Chapter 3, Problem 3.44P

The force-couple system consists of the force $\text{R}=250\text{i}+360\text{j}-400\text{k}\text{N}$ and the couple-vector ${\text{C}}^R=1200\text{i}+750\text{j}+560\text{k}\text{N}\cdot \text{m}\text{.}$ Determine the equivalent wrench and find the coordinates of the point where the axis of the wrench crosses the xy-plane.

To determine

The equivalent wrench and the coordinates of the point where it crosses the x-y plane.

Answer to Problem 3.44P

The equivalent wrench consists of:

Force $R=\left(250\stackrel{\wedge }{i}+360\stackrel{\wedge }{j}-400\stackrel{\wedge }{k}\right)N$

Couple $C^R=\left(245.7\stackrel{\wedge }{i}+353.8\stackrel{\wedge }{j}-393.1\stackrel{\wedge }{k}\right)N.m$

And axis of the wrench passes through the point (0.991, -2.31, 0) m.

Explanation of Solution

Given information:

Force couple system with:

  $\begin{array}{l}R=\left(250\stackrel{\wedge }{i}+360\stackrel{\wedge }{j}-400\stackrel{\wedge }{k}\right)N\\ C^R=\left(1200\stackrel{\wedge }{i}+750\stackrel{\wedge }{j}+560\stackrel{\wedge }{k}\right)N\cdot m\end{array}$

Calculations:

Unit vector in the direction of R:

  $\begin{array}{l}\lambda =\frac{R}{\left|R\right|}\\ \lambda =\left(\frac{250\stackrel{\wedge }{i}+360\stackrel{\wedge }{j}-400\stackrel{\wedge }{k}}{\sqrt{{250}^2+{360}^2+{\left(-400\right)}^2}}\right)\\ \lambda =\left(0.4213\stackrel{\wedge }{i}+0.6067\stackrel{\wedge }{j}-0.6741\stackrel{\wedge }{k}\right)\end{array}$

Component of C^R in the direction of $\lambda$ is:

  $\begin{array}{l}{C}_t^R=C^R\cdot \lambda =1200\left(0.4213\right)+750\left(0.6067\right)+560\left(-0.6741\right)\\ C_t^R=583.1N\cdot m\end{array}$

Expressing in vector form:

  $\begin{array}{l}{C}_t^R=C_t^R\cdot \lambda =583.1\left(0.4213\stackrel{\wedge }{i}+0.6067\stackrel{\wedge }{j}-0.6741\stackrel{\wedge }{k}\right)\\ C_t^R=\left(245.7\stackrel{\wedge }{i}+353.8\stackrel{\wedge }{j}-393.1\stackrel{\wedge }{k}\right)N\cdot m\end{array}$

Hence, the equivalent wrench will consists of:

Force $R=\left(250\stackrel{\wedge }{i}+360\stackrel{\wedge }{j}-400\stackrel{\wedge }{k}\right)N$, and

Couple $C^R=\left(245.7\stackrel{\wedge }{i}+353.8\stackrel{\wedge }{j}-393.1\stackrel{\wedge }{k}\right)N.m$

Now let the wrench intersects the x-y plane at A (x, y, 0).

  $r_{OA}\times R=C_n^R$

Where, $C_n^R$ - Component of C^R that is normal to $\lambda$ :

  $\begin{array}{l}{C}_n^R=C^R-C_t^R\\ C_n^R=\left(1200\stackrel{\wedge }{i}+750\stackrel{\wedge }{j}+560\stackrel{\wedge }{k}\right)-\left(245.7\stackrel{\wedge }{i}+353.8\stackrel{\wedge }{j}-393.1\stackrel{\wedge }{k}\right)\\ C_n^R=\left(954.3\stackrel{\wedge }{i}+396.2\stackrel{\wedge }{j}+953.1\stackrel{\wedge }{k}\right)N\cdot m\end{array}$

And, $r_{OA}=x\stackrel{\wedge }{i}+y\stackrel{\wedge }{j}$

Substituting,

  $\begin{array}{l}\left(\begin{array}{ccc}\stackrel{\wedge }{i}& \stackrel{\wedge }{j}& \stackrel{\wedge }{k}\\ x& y& 0\\ 250& 360& -400\end{array}\right)=954.3\stackrel{\wedge }{i}+396.2\stackrel{\wedge }{j}+953.1\stackrel{\wedge }{k}\\ or,\\ -400\left(y\right)\stackrel{\wedge }{i}+400\left(x\right)\stackrel{\wedge }{j}+\left(360x-250y\right)\stackrel{\wedge }{k}=954.3\stackrel{\wedge }{i}+396.2\stackrel{\wedge }{j}+953.1\stackrel{\wedge }{k}\end{array}$

Equating $\stackrel{\wedge }{i}$, $\stackrel{\wedge }{j}$ and $\stackrel{\wedge }{k}$ components:

  $\begin{array}{l}-400\left(y\right)=954.3\Rightarrow y=-2.39m\\ 400\left(x\right)=396.2\Rightarrow x=0.991m\\ \end{array}$

Conclusion:

The equivalent wrench consists of force $R=\left(250\stackrel{\wedge }{i}+360\stackrel{\wedge }{j}-400\stackrel{\wedge }{k}\right)N$ and couple $C^R=\left(245.7\stackrel{\wedge }{i}+353.8\stackrel{\wedge }{j}-393.1\stackrel{\wedge }{k}\right)N.m$

And this wrench intersects the x-y plane at the point (0.991, -2.31, 0) m.