Question 3.SP.4: A rectangular plate is supported by brackets at A and B and ...
A rectangular plate is supported by brackets at A and B and by a wire CD. If the tension in the wire is 200 N, determine the moment about A of the force exerted by the wire on point C.
STRATEGY: The solution requires resolving the tension in the wire and the position vector from A to C into rectangular components. You will need a unit vector approach to determine the force components.
MODELING and ANALYSIS: Obtain the moment \pmb{M_A} about A of the force F exerted by the wire on point C by forming the vector product
\pmb{M}_A=\pmb{r}_{C / A} \times \pmb{F} (1)
where \pmb{r}_{C / A} is the vector from A to C
\pmb{r}_{C / A}=\overrightarrow{AC}=(0.3 m) \pmb{i}+(0.08 m) \pmb{k} (2)
and F is the 200-N force directed along CD (Fig. 1). Introducing the unit vector
\pmb{\lambda}=\overrightarrow{CD} / CD,you can express F as
\pmb{F}=F \lambda=(200 N) \frac{\overrightarrow{CD}}{C D} (3)
Resolving the vector \overrightarrow{CD} into rectangular components, you have
\overrightarrow{CD}=-(0.3 m) \pmb{i}+(0.24 m) \pmb{j}-(0.32 m) \pmb{k} \quad C D=0.50 mSubstituting into (3) gives you
\begin{aligned}\pmb{F} &=\frac{200 N}{0.50 m}[-(0.3 m) \pmb{i}+(0.24 m) \pmb{j}-(0.32 m) \pmb{k}] \\&=-(120 N) \pmb{i}+(96 N) \pmb{j}-(128 N) \pmb{k}\end{aligned} (4)
Substituting for \pmb{r}_{C / A} and F from (2) and (4) into (1) and recalling the relations in Eq. (3.7) of Sec. 3.1D, you obtain (Fig. 2)
i × i = 0 j × i = -k k × i = j
i × j = k j × j = 0 k × j = -i (3.7)
i × k = -j j × k = i k × k = 0
\pmb{M}_A=\pmb{r}_{C / A} \times \pmb{F}=(0.3 \pmb{i}+0.08 \pmb{k}) \times(-120 \pmb{i}+96 \pmb{j}-128 \pmb{k})= (0.3)(96)k +(0.3)(−128)(−j)+(0.08)(−120)j +(0.08)(96)(−i)
\pmb{M}_A=-(7.68 N \cdot m) \pmb{i}+(28.8 N \cdot m) \pmb{j}+(28.8 N \cdot m) \pmb{k}Alternative Solution. As indicated in Sec. 3.1F, you can also express the moment \pmb{M}_A in the form of a determinant:
\pmb{M}_A=\left|\begin{array}{ccc}\pmb{i} & \pmb{j} & \pmb{k} \\x_c-x_A & y_c-y_A & z_c-z_A \\F_x & F_y & F_z\end{array}\right|=\left|\begin{array}{ccc}\pmb{i} & \pmb{j} & \pmb{k} \\0.3 & 0 & 0.08 \\-120 & 96 & -128\end{array}\right|\pmb{M}_A=-(7.68 N \cdot m) \pmb{i}+(28.8 N \cdot m) \pmb{j}+(28.8 N \cdot m) \pmb{k}
REFLECT and THINK: Two-dimensional problems often are solved easily using a scalar approach, but the versatility of a vector analysis is quite apparent in a three-dimensional problem such as this.
