# Finding the Acceleration of a Bag Being Raised by a Rope on a Moving Kid

• Santilopez10
In summary, the question asks for the total acceleration of the bag in reference to the boy's position, which can be found by expressing the position of the bag as a function of the boy's position and differentiating twice with respect to time. The gravitational acceleration is irrelevant in this purely geometrical problem.
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?

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?

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.

Orodruin said:
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.
$$y(x)=\frac{(a-g)(\frac{x}{v_0})^2}{2}$$ then ##y''(x)=\frac{a-g}{{v_0}^2}##?

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.

Orodruin said:
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?

Santilopez10 said:
##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?

haruspex said:
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}##

Santilopez10 said:
##L=(h-y)+\sqrt{x^2+(h-y)^2}##
This is incorrect. Please explain your reasoning. Drawing a picture might help.

Orodruin said:
This is incorrect. Please explain your reasoning. Drawing a picture might help.
you are correct, ##L=(h-y)+\sqrt{x^2+h^2}##

haruspex said:
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?).

Santilopez10 said:
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?

haruspex said:
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!

## 1. How do I securely tie the rope to the bag?

The best way to securely tie the rope to the bag is to use a double knot. Start by tying a regular knot with the two ends of the rope. Then, repeat the process by tying another knot on top of the first one. This will create a strong and secure hold on the bag.

## 2. What type of rope is best for raising a bag?

The type of rope you should use for raising a bag depends on the weight of the bag. For lighter bags, a nylon rope is recommended as it is strong and durable. For heavier bags, a thicker and stronger rope such as a polypropylene rope may be necessary.

## 3. How high can I safely raise a bag with a rope?

The safe height for raising a bag with a rope varies depending on the type of rope and the weight of the bag. Generally, a rope can safely support up to 10 times its own weight. It is important to always check the weight limit of the rope and ensure it is securely tied before raising the bag.

## 4. Can I use the same rope for multiple bag raises?

Yes, you can use the same rope for multiple bag raises as long as it is in good condition and can support the weight of the bag. It is important to regularly check the rope for any signs of wear and tear and replace it if necessary.

## 5. How do I safely lower the bag after raising it with a rope?

To safely lower the bag, slowly release the rope while maintaining a firm grip on it. Make sure the bag is securely attached to the rope before lowering it. It is also helpful to have someone assist you in guiding the bag down to ensure a controlled descent.

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