of M_{c}\,=\,\sqrt{(-0.2F)^2+(0.35F)^2}

(Remember that the question tells us the magnitude of M_c is 400 N•m)

400=\sqrt{(-0.2F)^2+(0.35F)^2}

(square both sides)

160000=(-0.2F)^2+(0.35F)^2

160000=0.04F^2+0.1225F^2

160000=0.1625F^2

(divide both sides by 0.1625)

F^2=984615.38

(take the square root of both sides)

F=992.28 N

 

Final Answer:

F=992.28 N

 

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 4, question 4-84.

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