Boxes A and B are at rest on a conveyor belt that is initially at rest. The belt is suddenly started in an upward direction so that slipping occurs between the belt and the boxes. Knowing that the coefficients of kinetic friction between the belt and the boxes are (μk)A = 0.30 and (μk)B = 0.32,

Question

Boxes A and B are at rest on a conveyor belt that is initially at rest. The belt is suddenly started in an upward direction so that slipping occurs between the belt and the boxes. Knowing that the coefficients of kinetic friction between the belt and the boxes are [latex]{ \left( { \mu }_{ k } \right) }_{ A }=0.30[/latex] and [latex]{ { (\mu }_{ k }) }_{ B }=0.32[/latex] determine the initial acceleration of each box.

Accepted Answer

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

[latex]A: \quad+ warrow \Sigma F_{y}=0: \quad N_{A}-W_{A} \cos 15^{\circ}=0[/latex]

or [latex]\quad N_{A}=W_{A} \cos 15^{\circ}[/latex]

Slipping:

[latex]\begin{aligned}F_{A} & =\left(\mu_{k} ight)_{A} N_{A} \& =0.3 W_{A} \cos 15^{\circ} \+ earrow \Sigma F_{x} & =m_{A} a_{A}: F_{A}-W_{A} \sin 15^{\circ}=m_{A} a_{A}\end{aligned}[/latex]

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

or

[latex]\begin{aligned}a_{A} & =\left(32.2 \mathrm{\ ft} / \mathrm{s}^{2} ight)\left(0.3 \cos 15^{\circ}-\sin 15^{\circ} ight) \& =0.997 \mathrm{\ ft} / \mathrm{s}^{2}\end{aligned}[/latex]

[latex]B: \quad+ warrow \Sigma F_{y}=0: N_{B}-W_{B} \cos 15^{\circ}=0[/latex]

or [latex]\quad N_{B}=W_{B} \cos 15^{\circ}[/latex]

Slipping: [latex]\quad F_{B}=\left(\mu_{k} ight)_{B} N_{B}[/latex]

[latex]=0.32 W_{B} \cos 15^{\circ}[/latex]

[latex]+ earrow \Sigma F_{x}=m_{B} a_{B}: \quad F_{B}-W_{B} \sin 15^{\circ}=m_{B} a_{B}[/latex]

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

or

[latex]\begin{aligned}a_{B} & =\left(32.2 \mathrm{\ ft} / \mathrm{s}^{2} ight)\left(0.32 \cos 15^{\circ}-\sin 15^{\circ} ight)=1.619 \mathrm{\ ft} / \mathrm{s}^{2} \a_{B} & >a_{A} \Rightarrow ext { assumption is correct }\end{aligned}[/latex]

[latex]\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}[/latex]

Note: If it is assumed that the boxes remain in contact [latex]\left(N_{A B} eq 0 ight)[/latex], then assuming [latex]N_{A B}[/latex] to be compression,

[latex]a_{A}=a_{B} \quad ext { and find }\left(\Sigma F_{x}=m a ight) ext { for each box. }[/latex]

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

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

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

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

This Question has been Answered.

Already have an account?

Login

Related Answered Questions