Turn on thread page Beta
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by sienna2266)
    Actually here is a better diagram to my previous post for ii):
    Could you kindly let me know what would be the next step? Would the resultant velocity be different from part i)? Many thanks
    This isn't the diagram I had in mind, but nevertheless, you may label the angle between 10 and R as \theta, then equate the magnitudes of the vertical components of the 3 and 10, and hence determine \theta, hence determine the leftover angle of the triangle. The overall bearing is the 150+(leftover angle)

    The diagram I had in mind was the following, where application of the sine rule is a more obvious approach:

    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by RDKGames)
    This isn't the diagram I had in mind, but nevertheless, you may label the angle between 10 and R as \theta, then equate the magnitudes of the vertical components of the 3 and 10, and hence determine \theta, hence determine the leftover angle of the triangle. The overall bearing is the 150+(leftover angle)

    The diagram I had in mind was the following, where application of the sine rule is a more obvious approach:

    Thank you!! In your method, how do you find the angle between the 3 component and 10component?
    Name:  64.PNG
Views: 8
Size:  150.9 KB
    I know it looks like a right angle but then again it's a sketch so how do find the angle between the 3 component and the 10 component?
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by sienna2266)
    Thank you!! In your method, how do you find the angle between the 3 component and 10component?
    I know it looks like a right angle but then again it's a sketch so how do find the angle between the 3 component and the 10 component?
    That angle is in terms of \theta, but you don't need to worry about it because you can find the angle between 3 and R instead.
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by RDKGames)
    That angle is in terms of \theta, but you don't need to worry about it because you can find the angle between 3 and R instead.
    I tried to find the angle between 3 and R but the sine and cosine rule didn't seem to work as we don't have enough information. Btw I used your diagram, so the diagram below:
    Also, R from i) is not the same as R from ii),right?
    Name:  64.PNG
Views: 8
Size:  150.9 KB
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by sienna2266)
    I tried to find the angle between 3 and R but the sine and cosine rule didn't seem to work as we don't have enough information. Btw I used your diagram, so the diagram below:
    Also, R from i) is not the same as R from ii),right?
    Yes you do have enough information.

    The angle between 3 and R is 60 degrees and so you employ the sine rule. \dfrac{\sin (270-\theta)}{3} = \dfrac{\sin 60}{10}

    R is indeed different in both both parts.
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by RDKGames)
    Yes you do have enough information.

    The angle between 3 and R is 60 degrees and so you employ the sine rule. \dfrac{\sin (270-\theta)}{3} = \dfrac{\sin 60}{10}

    R is indeed different in both both parts.
    How did you get the angle between 3 and R being 60 degrees?
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by sienna2266)
    How did you get the angle between 3 and R being 60 degrees?
    Mark down a north line at the other end of the resultant vector. The angle between this and 3 is 150. The angle between this north line and R is 90. Subtract one from the other.
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by RDKGames)
    Mark down a north line at the other end of the resultant vector. The angle between this and 3 is 150. The angle between this north line and R is 90. Subtract one from the other.
    Brilliant!Thank you! I got to the answer using your method.
    If we consider the alternative diagram for part ii):
    Attachment 726494
    You kindly advised that if I go about this way: "you may label the angle between 10 and R as , then equate the magnitudes of the vertical components of the 3 and 10, and hence determine , hence determine the leftover angle of the triangle. The overall bearing is the 150+(leftover angle)"
    I don't understand why you equate the magnitudes of the vertical components of the 3 and 10 and how do you determine theta from this?
    Attached Images
     
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by sienna2266)
    Brilliant!Thank you! I got to the answer using your method.
    If we consider the alternative diagram for part ii):
    You kindly advised that if I go about this way: "you may label the angle between 10 and R as , then equate the magnitudes of the vertical components of the 3 and 10, and hence determine , hence determine the leftover angle of the triangle. The overall bearing is the 150+(leftover angle)"
    I don't understand why you equate the magnitudes of the vertical components of the 3 and 10 and how do you determine theta from this?
    \theta is not the angle that answers the question, it is simply a step to get the angle between 10 and 3 which is what we want essentially - this is the leftover angle I'm talking about.

    Note that the resultant motion has no vertical component, so the vertical components of 10 and 3 must cancel each other. Indeed, the vertical components of both are going in opposite directions as required. The other requirement is that their magnitudes are the same. The expression for the vertical component of the 10 contains \theta
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by RDKGames)
    \theta is not the angle that answers the question, it is simply a step to get the angle between 10 and 3 which is what we want essentially - this is the leftover angle I'm talking about.

    Note that the resultant motion has no vertical component, so the vertical components of 10 and 3 must cancel each other. Indeed, the vertical components of both are going in opposite directions as required. The other requirement is that their magnitudes are the same. The expression for the vertical component of the 10 contains \theta
    Thanks!I got to the answer this way too!! Here is my working out:
    Name:  66.PNG
Views: 8
Size:  336.8 KB
    Is my working out appropriate?
    Could you have also considered the horizontal components only to work out theta, if we knew what R is? And could we have considered both the vertical and the horizontal components as a whole to work out theta, if we knew what R is?
    The top bit again clearer:
    Attachment 726502
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by sienna2266)
    Thanks!I got to the answer this way too!! Here is my working out:
    Is my working out appropriate?
    Yes.

    Could you have also considered the horizontal components only to work out theta, if we knew what R is?
    Yes.

    And could we have considered both the vertical and the horizontal components as a whole to work out theta, if we knew what R is?
    Yes, though this is just more work and the alternatives are quicker.
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by RDKGames)
    Yes.



    Yes.



    Yes, though this is just more work and the alternatives are quicker.
    Thank you so much for the help!!!!Really appreciate your time!!!
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: February 20, 2018

University open days

  1. Norwich University of the Arts
    Postgraduate Open Days Postgraduate
    Thu, 19 Jul '18
  2. University of Sunderland
    Postgraduate Open Day Postgraduate
    Thu, 19 Jul '18
  3. Plymouth College of Art
    All MA Programmes Postgraduate
    Thu, 19 Jul '18
Poll
How are you feeling in the run-up to Results Day 2018?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.