Raising a bag with a rope

[Santilopez10]

Homework Statement:

A boy decides to hang its bag from a tree branch, with the purpose of raising it. The boy walks with constant velocity in the x axis, and the bag moves in the y axis only. Taking into account that the lenght of the rope is constant and that we know the height h of the branch respect to the floor:
a) find the acceleration of the bag

Relevant Equations:

kinematic equations

So what I did was at first consider the case the kid is below the branch, so that x=0,t=0, then I thought that the length L of the rope should be $L=2h$ because we know the radius from the branch to the kid is just $x^2+y^2=r^2$ and when x=0, y=h. So then I wrote the motion equations for the bag: $$a(t)=a-g$$ $$x(t)=\frac{(a-g)t^2}{2}$$
And I want to find the time t so that the bag is at the branch, then x(t)=h: $$h=\frac{(a-g)t^2}{2} \rightarrow \sqrt{\frac{2h}{a-g}}=t$$
At this moment, $x=v_0 \sqrt{\frac{2h}{a-g}}$ then we have that the length of the rope is $h^2+(v_0 \sqrt{\frac{2h}{a-g}})^2=4h^2$ and we find that $$a= \frac{2{v_0}^2}{3h}+g$$
Is this correct?

 

[Emmo Amaranth]

If this is all the information you are given in the problem, then you cannot assume the length of the rope. Furthermore, the question appears to be asking you for the total acceleration of the bag in this reference frame. Think: if the boy stands still (holding the rope), what is the total acceleration? Then, as the boy starts moving, where does the change in motion come from? Is it necessary to account for gravity when calculating the total acceleration the bag undergoes?

 

[Orodruin]

No. This is a purely geometrical problem. It has nothing to do with gravitational acceleration. What you need to do is to express the position of the bag (which is y, not x) as a function of the position of the boy, which is x. Differentiating that twice with reapect to time will give you the acceleration of the bag.

 

[Santilopez10]

No. This is a purely geometrical problem. It has nothing to do with gravitational acceleration. What you need to do is to express the position of the bag (which is y, not x) as a function of the position of the boy, which is x. Differentiating that twice with reapect to time will give you the acceleration of the bag.

Using your notation:
$$y(x)=\frac{(a-g)(\frac{x}{v_0})^2}{2}$$ then $y''(x)=\frac{a-g}{{v_0}^2}$?

 

[Orodruin]

You have not defined $a$, what is it supposed to be? And again, the gravitational acceleration is completely irrelevant here - it is not a dynamics problem but dow to pure geometry. You have also not provided your argument so it is impossible to know where your misunderstanding starts.

 

[Santilopez10]

You have not defined $a$, what is it supposed to be? And again, the gravitational acceleration is completely irrelevant here - it is not a dynamics problem but dow to pure geometry. You have also not provided your argument so it is impossible to know where your misunderstanding starts.

$a$ is how I label the acceleration of the bag. I know that the position of the kid $x(t)=v_0t \rightarrow t=\frac{x}{v_0}$ and substituting in the y(t) equation I get $$y(x)=\frac{a x^2}{2 {v_o}^2}$$
This is what you mean?

 

[haruspex]

$a$ is how I label the acceleration of the bag. I know that the position of the kid $x(t)=v_0t \rightarrow t=\frac{x}{v_0}$ and substituting in the y(t) equation I get $$y(x)=\frac{a x^2}{2 {v_o}^2}$$
This is what you mean?

That cannot be right.
It is always a good idea to check whether a formula works in easy special cases. At x=0 you should end up with centripetal acceleration, v²/h.

Let the rope length be L. What is the relationship between x, L, h and y?

 

[Santilopez10]

That cannot be right.
It is always a good idea to check whether a formula works in easy special cases. At x=0 you should end up with centripetal acceleration, v²/h.

Let the rope length be L. What is the relationship between x, L, h and y?

$L=(h-y)+\sqrt{x^2+(h-y)^2}$

 

[Orodruin]

$L=(h-y)+\sqrt{x^2+(h-y)^2}$

This is incorrect. Please explain your reasoning. Drawing a picture might help.

 

[Santilopez10]

This is incorrect. Please explain your reasoning. Drawing a picture might help.

you are correct, $L=(h-y)+\sqrt{x^2+h^2}$

 

[Santilopez10]

That cannot be right.
It is always a good idea to check whether a formula works in easy special cases. At x=0 you should end up with centripetal acceleration, v²/h.

Let the rope length be L. What is the relationship between x, L, h and y?

Okay, using the relationship between L and the given data I found what you were saying, at x=0 $a=\frac{{v_0}^2}{h}$ still I am confused, because the acceleration changes as a function of time when it is supposed to be constant all along (Right?).

 

[haruspex]

it is supposed to be constant all along (Right?).

No, the acceleration is certainly not constant. I'm not sure whether you are supposed to find it as a function of time, as a function of x, or just its initial value.
Have you quoted the question exactly as given to you? Was there a diagram?

 

[Santilopez10]

No, the acceleration is certainly not constant. I'm not sure whether you are supposed to find it as a function of time, as a function of x, or just its initial value.
Have you quoted the question exactly as given to you? Was there a diagram?

I guess you are right, it only specifies that the kid moves with constant velocity. I thought that because the kid had 0 acceleration then the bag would too but it seems like it does not!