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

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

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

\pmb{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. 3.5), you obtain

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}_B=\pmb{r}_{A / B} × 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

The moment \pmb{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 \pmb{r}_{A / B}. Following the right-hand rule for assigning signs, we have

+\circlearrowleft M_B=\Sigma M_B =\Sigma Fd = −(400 N)(0.16 m)−(693 N)(0.2 m)= −202.6 N⋅m