Determine the wrench that is equivalent to the force system described in Sample Problem 3.1. Find the coordinates of the point where the axis of the wrench crosses the xy-plane.

Question

Determine the wrench that is equivalent to the force system described in Sample Problem 3.1. Find the coordinates of the point where the axis of the wrench crosses the [latex]xy[/latex]-plane.

Accepted Answer

As explained in the solution to Sample Problem 3.1, the original force system can be reduced to the force-couple system shown in Fig. (a): the force [latex]\mathbf{R} [/latex], acting at the origin [latex]O[/latex], and the couple [latex]C^R[/latex], where

[latex]\mathbf{R} =38.8\mathbf{i} +43.2\mathbf{j} +38.4\mathbf{k} \:lb[/latex] (a)

[latex]C^R=390\mathbf{i} +116\mathbf{j} +360\mathbf{k} \:lb\cdot ft[/latex]

The magnitude of [latex]\mathbf{R} [/latex] is

[latex]R=\sqrt{(38.8)^2+(43.2)^2+(38.4)^2}=69.6\:lb[/latex]

We begin by determining the axis of the wrench, defined by the unit vector [latex]\mathbf{ λ}[/latex] in the direction of [latex]\mathbf{R} [/latex]:

[latex]\mathbf{λ} =\frac{\mathbf{R} }{R}=\frac{38.8\mathbf{i} +43.2\mathbf{j} +38.4\mathbf{k} }{69.6}=0.557\mathbf{i} +0.621\mathbf{j} +0.552\mathbf{k} [/latex]

The component of [latex]C^R[/latex] in the direction of [latex]\mathbf{λ}[/latex] can now be obtained from Eq. (3.12):

[latex]C^{R}_{t}=(C^R\cdot \mathbf{λ})\mathbf{λ}[/latex]. The magnitude of this vector is

[latex]C^R_{t}=C^R\cdot λ=(390\mathbf{i} +116\mathbf{j} +360\mathbf{k} )\cdot (0.557\mathbf{i} +0.621\mathbf{j} +0.552\mathbf{k} )=488\:lb\cdot ft[/latex]

which gives

[latex]C^R_{t}=C^R_{t}\mathbf{λ}=488(0.557\mathbf{i} +0.621\mathbf{j} +0.552\mathbf{k} )=272\mathbf{i} +303\mathbf{j} +269\mathbf{k} \:lb\cdot ft[/latex]

Therefore, the wrench consists of the force-couple system

[latex]\mathbf{R}=38.8\mathbf{i} +43.2\mathbf{j} +38.4\mathbf{k} \:lb[/latex]

[latex]C^R_{t}=272\mathbf{i} +303\mathbf{j} +269\mathbf{k} \:lb\cdot ft[/latex]

To find the coordinates of the point where the axis of the wrench intersects the [latex]xy[/latex]-plane, we must find [latex]C^R_{n}[/latex], the component of [latex]C^R[/latex] that is normal to [latex]\mathbf{λ}[/latex]. From [latex]Eq.\:(3.13)[/latex], we obtain

[latex]C^R_{n}=C^R−C^R_{t}=(390\mathbf{i} +116\mathbf{j} +360\mathbf{k} )−(272\mathbf{i} +303\mathbf{j} +269\mathbf{k} )=118\mathbf{i} −187\mathbf{j} +91\mathbf{k} \:lb\cdot ft[/latex]

Referring to Fig. (a), we let [latex]r=x\mathbf{i} +y\mathbf{j} [/latex] be the vector from the origin [latex]O[/latex] to [latex]A[/latex], the point where the wrench intersects the [latex]xy[/latex]-plane. Using [latex]Eq.\:(3.14)[/latex], we have

[latex]\mathbf{r}×\mathbf{R}=C^R_{n}[/latex]

[latex]\left|\begin{matrix}\mathbf{i} &\mathbf{j} &\mathbf{k} \x&y&0\38.8&43.2&38.4\end{matrix} ight|=118i−187\mathbf{j} +91\mathbf{k} [/latex]

After expanding the determinant, we get

[latex]38.4y\mathbf{i} −38.4x\mathbf{j} +(43.2x−38.8y)\mathbf{k} =118i−\mathbf{j} +91\mathbf{k} [/latex]

Equating the coefficients of [latex]\mathbf{i} [/latex] and [latex]\mathbf{j} [/latex] yields

[latex]38.4y=118[/latex] [latex]y=3.07\:ft[/latex]

[latex]−38.4x=−187[/latex] [latex]x=4.87\:ft[/latex]

The third equation, obtained by equating the coefficients of [latex]\mathbf{k} [/latex], is not independent of the preceding two equations, as can be easily verified.

The resultant wrench is depicted in Fig. (b), which shows the magnitudes of the force and the couple-vector.

Question 3.10: Determine the wrench that is equivalent to the force system ...

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