Question 3.SP.2: A force of 800 N acts on a bracket as shown. Determine the m......

MODELING and ANALYSIS: Obtain the moment \mathbf{M}_B of the force F about B by forming the vector product

\mathbf{M}_B=\mathbf{r}_{A / B} \times \mathbf{F}

where \mathbf{r}_{A / B} is the vector drawn from B to A (Fig. 1). Resolving \mathbf{r}_{A / B} and F into rectangular components, you have

\begin{aligned}\mathbf{r}_{A / B} & =-(0.2 \mathrm{~m}) \mathbf{i}+(0.16 \mathrm{~m}) \mathbf{j} \\\mathbf{F} & =(800 \mathrm{~N}) \cos 60^{\circ} \mathbf{i}+(800 \mathrm{~N}) \sin 60^{\circ} \mathbf{j} \\& =(400 \mathrm{~N}) \mathbf{i}+(693 \mathrm{~N}) \mathbf{j}\end{aligned}

Recalling the relations in Eq. (3.7) for the cross products of unit vectors (Sec. 3.5), you obtain

\begin{array}{l}\mathbf{i} \times \mathbf{i}=\mathbf{0} \quad \mathbf{j} \times \mathbf{i}=-\mathbf{k} \quad \mathbf{k} \times \mathbf{i}=\mathbf{j} \\\mathbf{i} \times \mathbf{j}=\mathbf{k} \quad \mathbf{j} \times \mathbf{j}=\mathbf{0} \quad \mathbf{k} \times \mathbf{j}=-\mathbf{i} \\\mathbf{i} \times \mathbf{k}=-\mathbf{j} \quad \mathbf{j} \times \mathbf{k}=\mathbf{i} \quad \mathbf{k} \times \mathbf{k}=\mathbf{0} \\\end{array}                (3.7)

The moment \mathbf{M}_B is a vector perpendicular to the plane of the figure and pointing into the page.

REFLECT and THINK: We can also use a scalar approach to solve this problem using the components for the force F and the position vector \mathbf{r}_{A / B}.

Following the right-hand rule for assigning signs, we have

+ \circlearrowleft \mathbf{M}_B = \sum{\mathbf{M}_B }= \sum{Fd} = -(400  N)(0.16  m) – (693  N)(0.2  m)= -202.6 \mathrm{~N} \cdot \mathrm{m}