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HoldingOn · Jul 28, 2018

http://prntscr.com/kc3cpc

I get that you apply the chain rule twice, and use arc length of circle on the winch, but can't get seem to get the second component of that rule.

Also how would you do part iv. In this- http://prntscr.com/kc3pn0

Thanks in advance

integral95 · Jul 28, 2018

HoldingOn said:
http://prntscr.com/kc3cpc

I get that you apply the chain rule twice, and use arc length of circle on the winch, but can't get seem to get the second component of that rule.

Also how would you do part iv. In this- http://prntscr.com/kc3pn0

Thanks in advance

I have no idea what the heck is $\theta$ in your first part

I know that you should to consider $\frac{d \theta}{dt} = \frac{dh}{dt} \frac{dx}{dh} \frac{d \theta}{dx}$

But I can't seem to get an equation of theta in terms of x....

But for your 2nd screenshot, you basically sub x = 12 and y = 0 in ii) then use the inequality in iii) to simplify.

$h + 12tan \theta = \frac{144g}{2V^2cos^2 \theta}$

Hint: $\frac{1}{V^2} < \frac{2sin \theta cos \theta}{9g}$

dan964 · Jul 28, 2018

integral95 said:
I have no idea what the heck is $\theta$ in your first part

I know that you should to consider $\frac{d \theta}{dt} = \frac{dh}{dt} \frac{dx}{dh} \frac{d \theta}{dx}$

But I can't seem to get an equation of theta in terms of x....

But for your 2nd screenshot, you basically sub x = 12 and y = 0 in ii) then use the inequality in iii) to simplify.

$h + 12tan \theta = \frac{144g}{2V^2cos^2 \theta}$

Hint: $\frac{1}{V^2} < \frac{2sin \theta cos \theta}{9g}$

Theta would be the angle of turning of the winch.

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HoldingOn

Active Member

integral95

Well-Known Member

dan964

what

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