2 SUPER QUICK M2 Moments questions

Original post by creativebuzz

And why would there be a reaction force at O?

Something's got to support the whole structure; without it the whole construction would fall down, i.e. no equilibrium. :tongue:

Reply 8

OP

Original post by ghostwalker

Something's got to support the whole structure; without it the whole construction would fall down, i.e. no equilibrium. :tongue:

I see!

In this question I did sin45 instead of cos45 (I put a purple line in so you can see why I put sin45 instead of cos45) but the mark scheme says it should be cos45, why is that?

Reply 9

Zacken

22

Original post by creativebuzz

I see!

In this question I did sin45 instead of cos45 (I put a purple line in so you can see why I put sin45 instead of cos45) but the mark scheme says it should be cos45, why is that?

Does it matter?

\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}

Reply 10

Original post by creativebuzz

I see!

In this question I did sin45 instead of cos45 (I put a purple line in so you can see why I put sin45 instead of cos45) but the mark scheme says it should be cos45, why is that?

As Zacken said - it doesn't matter.

However, since they've marked an angle as 45, I would assume they're using that same angle in their calculation. Sin45 would have been the better choice.

If the angle had been 30 degrees, say, then cos30 would be incorrect.

Reply 11

OP

Original post by ghostwalker

As Zacken said - it doesn't matter.

However, since they've marked an angle as 45, I would assume they're using that same angle in their calculation. Sin45 would have been the better choice.

If the angle had been 30 degrees, say, then cos30 would be incorrect.

Oh yeah I knew that it didn't matter!

I just wanted to make sure that I didn't get the answer right because of a fluke/my method was legit

Reply 12

OP

Original post by ghostwalker

As Zacken said - it doesn't matter.

However, since they've marked an angle as 45, I would assume they're using that same angle in their calculation. Sin45 would have been the better choice.

If the angle had been 30 degrees, say, then cos30 would be incorrect.

Also, why did they give the lamina's mass as 12 (+15 of course) because isn't the mass of a lamina usually found by doing the area (so 5x8 and then add the particles' masses) in centre of mass questions? This is literally the only thing I don't understand about CoM.

Reply 13

Original post by creativebuzz

Also, why did they give the lamina's mass as 12 (+15 of course) because isn't the mass of a lamina usually found by doing the area (so 5x8 and then add the particles' masses) in centre of mass questions? This is literally the only thing I don't understand about CoM.

The area by itself is insufficient. Assuming there is some thickness to the lamina, you'd need to know the mass per unit area, to then work out the mass of the lamina. 5x8 is not a mass.

Reply 14

OP

Original post by ghostwalker

The area by itself is insufficient. Assuming there is some thickness to the lamina, you'd need to know the mass per unit area, to then work out the mass of the lamina. 5x8 is not a mass.

Ah thickness of course, that makes sense! But how would I know when finding the area is sufficient because in quite a lot of CoM questions (particularly composite shapes) tend to involve finding the area!

Reply 15

Original post by creativebuzz

Ah thickness of course, that makes sense! But how would I know when finding the area is sufficient because in quite a lot of CoM questions (particularly composite shapes) tend to involve finding the area!

If you're given any masses, then you need to work in terms of mass.

If no masses are given, then it's areas; assuming density is constant.

Reply 16

OP

Original post by ghostwalker

If you're given any masses, then you need to work in terms of mass.

If no masses are given, then it's areas; assuming density is constant.

Gotcha.

I managed to get full marks on part a, but in part b what do they mean by "give the direction"?
Also, is there some reason/theory behind how we can find what direction the X and Y reaction forces are working in without plugging in your values and just seeing if your answer is negative or not?

Reply 17

Original post by creativebuzz

I managed to get full marks on part a, but in part b what do they mean by "give the direction"?

Force is a vector quantity, it has a magnitude and direction.

E.g. a force represented by i+j has magntude of root(2), and a direction 45 degrees to i vector going anticlcockwise.

Also, is there some reason/theory behind how we can find what direction the X and Y reaction forces are working in without plugging in your values and just seeing if your answer is negative or not?

Logic and common sense. Can't really give a detailed description, but take your example.

If the rod wasn't attached at A, would it move to the right or left? Force at A must be in the opposite direction to hold it in equilibrium.

Then consider D. If the rod wasn't attached at A, would it rotate clockwise or anticlockwise about D? Vertical force at A must act to stop that, hence....

Reply 18

OP