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

STRATEGY: You can resolve both the force and the position vector from B to A into rectangular components and then use a vector approach to complete the solution.

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

\mathbf{r}_{A / B} = − (0.2 m)i + (0.16 m)j

F = ( 800 N ) cos 60°i + ( 800 N ) sin 60°j
= ( 400 N )i + ( 693 N )j

Recalling the relations in Eq. (3.7) for the cross products of unit vectors (Sec. 3.1D), you obtain

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

\mathbf{M}_{B}=\mathbf{r}_{A / B}\times\mathbf{F} = [−(0.2 m)i + (0.16 m)j] × [(400 N)i + (693 N)j]

= − (138.6 N·m)k − (64.0 N·m)k

 = − (202.6 N·m)k

 \mathbf{M}_{B} = 203 N·m ↻ ◂

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

+↺ \mathbf{M}_{B} = Σ \mathbf{M}_{B} = ΣFd = − ( 400 N )( 0.16 m ) − ( 693 N )( 0.2 m ) = − 202.6 N·m

\mathbf{M}_{B} = 203 N·m ↻ ◂