You are Here: Home >< Maths

# Easy Mechanics 1 question help Watch

1. Just haven't learned it yet, so Instead of waiting to ask teachers, the amazing TSR forums can help xD

9) A force F acts in the direction of -6i-8j and has a magnitude of 20 N.
Force F is
(a) (-12i-16j)N
(b) (12i+16j)N
(c) (16i-12j)N
(d) (14i-2j)N
Absolutely no idea as I've not got to that part at school yet, and easier to ask you guys. Thanks! Please explain how you got the answer too please!
2. (Original post by joe12345marc)
Just haven't learned it yet, so Instead of waiting to ask teachers, the amazing TSR forums can help xD

9) A force F acts in the direction of -6i-8j and has a magnitude of 20 N.
Force F is
(a) (-12i-16j)N
(b) (12i+16j)N
(c) (16i-12j)N
(d) (14i-2j)N
Absolutely no idea as I've not got to that part at school yet, and easier to ask you guys. Thanks! Please explain how you got the answer too please!
You can find the angle that F acts in by taking arctan(-8/-6) = arctan(4/3).

(Note: this is the angle anticlockwise from the horizontal.)

Using the magnitude, you can use Pythagoras' Theorem to find the i and j components. See?
3. (Original post by SH0405)
You can find the angle that F acts in by taking arctan(-8/-6) = arctan(4/3).

(Note: this is the angle anticlockwise from the horizontal.)

Using the magnitude, you can use Pythagoras' Theorem to find the i and j components. See?
Ah so I work out the resultant direction of the force as a bearing, then use that to work out the i and j components in terms of N? Aah I understand, brilliant thanks! Really appreciate the hints instead of giving the answer
4. (Original post by joe12345marc)
Ah so I work out the resultant direction of the force as a bearing, then use that to work out the i and j components in terms of N? Aah I understand, brilliant thanks! Really appreciate the hints instead of giving the answer
Yep. No problem. What answer do you get?....
5. (Original post by SH0405)
Yep. No problem. What answer do you get?....
Ended up with (b) (12i + 16j)N as I'm not so sure about negatives and positives... But I know the i and j are +/- 12 and +/- 16 respectively.
6. (Original post by joe12345marc)
Ended up with (b) (12i + 16j)N as I'm not so sure about negatives and positives... But I know the i and j are +/- 12 and +/- 16 respectively.
It's the negative one: (-12i-16j)N. This is because we are told the direction in which F acts, which, in layman's terms, is to the left (negative i direction) and down (negative j direction). (Third quadrant).

Hope that's clear! :/
7. (Original post by SH0405)
It's the negative one: (-12i-16j)N. This is because we are told the direction in which F acts, which, in layman's terms, is to the right (negative i direction) and down (negative j direction).

Hope that's clear! :/
Aah, yeah I understand why (I'd think of it better as being in the 3rd quadrant when you put it like that) but thanks I understand now! Really appreciate the help!
8. (Original post by joe12345marc)
Aah, yeah I understand why (I'd think of it better as being in the 3rd quadrant when you put it like that) but thanks I understand now! Really appreciate the help!
I meant to the left by the way. Whoops.
9. Tip: Make a triangle with appropriate side lengths
10. (Original post by SH0405)
I meant to the left by the way. Whoops.
Yeah I figured, as obviously to the right isn't the negative i direction. As soon as you said anything about direction, I got it tbh
11. (Original post by joe12345marc)
Yeah I figured, as obviously to the right isn't the negative i direction. As soon as you said anything about direction, I got it tbh
Ah, good. If there's one thing you need to know in Mechanics, it's you're lefts and rights.
12. (Original post by SH0405)
Ah, good. If there's one thing you need to know in Mechanics, it's you're lefts and rights.
Aha yeah I can tell, such a big factor. It seems to be, no matter how good you are at maths, drawing out the situation always helps xD Do love mechanics so far, hopefully it gets really difficult and interesting before the end of M3!
13. (Original post by joe12345marc)
Aha yeah I can tell, such a big factor. It seems to be, no matter how good you are at maths, drawing out the situation always helps xD Do love mechanics so far, hopefully it gets really difficult and interesting before the end of M3!
You title it easy mechanics yet you need help with it , so is that not a contradiction.
You title it easy mechanics yet you need help with it , so is that not a contradiction.
So I'm not allowed to not know something that is easy? I titled it easy because I know anybody that has done M1 would be able to help. I just haven't learned it yet, I'd just rather learn the proper way of doing it. Sorry, I'll not ask for help next time
You title it easy mechanics yet you need help with it , so is that not a contradiction.

You phrase your post as a question, yet you do not end it with a question mark. So is that not a contradiction?
16. (Original post by SH0405)
You phrase your post as a question, yet you do not end it with a question mark. So is that not a contradiction?
Ooh, I would rep this but I reached my rep limit today.
Ooh, I would rep this but I reached my rep limit today.
There's always tomorrow, my friend. Always tomorrow.

18. (Original post by joe12345marc)
Yeah I figured, as obviously to the right isn't the negative i direction. As soon as you said anything about direction, I got it tbh
There's no need for any angles or trig or drawing triangles in this question

The required force is acting in the direction of the original vector you're given, so it must be a multiple of that force. The magnitude of the original is sqrt(6^2 + 8^2) = 10 and you want a vector of magnitude 20, so all you need to do is multiply the original force by 2 and there's your answer
19. (Original post by davros)
There's no need for any angles or trig or drawing triangles in this question

The required force is acting in the direction of the original vector you're given, so it must be a multiple of that force. The magnitude of the original is sqrt(6^2 + 8^2) = 10 and you want a vector of magnitude 20, so all you need to do is multiply the original force by 2 and there's your answer
Oooohhh my word that makes so much logical sense as well! Thanks a ton man! It's really useful to know both a shortcut and the safe way of doing something like this, forces can be complex later on. Thanks!
20. (Original post by davros)
There's no need for any angles or trig or drawing triangles in this question

The required force is acting in the direction of the original vector you're given, so it must be a multiple of that force. The magnitude of the original is sqrt(6^2 + 8^2) = 10 and you want a vector of magnitude 20, so all you need to do is multiply the original force by 2 and there's your answer
True, but I think it takes away the understanding of the question, especially with regard to the direction. Still a perfectly sound method though.

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: January 26, 2015
Today on TSR

### Should I ask for his number?

Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• 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
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• 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