The values of Rx = ΣFx, Ry = ΣFy, and ΣMO for five force systems lying in the xy-plane are listed in the following table. Point O is the origin of the coordinate system, and positive moments are counterclockwise. Determine the resultant for each force system, and show it on a sketch of

Question

The values of [latex]R_{x}=ΣF_{x}[/latex], [latex]R_{y}=ΣF_{y}[/latex], and [latex]ΣM_{O}[/latex] for five force systems lying in the [latex]xy[/latex]-plane are listed in the following table. Point [latex]O[/latex] is the origin of the coordinate system, and positive moments are counterclockwise. Determine the resultant for each force system, and show it on a sketch of the coordinate system.

Part	[latex]R_{x}[/latex]	[latex]R_{y}[/latex]	[latex]ΣM_{O}[/latex]
1	[latex]0[/latex]	[latex]200\:N[/latex]	[latex]400\:N\cdot m[/latex]
2	[latex]0[/latex]	[latex]200\:N[/latex]	[latex]-400\:N\cdot m[/latex]
3	[latex]300\:lb[/latex]	[latex]400\:lb[/latex]	[latex]600\:lb\cdot ft[/latex]
4	[latex]400\:N[/latex]	[latex]-600\:N[/latex]	[latex]-900\:N\cdot m[/latex]
5	[latex]0[/latex]	[latex]0[/latex]	[latex]-200\:lb\cdot ft[/latex]

Accepted Answer

Part 1

[latex]R_{x}=0[/latex] [latex]R_{y}=200\:N[/latex] [latex]ΣM_{O}=400\:N\cdot m[/latex]

The resultant is a [latex]200-N[/latex] force that is parallel to the [latex]y[/latex]-axis, as shown in Fig. (a).

Letting [latex]x[/latex] be the distance from point [latex]O[/latex] to the line of action of the resultant, as shown in Fig. (a), and using Eq. (3.7), we have

[latex]ΣM_{O}=R_{x}\:\:\:\:\underset{+}{\curvearrowleft}\:\:\:\:400=200x[/latex]

which gives

[latex]x =2\:m[/latex]

Part 2

[latex]R_{x}=0[/latex] [latex]R_{y}=200\:N[/latex] [latex]ΣM_{O}=-400\:N\cdot m[/latex]

The resultant is the same [latex]200-N[/latex] force as in Part 1, but here the moment equation gives

[latex]ΣM_{O}=R_{x}\:\:\:\:\underset{+}{\curvearrowleft}\:\:\:\:-400=200x[/latex]

or

[latex]x =-2\:m[/latex]

The negative sign indicates that [latex]x[/latex] lies to the left of point [latex]O[/latex], as shown in Fig. (b).

Part 3

[latex]R_{x}=300\:lb[/latex] [latex]R_{y}=400\:lb[/latex] [latex]ΣM_{O}=600\:lb\cdot ft[/latex]

The resultant is the force [latex]R=300i+400j\:lb[/latex]. Its magnitude is [latex]R=\sqrt{(300)^2+(400)^2}=500\:lb[/latex]. The moment equation of Eqs. (3.5) must be used to determine the line of action of [latex]R[/latex]. Letting [latex]d[/latex] be the perpendicular distance from point [latex]O[/latex] to the line of action of [latex]R[/latex], as shown in Fig. (c), we have

[latex]ΣM_{O}=R_{d}\:\:\:\:\underset{+}{\curvearrowleft}\:\:\:\:600=500d[/latex]

which yields

[latex]d =1.2\:ft[/latex]

The points where [latex]R[/latex] intersects the coordinate axes can then be determined by trigonometry or by using the principle of moments, as follows.

With [latex]R[/latex] placed at [latex]A[/latex], as in Fig. (c): With [latex]R[/latex] placed at [latex]B[/latex], as in Fig. (d):

[latex]ΣM_{O}=R_{y}x[/latex] [latex]ΣM_{O}=R_{x}y[/latex]

[latex]\underset{+}{\curvearrowleft}\:\:\:\:600=400x[/latex] [latex]\underset{+}{\curvearrowleft}\:\:\:\:600=300y[/latex]

[latex]x =1.5\:ft[/latex] [latex]y =2\:ft[/latex]

Part 4

[latex]R_{x}=400\:N[/latex] [latex]R_{y}=-600\:N[/latex] [latex]ΣM_{O}=-900\:N\cdot m[/latex]

The resultant is the force [latex]R=400i-600j\:N[/latex]; its magnitude is [latex]R=\sqrt{(400)^2+(600)^2}=721.1\:N[/latex]. Letting [latex]d[/latex] be the perpendicular distance from point [latex]O[/latex] to the line of action of [latex]R[/latex], as shown in Fig. (e), we have

[latex]ΣM_{O}=Rd\:\:\:\:\underset{+}{\curvearrowleft}\:\:\:\:-900=-721.1d[/latex]

which gives

[latex]d=1.248\:m[/latex]

Note that the line of action of [latex]R[/latex] must be placed to the right of the origin, so that its moment about point [latex]O[/latex] has the same sense as [latex]ΣM_{O}[/latex]—that is, clockwise.

Part 5

[latex]R_{x}=0[/latex] [latex]R_{y}=0[/latex] [latex]ΣM_{O}=-200\:lb\cdot ft[/latex]

Because the sum of the forces is zero, the resultant of this force system is a [latex]200-lb\cdot ft[/latex] clockwise couple, as shown in Fig. (f).

Part	$R_{x}$	$R_{y}$	$ΣM_{O}$
1	$0$	$200\:N$	$400\:N\cdot m$
2	$0$	$200\:N$	$-400\:N\cdot m$
3	$300\:lb$	$400\:lb$	$600\:lb\cdot ft$
4	$400\:N$	$-600\:N$	$-900\:N\cdot m$
5	$0$	$0$	$-200\:lb\cdot ft$

Question 3.3: The values of Rx = ΣFx, Ry = ΣFy, and ΣMO for five force sys...

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