Problem

The elements of a simplified clam-shell bucket for a dredge are shown. With the block at...

The elements of a simplified clam-shell bucket for a dredge are shown. With the block at O considered fixed and with the constant velocity of the control cable at C equal to 0.5 m/s determine the angular acceleration of the right-hand bucket jaw when as the bucket jaws are closing.

Step-by-Step Solution

Solution 1

In \(\triangle \mathrm{OBC}\), apply sine law and calculate the angle made by link \(O B\).

$$ \begin{aligned} \frac{\overline{B C}}{\sin \beta} &=\frac{\overline{O B}}{\sin 67.5^{\circ}} \\ \sin \beta &=\frac{0.5 \times \sin 67.5^{\circ}}{0.6} \\ \beta &=\sin ^{-1}\left(\frac{0.5 \times \sin 67.5^{\circ}}{0.6}\right) \\ &=50.34^{\circ} \end{aligned} $$

Determine the angle \(\angle O B C\) using relation,

$$ \begin{aligned} \angle O B C &=180^{\circ}-\left(67.5^{\circ}+50.34^{\circ}\right) \\ &=62.16^{\circ} \end{aligned} $$

Apply sine law and calculate the magnitude \(\overline{O C}\).

$$ \begin{aligned} \frac{\overline{O C}}{\sin 62.16^{\circ}} &=\frac{O B}{\sin 67.5^{\circ}} \\ \overline{O C} &=\frac{0.6 \times \sin 62.16^{\circ}}{\sin 67.5^{\circ}} \\ &=0.5742 \mathrm{~m} \end{aligned} $$

Calculate the magnitude for \(\overline{C C_{1}}\) and \(\overline{O C_{1}}\) from \(\Delta O C C_{1}\) as follows:

$$ \begin{aligned} \overline{C C_{1}} &=\overline{O C} \tan \beta \\ &=0.5742 \times \tan 50.34^{\circ} \\ &=0.6926 \mathrm{~m} \end{aligned} $$

Determine the magnitude of \(\left(\overline{O C_{1}}\right)\).

$$ \begin{aligned} \overline{O C_{1}} &=\frac{\overline{O C}}{\cos \beta} \\ &=\frac{0.5742}{\cos 50.34^{\circ}} \\ &=0.8996 \mathrm{~m} \end{aligned} $$

From geometry calculate the length of \(\overline{B C_{1}}\) as follows:

$$ \begin{aligned} \overline{B C_{1}} &=\overline{O C_{1}}-\overline{O B} \\ &=(0.8996-0.6) \\ &=0.2996 \mathrm{~m} \end{aligned} $$

Calculate the angular velocity of bucket \(\left(\omega_{B C}\right)\) using relation,

$$ \omega_{B C}=\frac{v_{e}}{\overline{C C_{1}}} $$

Substitute, \(0.5 \mathrm{~m} / \mathrm{s}\) for \(v_{C}\) and \(0.6926 \mathrm{~m}\) for \(\overline{C C_{1}}\).

$$ \begin{aligned} \omega_{B C} &=\frac{0.5}{0.6926} \\ &=0.7219 \mathrm{rad} / \mathrm{s} \end{aligned} $$

Determine the linear velocity at point \(B\) as follows:

$$ \begin{aligned} v_{\bar{B}} &=\omega_{B C} \times \overline{B C_{1}} \\ &=0.7219 \times 0.2996 \\ &=0.2162 \mathrm{~m} / \mathrm{s} \end{aligned} $$

Determine the normal component of acceleration \(\left(a_{B / C}\right)_{n}\) at \(B\) relative to \(C\) using relation,

$$ \left(a_{B / C}\right)_{n}=\left(\omega_{B C}\right)^{2} \times \overline{B C} $$

Substitute, \(0.7219 \mathrm{rad}\) for \(\omega_{B C}\) and \(0.5 \mathrm{~m}\) for \(\overline{B C}\).

$$ \begin{aligned} \left(a_{B C C}\right)_{n} &=(0.7219)^{2} \times 0.5 \\ &=0.2605 \mathrm{~m} / \mathrm{s}^{2} \end{aligned} $$

Determine the normal component of acceleration \(\left(a_{B \dots O}\right)_{n}\) at \(B\) relative to \(O\) using relation,

$$ \begin{aligned} \left(a_{B / O}\right)_{n} &=\frac{v_{B}^{2}}{O B} \\ &=\frac{(0.2162)^{2}}{0.6} \\ &=0.078 \mathrm{~m} / \mathrm{s}^{2} \end{aligned} $$

