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 \mathbf{M}_{A} about A of the force F exerted by the wire on point C by forming the vector product
\mathbf{M}_{A}=\mathbf{r}_{C/A}\times\mathbf{F} (1)
where \mathbf{r}_{C/A} is the vector from A to C
\mathbf{r}_{C/A} =\overrightarrow{AC} = (0.3 m)i + (0.08 m)k (2)
and F is the 200-N force directed along CD (Fig. 1). Introducing the unit vector
λ = \overrightarrow{CD} / CD,
you can express F as
\mathbf{F}=F\lambda=(200\,\mathbf{N}){\frac{{\overrightarrow{{C D}}}}{C D}} (3)
Resolving the vector \overrightarrow{CD} into rectangular components, you have
\overrightarrow{CD}= − (0.3 m)i + (0.24 m)j − (0.32 m)k CD = 0.50 m
Substituting into (3) gives you
\mathbf{F}={\frac{200\;\mathbf{N}}{0.50\;\mathbf{m}}}\left[-(0.3\;\mathbf{m})\mathbf{i}+(0.24\;\mathbf{m})\mathbf{j}-(0.32\;\mathbf{m})\mathbf{k}\right]
= − (120 N)i + (96 N)j − (128 N)k (4)
Substituting for \mathbf{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)
\begin{matrix} i \times i = 0 & j × i = −k& k × i = j\\i × j = k & j × j = 0 & k × j = − i\\ i × k = − j & j × k = i & k × k = 0 \end{matrix}\qquad (3.7)\\
\mathbf{M}_{A}=\mathbf{r}_{C/A}\times\mathbf{F}
= (0.3i + 0.08k) × (−120i + 96j − 128k)
= ( 0.3 )( 96 )k + ( 0.3 )( −128 )( −j) + ( 0.08 )( −120 )j + ( 0.08 )( 96 )( −i)
\mathbf{M}_{A} = − ( 7.68 N·m )i + ( 28.8 N·m )j + ( 28.8 N·m )k ◂
ALTERNATIVE SOLUTION. As indicated in Sec. 3.1F, you can also express the moment M_A in the form of a determinant:
M_{A}= \Bigg| \begin{matrix}i & j & k \\ x_C-x_A & y_C-y_A &z_C – z_A \\ F_x & F_y & F_z \end{matrix} \Bigg| =\Bigg|\begin{array}{c c c}{{{\bf i}}}&{{{\bf j}}}&{{\bf k}}\\ {{0.3}}&{{0}}&{{0.08}}\\ {{-120}}&{{96}}&{{-128}}\end{array}\Bigg|
\mathbf{M}_{A} = − (7.68 N·m)i + (28.8 N·m)j + (28.8 N·m)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.
