Question 12.16: Boxes A and B are at rest on a conveyor belt that is initial...

Assume that a_{B}>a_{A} so that the normal force N_{A B} between the boxes is zero.

A: \quad+\nwarrow \Sigma F_{y}=0: \quad N_{A}-W_{A} \cos 15^{\circ}=0

or \quad N_{A}=W_{A} \cos 15^{\circ}

Slipping:

or \quad 0.3 W_{A} \cos 15^{\circ}-W_{A} \sin 15^{\circ}=\frac{W_{A}}{g} a_{A}

or 

B: \quad+\nwarrow \Sigma F_{y}=0: N_{B}-W_{B} \cos 15^{\circ}=0

or \quad N_{B}=W_{B} \cos 15^{\circ}

Slipping: \quad F_{B}=\left(\mu_{k}\right)_{B} N_{B}

=0.32 W_{B} \cos 15^{\circ}

+\nearrow \Sigma F_{x}=m_{B} a_{B}: \quad F_{B}-W_{B} \sin 15^{\circ}=m_{B} a_{B}

or \quad 0.32 W_{B} \cos 15^{\circ}-W_{B} \sin 15^{\circ}  =\frac{W_{B}}{g} a_{B}

or

\begin{aligned}& \mathbf{a}_{A}=0.997 \mathrm{\ ft} / \mathrm{s}^{2} \ ⦨ 15^{\circ} \blacktriangleleft\\& \mathbf{a}_{B}=1.619 \mathrm{\ ft} / \mathrm{s}^{2} \ ⦨ 15^{\circ}\blacktriangleleft\end{aligned}

Note: If it is assumed that the boxes remain in contact \left(N_{A B} \neq 0\right), then assuming N_{A B} to be compression,

A: \quad0.3 W_{A} \cos 15^{\circ}-W_{A} \sin 15^{\circ}-N_{A B}=\frac{W_{A}}{g} a

B: \quad 0.32 W_{B} \cos 15^{\circ}-W_{B} \sin 15^{\circ}+N_{A B}=\frac{W_{B}}{g} a

Solving yields a=1.273 \mathrm{\ ft} / \mathrm{s}^{2} and N_{A B}=-0.859 \mathrm{\ lb}, which contradicts the assumption.