A semicircular slot of 10-in. radius is cut in a flat plate which rotates about the vertical AD at a constant rate of 14 rad/s. A small, 0.8-lb block E is designed to slide in the slot as the plate rotates. Knowing that the coefficients of friction are μs = 0.35 and μk = 0.25, determine whether the

Question

A semicircular slot of 10-in. radius is cut in a flat plate which rotates about the vertical AD at a constant rate of 14 rad/s. A small, 0.8-lb block E is designed to slide in the slot as the plate rotates. Knowing that the coefficients of friction are [latex]{ \mu }_{ s }=0.35[/latex] and [latex]{ \mu}_{ k }=0.25[/latex], determine whether the block will slide in the slot if it is released in the position corresponding to (a) [latex] heta =80°[/latex], (b)[latex] heta =40°[/latex]. Also determine the magnitude and the direction of the friction force exerted on the block immediately after it is released.

Accepted Answer

First note

[latex]\begin{aligned} ho & =\frac{1}{12}(26-10 \sin heta) \mathrm{\ ft} \ u_{E} & = ho \dot{\phi}_{A B C D}\end{aligned}[/latex]

Then

[latex]\begin{aligned}a_{n} & =\frac{ u_{E}^{2}}{ ho}= ho\left(\dot{\phi}_{A B C D} ight)^{2} \& =\left[\frac{1}{12}(26-10 \sin heta) \mathrm{ft} ight](14 \mathrm{\ rad} / \mathrm{s})^{2} \& =\frac{98}{3}(13-5 \sin heta) \mathrm{ft} / \mathrm{s}^{2}\end{aligned}[/latex]

Assume that the block is at rest with respect to the plate.

[latex]\begin{aligned}& \quad\quad +\swarrow \Sigma F_{x}=m a_{x}: \quad N+W \cos heta =m \frac{ u_{E}^{2}}{ ho} \sin heta \ ext{or}& \quad\quad\quad\quad\quad\quad N =W\left\lgroup-\cos heta+\frac{ u_{E}^{2}}{g ho} \sin heta ight group \& \quad\quad +\searrow \Sigma F_{y}=m a_{y}:-F+W \sin heta =-m \frac{ u_{E}^{2}}{ ho} \cos heta \ ext{or} & \quad\quad\quad\quad\quad\quad F =W\left\lgroup\sin heta+\frac{ u_{E}^{2}}{g ho} \cos heta ight group\end{aligned}[/latex]

(a) We have [latex]\quad heta=80^{\circ}[/latex]

Then

[latex]\begin{aligned}N & =(0.8 \mathrm{\ lb})\left[-\cos 80^{\circ}+\frac{1}{32.2 \mathrm{~ft} / \mathrm{s}^{2}} imes \frac{98}{3}\left(13-5 \sin 80^{\circ} ight) \mathrm{\ ft} / \mathrm{s}^{2} imes \sin 80^{\circ} ight] \& =6.3159 \mathrm{\ lb} \F & =(0.8 \mathrm{\ lb})\left[\sin 80^{\circ}+\frac{1}{32.2 \mathrm{~ft} / \mathrm{s}^{2}} imes \frac{98}{3}\left(13-5 \sin 80^{\circ} ight) \mathrm{\ ft} / \mathrm{s}^{2} imes \cos 80^{\circ} ight] \& =1.92601 \mathrm{\ lb}\end{aligned}[/latex]

Now [latex]\quad F_{\max }=\mu_{s} N=0.35(6.3159 \mathrm{\ lb})=2.2106 \mathrm{\ lb}[/latex]

The block does not slide in the slot, and

[latex]\mathbf{F}=1.926 \mathrm{\ lb} \ ⦩ 80^{\circ}\blacktriangleleft[/latex]

(b) We have [latex]\quad heta=40^{\circ}[/latex]

Then

[latex]\begin{aligned}N & =(0.8 \mathrm{\ lb})\left[-\cos 40^{\circ}+\frac{1}{32.2 \mathrm{\ ft} / \mathrm{s}^{2}} imes \frac{98}{3}\left(13-5 \sin 40^{\circ} ight) \mathrm{\ ft} / \mathrm{s}^{2} imes \sin 40^{\circ} ight] \& =4.4924 \mathrm{\ lb} \F & =(0.8 \mathrm{\ lb})\left[\sin 40^{\circ}+\frac{1}{32.2 \mathrm{\ ft} / \mathrm{s}^{2}} imes \frac{98}{3}\left(13-5 \sin 40^{\circ} ight) \mathrm{\ ft} / \mathrm{s}^{2} imes \cos 40^{\circ} ight] \& =6.5984 \mathrm{\ lb}\end{aligned}[/latex]

Now

[latex]\begin{gathered}F_{\max }=\mu_{s} N, ext { from which it follows that } \F>F_{\max }\end{gathered}[/latex]

Block E will slide in the slot

and

[latex]\begin{aligned}\mathbf{a}_{E} & =\mathbf{a}_{n}+\mathbf{a}_{E / ext { plate }} \& =\mathbf{a}_{n}+\left(\mathbf{a}_{E / ext { plate }} ight)_{t}+\left(\mathbf{a}_{E / ext { plate }} ight)_{n}\end{aligned}[/latex]

At [latex]t=0[/latex], the block is at rest relative to the plate, thus [latex]\left(\mathbf{a}_{E / p ext { plate }} ight)_{n}=0[/latex] at t = 0, so that [latex]\mathbf{a}_{E / ext { plate }}[/latex] must be directed tangentially to the slot.

