Question 3.SP.8: A 4.80-m-long beam is subjected to the forces shown. Reduce ...
A 4.80-m-long beam is subjected to the forces shown. Reduce the given system of forces to (a) an equivalent force-couple system at A, (b) an equivalent force-couple system at B, (c) a single force or resultant. Note: Because the reactions at the supports are not included in the given system of forces, the given system will not maintain the beam in equilibrium.
STRATEGY: The force part of an equivalent force-couple system is simply the sum of the forces involved. The couple part is the sum of the moments caused by each force relative to the point of interest. Once you find the equivalent force-couple at one point, you can transfer it to any other point by a moment calculation.
MODELING and ANALYSIS:
a. Force-Couple System at A.
The force-couple system at A equivalent to the given system of forces consists of a force R and a couple \pmb{M}_A^R defined as (Fig. 1)
\begin{aligned}\pmb{R} &=\Sigma \pmb{F} \\&=(150 N) \pmb{j}-(600 N) \pmb{j}+(100 N) \pmb{j}-(250 N) \pmb{j}=-(600 N) \pmb{j} \\\pmb{M}_A^R &=\Sigma(\pmb{r} \times \pmb{F}) \\&=(1.6 \pmb{i}) \times(-600 \pmb{j})+(2.8 \pmb{i}) \times(100 \pmb{j})+(4.8 \pmb{i}) \times(-250 \pmb{j}) \\&=-(1880 N\cdot m) \pmb{k}\end{aligned}The equivalent force-couple system at A is thus
\pmb{R}=600 N \downarrow \quad \pmb{M}_A^R=1880 N\cdot m \circlearrowrightb. Force-Couple System at B.
You want to find a force-couple system at B equivalent to the force-couple system at A determined in part a. The force R is unchanged, but you must determine a new couple \pmb{M}_B^R, the moment of which is equal to the moment about B of the force-couple system determined in part a (Fig. 2). You have
\begin{aligned}\pmb{M}_B^R &=\pmb{M}_A^R+\overrightarrow{BA} \times \pmb{R} \\&=-(1880 N\cdot m) \pmb{k}+(-4.8 m) \pmb{i} \times(-600 N) \pmb{j} \\&=-(1880 N\cdot m) \pmb{k}+(2880 N\cdot m) \pmb{k}=+(1000 N\cdot m) \pmb{k}\end{aligned}The equivalent force-couple system at B is thus
\pmb{R}=600 N \downarrow \quad \pmb{M}_B^R=1000 N\cdot m \circlearrowrightc. Single Force or Resultant. The resultant of the given system of forces is equal to R, and its point of application must be such that the moment of R about A is equal to \pmb{M}_A^R (Fig. 3). This equality of moments leads to
\begin{aligned}\pmb{r} \times \pmb{R}&=\pmb{M}_A^R\\ x \pmb{i} \times(-600 N) \pmb{j}&=-(1880 N\cdot m) \pmb{k}\\-x(600 N) \pmb{k}&=-(1880 N\cdot m) \pmb{k}\end{aligned}Solving for x, you get x = 3.13 m. Thus, the single force equivalent to the given system is defined as
R = 600 N ↓ x = 3.13 m
REFLECT and THINK: This reduction of a given system of forces to a single equivalent force uses the same principles that you will use later for finding centers of gravity and centers of mass, which are important parameters in engineering mechanics.
