In Exercise 9, what would the distance across the top of the worm thread have to be to make d equal to 0 inches?

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In Exercise 9, what would the distance across the top of the worm thread have to be to make d equal to 0 inches?

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Answer 1 · 2018-01-25T08:04:18

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When $d = 0$ $d=0$ inches the situation will be represented by the modified figure as shown above. The distance across the top of the worm thread then will be $P R$ $PR$

Now $P R = 2 P Q = \frac{2 P Q}{O Q} \times O Q$ $PR =2PQ= (2PQ)/(OQ)xxOQ$

$\Rightarrow P R = \frac{2 P Q}{O Q} \times (Q C + C Q)$ $=>PR = (2PQ)/(OQ)xx(QC+CQ)$

$\Rightarrow P R = \frac{2 P Q}{O Q} \times (Q C + \frac{C Q}{C N} \times C N)$ $=>PR = (2PQ)/(OQ)xx(QC+(CQ)/(CN)xxCN)$

$\Rightarrow P R = \frac{2 P Q}{O Q} \times (C N + \frac{C Q}{C N} \times C N)$ $=>PR = (2PQ)/(OQ)xx(CN+(CQ)/(CN)xxCN)$

$\Rightarrow P R = \frac{2 P Q}{O Q} \times C N (1 + \frac{C Q}{C N})$ $=>PR = (2PQ)/(OQ)xxCN(1+(CQ)/(CN))$

$\Rightarrow P R = 2 tan θ \times 0.75 (1 + csc θ)$ $=>PR = 2tanthetaxx0.75(1+csctheta)$

$\Rightarrow P R = 1.5 (tan θ + tan θ \times csc θ)$ $=>PR = 1.5(tantheta+tanthetaxxcsctheta)$

$\Rightarrow P R = 1.5 (tan θ + sec θ)$ $=>PR = 1.5(tantheta+sectheta)$

$\Rightarrow P R = 1.5 ({tan 15}^{\circ} + {sec 15}^{\circ}) \approx 1.95$ $=>PR = 1.5(tan15^@+sec15^@)~~1.95$ inch

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In Exercise 9, what would the distance across the top of the worm thread have to be to make d equal to 0 inches?

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