Finding the direction on a second displacement

AI Thread Summary
The discussion revolves around solving a physics problem involving two displacements and a resultant vector. The first displacement is 150 cm at 130° to the positive x-axis, and the resultant displacement is 136 cm at 40° to the positive x-axis. Participants emphasize the importance of visualizing the vectors correctly and suggest drawing them from the same point to understand their relationship. The confusion primarily stems from determining the direction of the second displacement after calculating its magnitude. Ultimately, the conversation highlights the need for clarity in vector addition and the use of trigonometric relationships to find angles.
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Homework Statement



A man pushing a mop across a floor causes it to undergo two displacements. The first has a magnitude of 150 cm and makes an angle of 130° with the positive x axis. The resultant displacement has a magnitude of 136 cm and is directed at an angle of 40.0° to the positive x axis. Find the magnitude and direction of the second displacement.

The Attempt at a Solution



I was able to figure out the magnitude through trial and error. I thought because it says to find the second displacement, and not the resultant displacement that Pythagoreans formula would not work, but it did. sqrt(150^2+136^2) gave me 202.5cm and the online homework accepted it as the right answer.

Because it did accept it as the right answer, I'm very confused. I thought the resultant displacement was the net displacement of one and two (the hypotenuse), but since the Pythagorean formula worked, I can't visualize what the displacements should look like in my head. I tried creating an arbitrary triangle and setting opposite equal to 130 and adjacent to 150 and then using arctan to find the angle, but it came back as incorrect. Now I have no idea how to find the direction of the 202.5 magnitude displacement.

Please help! I refuse to stop for the night until I figure this problem out.
 
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First draw a picture... you should know what the vectors look like...

Did you draw the original vector... and the resultant vector?
 
I tried to draw a picture, but the problem description is really throwing me off. Also, I don't have a protractor.

But even with a picture, I don't know how to find the angle.
 
Imperator said:
I tried to draw a picture, but the problem description is really throwing me off. Also, I don't have a protractor.

But even with a picture, I don't know how to find the angle.

Yes, but you need the picture correctly, so that you know which angle to find...

You don't need a protractor... just sketch it with north south east west axes... Draw the two vectors given in the problem at 130 degrees and 40 degrees...
 
Here is what is throwing me off, we have learned to draw the vectors in class, starting with 1, and drawing the next vector off the first vectors tail. But as I understand the problem, it gives the first vector, and the resultant vector. So do they both originate from the same point?
 
Imperator said:
Here is what is throwing me off, we have learned to draw the vectors in class, starting with 1, and drawing the next vector off the first vectors tail. But as I understand the problem, it gives the first vector, and the resultant vector. So do they both originate from the same point?

If you draw the two vectors from the same point... then suppose you draw a vector, from where the original vector ends... to where the resultant ends... (ie the arrowhead of the original vector... to the arrowhead of the resultant)

Do you agree that the original vector plus this new vector... add to the resultant vector? So this new vector is the vector you need.

If you find this approach confusing... here's another way to go about it...

you have two vectors that add to the resultant: \vec{a} + \vec{b} = \vec{r}

You are given a and r... you need to find b. So solve for b in the equation:

\vec{b} = \vec{r} - \vec{a}

or in other words:

\vec{b} = \vec{r} + (-\vec{a})

So drawing \vec{r}... and adding to it -\vec{a} will also give you the result.
 
Ok, I have the picture, that makes sense now. Forgive me for being dense, but I still can't see how this helps me find the direction (angle) of the second vector in relation to the x-axis.
 
Imperator said:
Ok, I have the picture, that makes sense now. Forgive me for being dense, but I still can't see how this helps me find the direction (angle) of the second vector in relation to the x-axis.

You can calculate that angle... Have a look at this sketch:

http://www.imagevimage.com/gallery.php?entry=images/angle.gif

Does the picture make sense? can you find angle x?
 
I can't figure out how to find values for two sides of the triangle it creates, so I can't find the angle. Is there another way I am unaware of?
 
  • #10
Imperator said:
I can't figure out how to find values for two sides of the triangle it creates, so I can't find the angle. Is there another way I am unaware of?

What is the angle with the 2 squiggly lines...
 
  • #11
Hi.
I need some help with physics/motion and i wasn't sure where to post.
This was the first thing i saw when i used google.
I am having a really really hard time understanding motion and where to plug things into/ what the formulas are.
I would really appreciate some help if anyone could.
Also please try not to think I'm an idiot i am just learning all of this.
1) What is the formula for resultant displacement *lost my notes :(
2) If the formula is(for resultant displacement) dr= d1+d2 then where do i plug in the directions like north and south
The question is:
A soccer player leaves the bench and runs 35m[N] and then 25m. Find the resultant displacement.
 
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