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a) $F_{up}=1.10gM$
b) $W=1.1gMh$

Giancoli 6th Edition, Chapter 6, Problem 9

(2:30)

Chapter 6, Problem 9 is solved.

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Transcript for this Giancoli solution

Lets draw the helicopter first and then we'll be able to put some force vectors on there and we'll know how to calculate the net force relate the to acceleration. So there is the helicopter and we have the force up provided by the spinning rotors and gravity is pulling it down, gravity is going to be a little shorter because it has a net force up; its accelerating upwards. Let's write down the net force, it's the up force minus the down force, which is gravity, and that net force is equal mass times acceleration and we are told that's 'M', which is the mass of this helicopter times 0.1 'g's' so that's the acceleration. Then we can solve for the up force so 'Fup' equals 0.1 'g' times mass of the helicopter plus 'Mg' that's gravity. Then we can simplify that a little bit by factoring up the capital 'M' 0.1 'g' plus 'g' and we'll make that one look a little nicer even 1.10 'g' times 'M' that's what the force up is

For part B we'll find the work done by that force and it's going to be that force times the distance over which it acts which is 1.10 'g' times 'M', that's the up force and the distance that it [enacts] is the height 'h' that the helicopter is raised. So this is the work done by the rotors; the force created by the rotors.

Comments

the site is not working. It keeps saying server not available

Yes, sorry about that Smoothforu, we resized our server this morning to handle more visitors. The site was unresponsive for about 5min, but now it's a little bit faster.

Can you explain how M(0.10g) is equal to Force up - Force mg = ma?
I get the equation: F=ma and sum of F =Fup-Fmg=ma. I don't get how M(0.10g) is equal to all that.

Thanks

Hi trandzuyhieu, thanks for the question. $M(0.10g)$ replaces the $ma$ in $F_{up} - mg = ma$. Data from the question say the helicopter mass is $M$, and it's acceleration is $(0.10g)$, so we substitute for each of these in $ma$: $M$ replaces $m$, and $(0.10g)$ replaces $a$.

All the best,
Mr. Dychko