Representing Parallel Forces by a Single Force System 1 in Fig. 4.46 consists of parallel forces. Suppose you want to represent it by a force F (system 2). What is F, and where does its line of action intersect the x-z plane?

Question

Representing Parallel Forces by a Single Force

System 1 in Fig. 4.46 consists of parallel forces. Suppose you want to represent it by a force F (system 2). What is F, and where does its line of action intersect the x-z plane?

Strategy
We can determine F from the condition that the sums of the forces in the two systems must be equal. For the two systems to be equivalent, we must choose the point of application P so that the sums of the moments about a point are equal. This condition will tell us where the line of action intersects the x-z plane.

Accepted Answer

The sums of the forces must be equal:

[latex](\sum{F} )_{2} = (\sum{F} )_{1}[/latex]

F = 30j + 20j - 10j = 40j (lb).

The sums of the moments about an arbitrary point must be equal: Let the coordinates of point P be (x, y, z). The sums of the moments about the origin O must be equal.

[latex](\sum{M_{O} } )_{2} =(\sum{M_{O} } )_{1}[/latex] :

[latex]\left|\begin{matrix} i &j &k \ x & y& z \ 0 & 40 & 0 \end{matrix} ight| =\left|\begin{matrix} i &j &k \ 6 & 0 & 2 \ 0 & 30 & 0 \end{matrix} ight| + \left|\begin{matrix} i &j &k \ 2 & 0& 4 \0 & - 10 &0\end{matrix} ight| +\left|\begin{matrix} i &j &k \ - 3 & 0 & - 2 \ 0 & 20 & 0 \end{matrix} ight|[/latex]

Expanding the determinants, we obtain

(20 + 40z)i + (100 - 40x)k = 0.

The sums of the moments about the origin are equal if

x = 2.5 ft,

z = -0.5 ft.

The systems are equivalent if F = 40j (lb) and its line of action intersects the x-z plane at x = 2.5 ft and z = —0.5 ft. Notice that we did not obtain an equation for the y coordinate of P. The systems are equivalent if F is applied at any point along the line of action.

Discussion
We could have determined the x and z coordinates of point P in a simpler way. Since the sums of the moments about any point must be equal for the systems to be equivalent, the sums of the moments about any line must also be equal. Equating the sums of the moments about the x axis,

[latex]\left(\sum{M_{x axis} } ight) _{2} = \left(\sum{M_{x axis} } ight) _{1}[/latex] :

-40z = -(30)(2) + (10)(4) + (20)(2).

we obtain z = —0.5 ft, and equating the sums of the moments about the z axis,

[latex]\left(\sum{M_{z axis} } ight) _{2} = \left(\sum{M_{z axis} } ight) _{1}[/latex] :

40x = (30)(6) _ (10)(2) _ (20)(3),

we obtain x = 2.5 ft.

Question 4.20: Representing Parallel Forces by a Single Force System 1 in F...

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