The values of ΣFz, ΣMx, and ΣMy for three force systems that are parallel to the z-axis are as follows: Determine the resultant of each force system and show it on a sketch of the coordinate system.

Question

The values of [latex]ΣF_{z}[/latex], [latex]ΣM_{x}[/latex], and [latex]ΣM_{y}[/latex] for three force systems that are parallel to the [latex]z[/latex]-axis are as follows:

Part	[latex]ΣF_{z}[/latex]	[latex]ΣM_{x}[/latex]	[latex]ΣM_{y}[/latex]
1	[latex]50\:kN[/latex]	[latex]60\:kN\cdot m[/latex]	[latex]-125\:kN\cdot m[/latex]
2	[latex]-600\:lb[/latex]	[latex]0[/latex]	[latex]-1200\:lb\cdot ft[/latex]
3	[latex]0[/latex]	[latex]600\:lb\cdot in[/latex].	[latex]-800\:lb\cdot in[/latex].

Determine the resultant of each force system and show it on a sketch of the coordinate system.

Accepted Answer

Part 1

[latex]ΣF_{z}=50\:kN[/latex] [latex]ΣM_{x}=60\:kN\cdot m[/latex] [latex]ΣM_{y}=−125\:kN\cdot m[/latex]

The resultant is the force [latex]R=50k\:kN[/latex]. With [latex]ΣM_{O}=ΣM_{x}i+ΣM_{y}j[/latex] and [latex]r=xi+yj[/latex], Eq. (3.10) can be used to determine the line of action of [latex]R[/latex]:

[latex]ΣM_{O}=r×R[/latex]

[latex]60i−125j=\left|\begin{matrix}i&j&k\x&y&0\0&0&50\end{matrix} ight|=50yi−50xj[/latex]

Equating like components gives the following scalar equations:

[latex]60=50y[/latex] and [latex]−125=−50x[/latex]

which gives [latex]x=2.5\:m[/latex] and [latex]y=1.2\:m[/latex]. The sketch of the resultant is shown in Fig. (a).

Identical results for [latex]x[/latex] and [latex]y[/latex] are obtained if one uses Eqs. (3.11):

[latex]x=−\frac{ΣM_{y}}{R}=−\frac{−125}{50}=2.5\:m[/latex]

[latex]y=\frac{ΣM_{x}}{R}=\frac{60}{50}=1.2\:m[/latex]

Part 2

[latex]ΣF_{z}=−600\:lb[/latex] [latex]ΣM_{x}=0[/latex] [latex]ΣM_{y}=−1200\:lb\cdot ft[/latex]

The resultant is the force [latex]R =−600k\:lb\cdot In[/latex] this case, Eq. (3.10) yields

[latex]ΣM_{O}=r×R[/latex]

[latex]−1200j=\left|\begin{matrix}i&j&k\x&y&0\0&0&−600\end{matrix} ight|=−600yi+600xj[/latex]

Equating like components gives [latex]x=−2\:ft[/latex] and [latex]y=0[/latex]. The resultant is shown in Fig. (b).

Part 3

[latex]ΣF_{z}=0[/latex] [latex]ΣM_{x}=600\:lb\cdot in[/latex]. [latex]ΣM_{y}=-800\:lb\cdot in[/latex].

Because the sum of the forces is zero and the sum of the moments is not zero, the resultant is the couple-vector [latex]C^R=ΣM_{x}i+ΣM_{y}j=600i−800j\:lb\cdot in.[/latex], shown in Fig. (c). The magnitude of this couple-vector is [latex]1000\:lb\cdot in[/latex].

Part	$ΣF_{z}$	$ΣM_{x}$	$ΣM_{y}$
1	$50\:kN$	$60\:kN\cdot m$	$-125\:kN\cdot m$
2	$-600\:lb$	$0$	$-1200\:lb\cdot ft$
3	$0$	$600\:lb\cdot in$ .	$-800\:lb\cdot in$ .

Question 3.7: The values of ΣFz, ΣMx, and ΣMy for three force systems that...

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