The plate is acted on by four parallel forces, three of which are shown in Fig. (a). The fourth force P and its line of action are unknown. The resultant of this force system is the couple-vector C^R = −1100i + 1500j lb · ft. Determine P and its line of action.

Question

The plate is acted on by four parallel forces, three of which are shown in Fig. (a). The fourth force [latex]P[/latex] and its line of action are unknown. The resultant of this force system is the couple-vector [latex]C^R=−1100i+1500j\:lb\cdot ft[/latex]. Determine [latex]P[/latex] and its line of action.

Accepted Answer

Because the resultant is a couple, the sum of the forces must be zero:

[latex]{ iny+}\big\uparrow\:\:\:\:R=ΣF_{z}=P+300+400−200=0[/latex]

from which [latex]P=−500\:lb[/latex]. Therefore, the force [latex]P[/latex] is

[latex]P=−500k\:lb[/latex]

As shown in Fig. (b), we let [latex]A[/latex] be the point where [latex]P[/latex] intersects the [latex]xy[/latex]-plane. To determine the location of [latex]A[/latex], we equate the sum of the moments of the original forces about any point to the moment of the resultant about that point (in this case, the moment of the resultant about every point is simply [latex]C^R[/latex]). Choosing point [latex]O[/latex] as the moment center and noting that [latex]x_{A}i+y_{A}j[/latex] is the vector from point [latex]O[/latex] to point [latex]A[/latex], the moment equation becomes

[latex]C^R=ΣM_{O}=Σr×F[/latex]

[latex]−1100i+1500j=\left|\begin{matrix}i&j&k\3&0&0\0&0&−200\end{matrix} ight|+\left|\begin{matrix}i&j&k\2&3&0\0&0&300\end{matrix} ight|+\left|\begin{matrix}i&j&k\x_{A}&y_{A}&0\0&0&−500\end{matrix} ight|[/latex]

Expanding the above determinants and equating like components gives

[latex]−1100=900−500y_{A}[/latex]

[latex]1500=600−600+500x_{A}[/latex]

from which

[latex]x_{A}=3\:ft[/latex] and [latex]y_{A}=4\:ft[/latex]

Question 3.9: The plate is acted on by four parallel forces, three of whic...

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