Determine the acceleration at point \(B\) considering link \(C B\) using relation as follows:

$$ a_{b}=a_{C}+\left(a_{B / C}\right)_{n}+\left(a_{B / C}\right)_{t} $$

Substitute, 0 for \(a_{c}\) in relation.

$$ a_{i k}=\left(a_{B / C}\right)_{n}+\left(a_{B / C}\right)_{t} $$

Determine the acceleration at point \(B\) considering link \(O B\) using relation as follows:

$$ \overline{a_{B}}=\left(\overline{a_{B / O}}\right)_{n}+\left(\overline{a_{B I O}}\right), $$

Figure representing the vector diagram,

C:\Users\pavankishor\Desktop\36b.jpg

Determine the acceleration at point \(B\) for link \(B C\) along \(x\) direction as follows:

$$ \left(a_{B}\right)_{x}=\left(a_{B C C}\right)_{n} \sin 67.5^{\circ}+\left(a_{B C C}\right)_{t} \sin 22.5^{\circ} \ldots \ldots(1) $$

Determine the acceleration at point \(B\) for link \(O B\) along \(x\) direction as follows:

$$ \left(a_{B}\right)_{x}=\left(a_{B I O}\right)_{n} \sin 50.34^{\circ}+\left(a_{B / O}\right)_{t} \sin 39.66^{\circ} \ldots \ldots(2) $$

From equations (1) and (2) determine the relation for tangential acceleration at link \(C B\) and \(O B\).

$$ \left(a_{B / C}\right)_{n} \sin 67.5^{\circ}+\left(a_{B / C}\right), \sin 22.5^{\circ}=\left(a_{B / O}\right)_{n} \sin 50.34^{\circ}+\left(a_{B / O}\right)_{t} \sin 39.66^{\circ} $$

Substitute, \(0.078 \mathrm{~m} / \mathrm{s}^{2}\) for \(\left(a_{B / O}\right)_{n}\) and \(0.2605 \mathrm{~m} / \mathrm{s}^{2}\) for \(\left(a_{B i C}\right)_{n}\) in relation.

$$ \begin{aligned} &(0.2605 \times 0.9238)+\left(\left(a_{B / C}\right)_{t} \times 0.3826\right)=(0.078 \times 0.7698)+\left(a_{B / O}\right)_{t} \times 0.6382 \\ &\left(a_{B / O}\right)_{t}=0.5994\left(a_{B / C}\right)_{t}+0.2829 \end{aligned} $$

Determine the acceleration at point \(B\) for link \(B C\) along \(y\) direction as follows:

$$ \left(a_{B}\right)_{y}=\left(a_{B / C}\right)_{n} \cos 67.5^{\circ}-\left(a_{B / C}\right), \cos 22.5^{\circ} \ldots \ldots(4) $$

Determine the acceleration at point \(B\) for link \(O B\) along \(y\) direction as follows:

$$ \left(a_{B}\right)_{y}=\left(a_{B / O}\right), \cos 39.66^{\circ}-\left(a_{B B}\right)_{n} \cos 50.34^{\circ} \ldots \ldots(5) $$

From equations (4) and (5) determine the relation for tangential acceleration at link \(C B\) and \(O B\).

Substitute, \(0.078 \mathrm{~m} / \mathrm{s}^{2}\) for \(\left(a_{B / O}\right)_{n}\) and \(0.2605 \mathrm{~m} / \mathrm{s}^{2}\) for \(\left(a_{B / C}\right)_{n}\) in relation,

$$ \begin{aligned} &(0.2605 \times 0.3826)-\left(\left(a_{B / C}\right)_{t} \times 0.9238\right)=\left(\left(a_{B / O}\right)_{t} \times 0.7698\right)-(0.078 \times 0.6382) \\ &\left(a_{B / O}\right)_{t}=0.1940-1.2\left(a_{B / C}\right) \end{aligned} $$

From equations \((3)\) and \((6)\), calculate the tangential acceleration of link \(B C\).

$$ \begin{aligned} &0.5994\left(a_{B / C}\right),+0.2829=0.1940-1.2\left(a_{B / C}\right) \\ &\left(a_{B / C}\right)_{t}=-0.0494 \mathrm{~m} / \mathrm{s}^{2} \end{aligned} $$

Calculate the angular acceleration for link \(C B\),

$$ \begin{aligned} \alpha_{B C} &=\frac{\left(a_{B / C}\right)_{1}}{B C} \\ &=\frac{-0.0494}{0.5} \\ &=-0.0986 \end{aligned} $$

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