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M1 Mechanics - Forces at an angle (Edexcel)

Hello again!

I'm confused on this particular question. I'm lost in whether to resolve it either vertically, horizontally or even both!

The question:

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Reply 1

ghostwalker

Original post by KiddingMe

Hello again!

I'm confused on this particular question. I'm lost in whether to resolve it either vertically, horizontally or even both!

The question:

What are you trying to do - find the moment?

Reply 2

KiddingMe

Sorry, forgot to say. Yep, find the moment about point A.

Note: I kinda figured it out, if you 'rotate' the diagram, so the line of action (dashed line) is fully horizontal like this, which makes life a bit easier:

Just wondering, if you would do it this way or is there is a 'proper' way!

Reply 3

ghostwalker

Original post by KiddingMe

Just wondering, if you would do it this way or is there is a 'proper' way!

Not sure what you've done.

I'd extend the line of action of the force, and drop a perpendicular from A.

Perpendicular distance is 11.3 sin(180-112)

Then multiply by 14.5.

Which gives a different answer to the one you had, I think.

Reply 4

KiddingMe

Hmm, at the back of the book, the answer it gives is -152Nm. I got -151.9, I think I fluked it lol

Edit: I checked your calculation too, and it gives the same (151.9)!

(edited 11 years ago)

Reply 5

KiddingMe

Sorry to bother you again, but got another question regarding forces at an angle!

I've had a go, and got 2 other similar questions correct, but the book says something different for the answer for the total amount about point A regarding this question (Q.3) Cheers in advance!

The question:

My response:

If you could have a look, and can see where I went wrong, that would be much appreciated!

Reply 6

ghostwalker

Original post by KiddingMe

If you could have a look, and can see where I went wrong, that would be much appreciated!

Quite a few errors I'm afraid.

7.4N force - yep.
3.8N force - what's 180-130 degrees?
Both the other two forces will have horizontal and vertical components when taking moments about A.

Reply 7

KiddingMe

Ah right! I thought there might some resolving in both axis,but thought it was not possible, as the previous two parts (a. And b.) was more simple!

Btw, what did u mean by your second point about 180-130? For the 3.8N force, I thought it would be cos40 and sin40 for i and j respectively.

Sorry for wasting your time!

Reply 8

ghostwalker

Original post by KiddingMe

Btw, what did u mean by your second point about 180-130? For the 3.8N force, I thought it would be cos40 and sin40 for i and j respectively.

Sorry for wasting your time!

For the 3.8N force it's the horizontal component you're interested in, so Cos40.

Which is sin 50 - I was interpreting it via a different method.

Sorry for the confusion.

Reply 9

KiddingMe

Ah, I see! Dw, no confusion caused, I thought you might be doing cos :tongue:

Btw, I've retried it, using both i and j components for 2.9N and 9.3N but the answer I got doesn't match with the back. (Back of book says: +9.43Nm) I got -37.94 lol

Here is my working out (2nd attempt):

Reply 10

ghostwalker

Original post by KiddingMe

Ah, I see! Dw, no confusion caused, I thought you might be doing cos :tongue:

OK, the first thing that strikes me, is you've given them all a minus sign.

You need to assign one direction as positive (usually clockwise) and one as negative (usually anticlockwise).

You also need to take care when you resolve into horizontal and vertical components, as in the 2.9N force for example, the horizontal component is going anticlockwise, whereas the vertical component is going clockwise.

Reply 11

madfish

I have that textbook, it is so bad... I would get the main edexcel M1.. so much better

Reply 12

KiddingMe

I thought for the horizontal and vertical components of 2.9N both were clockwise, ( teacher taught us clockwise is negative and anticlockwise is positive) so negative symbol in front.

I saw both i and j axis were going clockwise of the point A:

How did you find out whether it was anticlockwise or clockwise : )

Haha, aww too many mistakes I'm making!

(edited 11 years ago)

Reply 13

KiddingMe

Original post by madfish

I have that textbook, it is so bad... I would get the main edexcel M1.. so much better

Ah right! It is a bit dodgy tbh! But we've nearly completed M1 so no point now, hehe. Cheers though : )

Reply 14

KiddingMe

I think I got it now:

I've looked at the clockwise/anticlockwise movement and altered signs and re-calculated... For once, I got +9.43Nm!

Cheers for the help, greatly appreciated : ). I'll do part b now, where you have to find out moment about B instead

Reply 15

KiddingMe

Damn, I tried part b, and you prob guessed, my answer didn't match the back of the book!

Here is my response:

Reply 16

ghostwalker

Original post by KiddingMe

Damn, I tried part b, and you prob guessed, my answer didn't match the back of the book!

Here is my response:

Some big angles there! And I'm not sure how you're doing them, but they're going to generate some minus signs when they're greater than 90, and that's going to interact with the +/- going anticlockwise/clockwise.

Have you been taught a method - if so, you'll need to elaborate.

Reply 17

KiddingMe

Ahh, i see what you mean about negative/positives affecting outcome.

Erm, for e.g. the 7.4N I did this:

Reply 18

ghostwalker

Original post by KiddingMe

Ahh, i see what you mean about negative/positives affecting outcome.

Erm, for e.g. the 7.4N I did this:

And you already have it as negative for going clockwise, and this (the cos) will make it positive - implying it's going anticlockwise, which it isn't.

I just draw (in my head) a little triangle, in this case the bottom left quadrant, and use an angle less than 90.