6.9: Force expressions and projections
Now we express the forces drawn in the FBD in the form of equations. For the forces in Figure 6.4, assuming a kinetic friction coefficient \(\mu_{k}\) during the sliding motion we get:
\[\begin{align} \overrightarrow{\boldsymbol{F}}_{A, g} & =-m_{A} g \hat{\boldsymbol{\jmath}} \tag{6.11} \label{6.11}\\[4pt] \overrightarrow{\boldsymbol{F}}_{A, N} & =F_{N} \hat{\boldsymbol{n}} \tag{6.12} \label{6.12}\\[4pt] \overrightarrow{\boldsymbol{F}}_{A, f} & =-\mu_{k} F_{N} \hat{\boldsymbol{t}} \tag{6.13} \label{6.13}\end{align}\]
Projecting the force vectors
After having drawn the object and force vectors in the FBD, we need their scalar components to evaluate Newton’s second law along each axis. We thus project the forces on the coordinate system (CS) that was selected in the previous section and was drawn in the sketch and/or FBD. This is done using the method described in Sec. 3.2.5, and shown in Figure 6.5. For example we get the \(x\) component of the vector \(\overrightarrow{\boldsymbol{F}}_{A, f}\) using \(F_{A, f, x}=\overrightarrow{\boldsymbol{F}}_{A, f} \cdot \hat{\boldsymbol{\imath}}\).
As projected components of \(\overrightarrow{\boldsymbol{F}}_{g}\) we obtain:
\[\begin{align} F_{A, g, t} & =m_{A} g \sin \alpha \tag{6.14} \label{6.14}\\[4pt] F_{A, g, n} & =-m_{A} g \cos \alpha \tag{6.15} \label{6.15}\end{align}\]