You are Here: Home >< Maths

# M1 😭😭 watch

1. Hey guys
Could someone please explain part c??

I understand how to get the angle 90-alpha
But don't understand what to do after that
2. (Original post by CrystalPark)

Hey guys
Could someone please explain part c??

I understand how to get the angle 90-alpha
But don't understand what to do after that
the resultant of the tension (the forces acting on the pulley) bisects the top right angle. Find half the top right angle, resolve the tensions on either side and add them together.

Eg. If the tension was 10 N and the angle at the top right was 60 degrees then you would resolve the tension twice which would give you 2 lots of Tcos30.

2Tcos(half of top right angle)

e.g:
3. (Original post by KloppOClock)
the resultant of the tension (the forces acting on the pulley) bisects the top right angle. Find half the top right angle, resolve the tensions on either side and add them together.

Eg. If the tension was 10 N and the angle at the top right was 60 degrees then you would resolve the tension twice which would give you 2 lots of Tcos30.

2Tcos(half of top right angle)

e.g:
Wow this is definitely much harder than I thought
But thank you once again

If you don't mind me asking, in vectors when do you equal the components and when do you set them to 0? Especially when it's north or west? Do you equate or set to 0??
4. (Original post by CrystalPark)
Wow this is definitely much harder than I thought
But thank you once again

If you don't mind me asking, in vectors when do you equal the components and when do you set them to 0? Especially when it's north or west? Do you equate or set to 0??
Ok lets say that A has velocity of (2i-3j) and starting position of (-2j)
Lets say that B has velocity of (3i+4j) and starting position of (i+j)

To find the difference between them both, find r=r0+vt for both of them then do
OB-OA as you can see why in this diagram. (at the bottom)

Okay so now you know the location of Q in relation to P.

If Q is northwest. south west, etc to P then the angle it makes between the i and j axis will be 45 degrees either side. Therefore when something is northwest, say, its i and j coordinates will be -i and j as they have to have the same length either way.

So if something is north west of something, the i components will be a multiple of -i and j component a multiple of j. they have the same multiple so set the i components = to -m and j components = to m then solve simultaneously to find t, which will be the time it is northwest

if something is north of something else, its I component must be the same. If something is travelling north it must only go up so it only has a j value so i = 0

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: June 7, 2016
Today on TSR

### Buying condoms for the first time - help!

Discussions on TSR

• Latest
Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest

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