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Easy Mechanics 1 question help

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!

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Reply 1
Avatar for SH0405
SH0405
15
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?
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 :smile:
Reply 3
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SH0405
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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 :smile:


Yep. No problem. What answer do you get?....
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.
Reply 5
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SH0405
15
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! :/
(edited 9 years ago)
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!
Reply 7
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SH0405
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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.
Tip: Make a triangle with appropriate side lengths
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
Reply 10
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SH0405
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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.
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!
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.
Original post by Kadak
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 :biggrin:
Reply 14
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SH0405
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Original post by Kadak
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?
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.
Reply 16
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SH0405
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Original post by Kadak
Ooh, I would rep this but I reached my rep limit today.


There's always tomorrow, my friend. Always tomorrow.

:wink:
Reply 17
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davros
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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 :smile:

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 :smile:
Original post by davros
There's no need for any angles or trig or drawing triangles in this question :smile:

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 :smile:


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!
Reply 19
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SH0405
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Original post by davros
There's no need for any angles or trig or drawing triangles in this question :smile:

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 :smile:


True, but I think it takes away the understanding of the question, especially with regard to the direction. Still a perfectly sound method though.

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