[latex]\begin{aligned}& +\swarrow \Sigma F_{x}=m a_{x}: \quad N+W \cos 40^{\circ} =m \frac{ u_{E}^{2}}{ ho} \sin 40^{\circ} \ ext{or } & \quad\quad\quad\quad N =W\left\lgroup-\cos 40^{\circ}+\frac{ u_{E}^{2}}{g ho} \sin 40^{\circ} ight group ext { (as above) } \& \quad\quad\quad\quad\quad =4.4924 \mathrm{\ lb}\end{aligned}[/latex]

Sliding:

[latex]\begin{aligned}F & =\mu_{k} N \& =0.25(4.4924 \mathrm{\ lb}) \& =1.123 \mathrm{\ lb}\end{aligned}[/latex]

Noting that [latex]\mathbf{F}[/latex] and [latex]\mathbf{a}_{E / ext { plane }}[/latex] must be directed as shown (if their directions are reversed, then [latex]\Sigma \mathbf{F}_{x}[/latex] is [latex]\searrow[/latex]while [latex]m \mathbf{a}_{x}[/latex] is [latex] warrow)[/latex], we have

the block slides downward in the slot and

[latex]\mathbf{F}=1.123\mathrm{\ lb}\ ⦩40^{\circ}\blacktriangleleft[/latex]

Alternative solutions.

(a) Assume that the block is at rest with respect to the plate.

[latex]\Sigma \mathbf{F}=m \mathbf{a}: \quad \mathbf{W}+\mathbf{R}=m \mathbf{a}_{n}[/latex]

Then

[latex]\begin{aligned} an \left(\phi-10^{\circ} ight) & =\frac{W}{m a_{n}}=\frac{W}{\frac{W}{g} \frac{ u_{E}^{2}}{ ho}}=\frac{g}{ ho\left(\dot{\phi}_{A B C D} ight)^{2}} \& =\frac{32.2 \mathrm{\ ft} / \mathrm{s}^{2}}{\frac{98}{3}\left(13-5 \sin 80^{\circ} ight) \mathrm{\ ft} / \mathrm{s}^{2}} & ext{(from above)}\end{aligned}[/latex]

or [latex]\quad\phi-10^{\circ}=6.9588^{\circ}[/latex]

and [latex]\quad\phi=16.9588^{\circ}[/latex]

Now [latex]\quad an \phi_{s}=\mu_{s} \quad \mu_{s}=0.35[/latex]

so that [latex]\quad\phi_{s}=19.29^{\circ}[/latex]

[latex]0<\phi<\phi_{s} \Rightarrow[/latex]Block does not slide and [latex]\mathbf{R}[/latex] is directed as shown.

Now [latex]\quad F=R \sin \phi \quad ext { and } \quad R=\frac{W}{\sin \left(\phi-10^{\circ} ight)}[/latex]

Then

[latex]\begin{aligned}F & =(0.8 \mathrm{\ lb}) \frac{\sin 16.9588^{\circ}}{\sin 6.9588^{\circ}} \& =1.926 \mathrm{\ lb}\end{aligned}[/latex]

The block does not slide in the slot and

[latex]\mathbf{F}=1.926\mathrm{\ lb}\ ⦩80^{\circ}\blacktriangleleft[/latex]

(b) Assume that the block is at rest with respect to the plate.

[latex]\Sigma \mathbf{F}=m \mathbf{a}: \quad \mathbf{W}+\mathbf{R}=m \mathbf{a}_{n}[/latex]

From Part a (above), it then follows that

[latex] an \left(\phi-50^{\circ} ight)=\frac{g}{ ho\left(\dot{\phi}_{A B C D} ight)^{2}}=\frac{32.2 \mathrm{\ ft} / \mathrm{s}^{2}}{\frac{98}{3}\left(13-5 \sin 40^{\circ} ight) \mathrm{\ ft} / \mathrm{s}^{2}}[/latex]

or [latex]\quad\phi-50^{\circ}=5.752^{\circ}[/latex]

and [latex]\quad\phi=55.752^{\circ}[/latex]

Now [latex]\quad\phi_{s}=19.29^{\circ}[/latex]

so that [latex]\quad \phi>\phi_{s}[/latex]

The block will slide in the slot and then

[latex]\begin{aligned}& \quad\quad\phi =\phi_{k}, \quad ext { where } \quad an \phi_{k}=\mu_{k} \quad \mu_{k}=0.25 \ ext{or} & \quad\quad \phi_{k} =14.0362^{\circ}\end{aligned}[/latex]

To determine in which direction the block will slide, consider the free-body diagrams for the two possible cases.

Now [latex]\quad\Sigma \mathbf{F}=m \mathbf{a}: \quad \mathbf{W}+\mathbf{R}=m \mathbf{a}_{n}+m \mathbf{a}_{E / ext { plate }}[/latex]

From the diagrams it can be concluded that this equation can be satisfied only if the block is sliding downward. Then

[latex]+\swarrow \Sigma F_{x}=m a_{x}: \quad W \cos 40^{\circ}+R \cos \phi_{k}=m \frac{ u_{E}^{2}}{ ho} \sin 40^{\circ}[/latex]

Now [latex]\quad F=R \sin \phi_{k}[/latex]

Then [latex]\quad W \cos 40^{\circ}+\frac{F}{ an \phi_{k}}=\frac{W}{g} \frac{ u_{E}^{2}}{ ho} \sin 40^{\circ}[/latex]

or

[latex]\begin{aligned}F & =\mu_{k} W\left\lgroup-\cos 40^{\circ}+\frac{ u_{E}^{2}}{g ho} \sin 40^{\circ} ight group \& =1.123 \mathrm{\ lb} \quad ext { (see the first solution) }\end{aligned}[/latex]

The block slides downward in the slot and

[latex]\mathbf{F}=1.123\mathrm{\ lb}\ ⦩40^{\circ}\blacktriangleleft[/latex]

Question 12.60: A semicircular slot of 10-in. radius is cut in a flat plate ...

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