Solving 10.0km + 5.0km [E15S] with Cos and Sin Laws

[rayj098]

Homework Statement

10.0km [N] + 5.0km [E15S]

Homework Equations

Cosine Law

The Attempt at a Solution

If you draw it out you see that 10km is 90 degrees north. From the 10km end, you do E15S for 5km.

Therefore, a = 10km, b = 5km, c = ?, C = 90 - 15 = 75.

Therefore:

c^2 = 10^2 + 5^2 - 2(10)(5) Cos 75
c^2 = 99.12
c = 9.9

Sin Law to find the angle of the resultant

Sin A = Sin B
----- ------
a = b

Sin 75 Sin B
------- = -------
9.9 5

B = 29

So final answer is 9.9km [N29E]


Somehow the answer is 4.2[N35E]...

 

[PeterO]

Homework Statement

10.0km [N] + 5.0km [E15S]

Homework Equations

Cosine Law

The Attempt at a Solution

If you draw it out you see that 10km is 90 degrees north. From the 10km end, you do E15S for 5km.

Therefore, a = 10km, b = 5km, c = ?, C = 90 - 15 = 75.

Therefore:

c^2 = 10^2 + 5^2 - 2(10)(5) Cos 75
c^2 = 99.12
c = 9.9

Sin Law to find the angle of the resultant

Sin A = Sin B
----- ------
a = b

Sin 75 Sin B
------- = -------
9.9 5

B = 29

So final answer is 9.9km [N29E]


Somehow the answer is 4.2[N35E]...

Your answer is the correct answer to the problem you presented.

Either you miss read [and posted] the question, or you have compared your answer to the answer of a different question.

Even if you went 10km North then 5km due South you would still be 5 km from you starting point. There is no way you finish within 4.2 km of your starting point!

 

[rayj098]

Your answer is the correct answer to the problem you presented.

Either you miss read [and posted] the question, or you have compared your answer to the answer of a different question.

Even if you went 10km North then 5km due South you would still be 5 km from you starting point. There is no way you finish within 4.2 km of your starting point!

Thanks. I also was confused how the displacement would be less than 5km.


Appreciate it for clarifying!


How would you do this one then?

10km [N25E] + 5km [E]?

 

[SammyS]

One vector is due north. The other is 15° South of due East. The angle made by the direction of the vectors is 105°.

cos(105°) = -cos(75°)

 

[PeterO]

One vector is due north. The other is 15° South of due East. The angle made by the direction of the vectors is 105°.

cos(105°) = -cos(75°)

When you add vectors you connect them head to to tail - that is why the angle is indeed 75° NOT 105°.

Sure the vectors separately will make an angle of 105°, if they have a common start point, but that is not how you add vectors.

 

[PeterO]

Thanks. I also was confused how the displacement would be less than 5km.


Appreciate it for clarifying!


How would you do this one then?

10km [N25E] + 5km [E]?

Do i the same way you did the previous question. You will be working with an angle bigger than 90° this time.

 

[PeterO]

Thanks. I also was confused how the displacement would be less than 5km.


Appreciate it for clarifying!


How would you do this one then?

10km [N25E] + 5km [E]?

You can use components of each vector if you wish.

10km [N25E] has a North component of about 9, and an East component about 4.2

When you add components your final vector will have components about 9 North and 9,2 East.

The final vector will be about 12.8 [N46E]

All those figures I approximated - I even guessed 46 for the angle, knowing that of the components had been 9 N and 9 East the angle would have been 45, so I just went a little bit bigger.
To get accurate components you use trig. and Pythagoras.

 

[SUCRALOSE]

How the hell do you find this angle? see attached pic

says: determine the angle for which the resultant of the three forces is vertical.

 

[PeterO]

How the hell do you find this angle? see attached pic

says: determine the angle for which the resultant of the three forces is vertical.

When you read the question did you realize that the statement in read means: "there is zero horizontal component to the resultant of the three forces?"

 

[SammyS]

How the hell do you find this angle? see attached pic

says: determine the angle for which the resultant of the three forces is vertical.

Hello SUCRALOSE . Welcome to PF !

You really should have posted this question in a new thread, rather than "hijacking" an existing link.

 

[SUCRALOSE]

was not my intention to take over this thread, I start a new one. and I still do not get it.

 

[PeterO]

was not my intention to take over this thread, I start a new one. and I still do not get it.

Can you answer my earlier question about what you saw in the wording please.