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# M4 2D moments watch

1. http://filestore.aqa.org.uk/sample-p...W-QP-JUN11.PDF

Can someone explain how to do question 5. I can't figure out which moments are negative and which are positive.
2. anyone know?
3. (Original post by AspiringUnderdog)
anyone know?
Consider the x, y components of the forces separately.

A diagram may help.

Assume "a" is positive - it doesn't matter positive or negative, but you need to be consistent.
4. (Original post by ghostwalker)
Consider the x, y components of the forces separately.

A diagram may help.

Assume "a" is positive - it doesn't matter positive or negative, but you need to be consistent.
I drew a diagram. I'm meant to have -a x 6 but I would've thought that this would be positive. Also to get the couple of -10 which you then say 10 is the magnitude you would have to have the positive and negatives a set way.
5. (Original post by AspiringUnderdog)
I drew a diagram. I'm meant to have -a x 6 but I would've thought that this would be positive. Also to get the couple of -10 which you then say 10 is the magnitude you would have to have the positive and negatives a set way.
If anticlockwise is positive, then you will get -ax6.

y-coordinate of position vector is 6, i.e. above the axis, and component of force is "a" in the direction of x-positive. That's clockwise, so minus a x 6.

Getting the magnitude of the couple is unaffected by whether you take clockwise or anticlockwise as positive.

As a different example:
If you get -6 anticlockwise, then this is the same as 6 clockwise.
6. (Original post by ghostwalker)
If anticlockwise is positive, then you will get -ax6.

y-coordinate of position vector is 6, i.e. above the axis, and component of force is "a" in the direction of x-positive. That's clockwise, so minus a x 6.

Getting the magnitude of the couple is unaffected by whether you take clockwise or anticlockwise as positive.

As a different example:
If you get -6 anticlockwise, then this is the same as 6 clockwise.
I know that it's the same but the mark scheme seems to want you to find - 10 first so it's probably using anticlockwise is positive.

Is there a good way to not get which forces are clockwise and which ones are anticlockwise confused as I think that that was my main issue.
7. (Original post by AspiringUnderdog)
I know that it's the same but the mark scheme seems to want you to find - 10 first so it's probably using anticlockwise is positive.

Is there a good way to not get which forces are clockwise and which ones are anticlockwise confused as I think that that was my main issue.
Both coordinate axes and forces are positive in the up direction and to the right.

Plot a general point in the top right quadrant, with a force going up and to the right.

Work out the moment of that:

If anticlockwise is positive, then

Moment of force = y component of force times x-coordinate - x component of force times y-coordinate.

Use that for all, regardless of quadrant, and the signs will take care of themselves.

Multiply the RHS of the equation by -1 if you want to use clockwise as positive.
8. (Original post by ghostwalker)
Both coordinate axes and forces are positive in the up direction and to the right.

Plot a general point in the top right quadrant, with a force going up and to the right.

Work out the moment of that:

If anticlockwise is positive, then

Moment of force = y component of force times x-coordinate - x component of force times y-coordinate.

Use that for all, regardless of quadrant, and the signs will take care of themselves.

Multiply the RHS of the equation by -1 if you want to use clockwise as positive.
Sorry if I'm being daft but does this work for all quadrants? Also if the force is down and right would you put both as negative?
9. (Original post by ghostwalker)
Both coordinate axes and forces are positive in the up direction and to the right.

Plot a general point in the top right quadrant, with a force going up and to the right.

Work out the moment of that:

If anticlockwise is positive, then

Moment of force = y component of force times x-coordinate - x component of force times y-coordinate.

Use that for all, regardless of quadrant, and the signs will take care of themselves.

Multiply the RHS of the equation by -1 if you want to use clockwise as positive.
Okay I tried pushing against my calculator from certain places and that seems to help.

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