# Homework Help: Relative frame

1. Oct 10, 2016

### ChloeYip

1. The problem statement, all variables and given/known data
When a cart travel to the left at v=-10m/s, the angle of rain come towards it is 50o from normal.
When the cart travel to the right at v=10m/s, the angle of rain water come is perpendicular to it.
What is the angle of rain come to the earth?

2. Relevant equations
relative frame,relative velosity, relative angle

3. The attempt at a solution
i have no idea...
thanks

2. Oct 10, 2016

### BvU

3. Oct 10, 2016

### ChloeYip

I dont even know how to draw except two separate diagram... which seems really meaningless...
How should i draw? Thanks.

4. Oct 10, 2016

### Empirical Wizardry

Hi Chloe, BvU is right, start by drawing a picture.
In this picture:

Define your ground as a stationary point of reference.

The rain will strikes the ground at some angle which we need to solve.

We first need to solve for the x and y velocity components of the rain relative to the ground (our stationary point). If we take the inverse tan of the ratio of these components we can get the angle we’re looking for, and this will be our last step.

So what information are we given? We know the angle of the rain striking an object (the cart) that is moving relative to the ground.
Since the cart moves either left or right, we must minus this velocity component of the cart to the horizontal (x) velocity component of the rain (we are finding the horizontal velocity of the rain with respect to the cart. This gives the general equation we need to use to solve for the velocity components of the rain relative to the ground:

Equation 1: tan(θ) = (vy, rain)/((vx, rain) - (vx, cart))
When moving at 10 m/s to the right we have, we have θ= 90 degrees, use this information to solve for (vx, rain).

See if you can use this information to solve the problem. Hope this helps!

Last edited: Oct 10, 2016
5. Oct 10, 2016

### ChloeYip

This is how i do with reference to Empirical Wizardry...
Hope i didn't misunderstand what u mean...
To be honest, it is quite abstract to imagine the tan theta part...
However, i got a different answer from you... can you share how you do in a more detail way please? Or point out what i have done wrong please? Thank you.

Ps I haven't got the model answer..

Last edited: Oct 10, 2016
6. Oct 10, 2016

### Empirical Wizardry

Ok no problem. Good attempt!
You have used 10cos(50) to get the horizontal velocity. This requires that the velocity of the rain vector to be 10 m/s. How did you determine the rain vector to be 10 m/s ?
The diagram below shows what you have calculated:

We don't know that the magnitude rain velocity vector is 10 m/s We know the cart has a velocity of -10 m/s, in this case where it is moving to the left.

Maybe we should try the question like this:

First of all let's ignore the ground entirely. All we worry about is the cart and the rain. We have 2 cases:

(1) The cart is moving to the left.
(2) The cart is moving to the right.

For case 1. If the cart is our reference point, we set the velocity of the cart to 0 m/s. For case 1 this means that we have to add 10 m/s to all horizontal velocities in our system (we get: -10 m/s + 10 m/s = 0 m/s). How does this affect the horizontal velocity of the rain vector?

For case 2 this means we have to subtract 10 m/s to all horizontal velocities in our system. Same idea as case 1. How does this affect the horizontal velocity of the rain vector?

Try and see if the equation: net horizontal velocity of rain = Vx, rain - Vx, cart makes sense (i got this wrong earlier sorry for that, I had Vx, rain + Vx, cart, I have edited the post since).

Try to understand this part first. Whenever we take something as a reference point we 'zero' it.

Using case 2 you should be able to obtain the horizontal component of the rain velocity vector, think about how you would do this from the angle given (90 degrees).

After you get the horizontal component try to solve for the vertical component of the rain velocity vector using case 1.

#### Attached Files:

File size:
1.7 KB
Views:
70
Last edited: Oct 10, 2016
7. Oct 10, 2016

### ChloeYip

I thought only the rain is always moving in the same Vx in same direction (positive) as the cart would give 90o strike, isn't it?

Do you mean both rain and cart both move to the right at v=20m/s at this case if use cart in case 1 as reference point?

I don't understand what you mean by here...

Thanks.

8. Oct 10, 2016

### Empirical Wizardry

I thought only the rain is always moving in the same Vx in same direction (positive) as the cart would give 90o strike, isn't it?

(1) Yep you're right, the rain moves with the same velocity as the cart in the 90 degree case. I may have made things more complicated than they need to be.
So if the cart has a velocity of 10 m/s, so does the rain. This is the horizontal component of the velocity of the rain only.

Do you mean both rain and cart both move to the right at v=20m/s at this case if use cart in case 1 as reference point?
(2) Yes the cart is the reference point. So if the rain is moving 10 m/s to the right, and the cart is moving 10 m/s to the left. Then the rate at which one cart moves (from one cart's perspective) from the other cart is 20 m/s (think of two people walking away from each other at the same speed.)

I don't understand what you mean by here...
(3) Yep you already figured it out in your first comment. Apologies again for over-complicating it! Basically if you plug in (vx, cart)) = 10 m/s into tan(90 degrees) = (vy, rain)/((vx, rain) - (vx, cart)) you will see that vx, rain) needs to equal 10 m/s. So don't worry about this last point

IMPORTANT NOTE: I totally missed that its the angle from the normal. You're right it should definitely be 40 degrees. That should replace the 50 degrees in the diagram I had earlier above. Apologies!!

9. Oct 10, 2016

### ChloeYip

So how can we find out the angle by this method?
thanks

10. Oct 10, 2016

### Empirical Wizardry

To clarify you want the angle of the rain with the ground at this point right?

11. Oct 10, 2016

### ChloeYip

yes, how can i find the angle of the rain with the ground?
thanks

12. Oct 10, 2016

### Empirical Wizardry

So we know that Vx, rain = 10 m/s, and θ = 40 degrees.

Using the formula:
tan(θ) = (vy, rain)/((vx, rain) - (vx, cart))
This equation relates the angle of the rain with respect to the car to the velocity components of the rain relative to the car.
We should get 20*tan(40 degrees) = (vy, rain)

After figuring out both (vx, rain) and (vy, rain), which are the velocities relative the the ground we can proceed to solve for theta. We take arctan or inverse tan of the ratio (vy, rain)/(vx, rain). For θ = 40 degrees this works out to be 59.2 degrees (note: I used 50 degrees earlier so I got 62.7 degrees before editing it out in my original response)

Hope this helps!

13. Oct 11, 2016

### ChloeYip

How can you come up with "
tan(θ) = (vy, rain)/((vx, rain) - (vx, cart)" ?
I thought v(rain)= sqrt [v(x,rain)^2 + v(y,rain)^2] by pyth theorem... why need to minus vx,cart?

14. Oct 11, 2016

### Empirical Wizardry

(1) List out all the variables, and fill out the values for the variables you know and don't know.
Write any equations relating the variables to each other. Specifically focus on writing equations relating unknown variables to known variables.

Then...
Draw a right angle triangle and make sure you know the general answers to:
sin(θ) =
cos(θ) =
tan(θ) =

(3) Use Pythagoras' theorem and then think about what information that gives you. What does it compute? How does this help answer the question?

(4) Review how vectors are added and subtracted, particular focus on vectors in the same direction.

(5) Try to define the velocities of the ground/earth, the cart and the rain on the Cartesian plane. What does the 'origin' correspond to?

If you can do all of that you should be able to come up with a solution.

15. Oct 11, 2016

### ChloeYip

Basically, i can answer most of the above question , except dont know how can you come up with How can you come up with " (vx, rain) - (vx, cart)" ?
Thank you