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How to solve that kind of problems? I know about clock&anticlockwise forces, but can't work out solutions, especially the first one.

Reply 1

Kevin De Bruyne

Original post by rizamadiyar

How to solve that kind of problems? I know about clock&anticlockwise forces, but can't work out solutions, especially the first one.

What are the clockwise and anti-clockwise moments in the first one?

Reply 2

rizamadiyar

Original post by SeanFM

What are the clockwise and anti-clockwise moments in the first one?

I guess the clockwise one is to the right off the pivot due to the larger distance(40 cm)

noot noot

Original post by rizamadiyar

I guess the clockwise one is to the right off the pivot due to the larger distance(40 cm)

Correct yes, but in terms of numbers and x?

Reply 5

rizamadiyar

Original post by SeanFM

Correct yes, but in terms of numbers and x?

If only I could know... Does it really matter? The answer is 30 cm, I can't figure it out

Reply 6

Kevin De Bruyne

Original post by rizamadiyar

If only I could know... Does it really matter? The answer is 30 cm, I can't figure it out

I am trying to guide you towards the solution.

Okay, to go one step back, it says in the question the beam is balanced. So what can you say about the clockwise and anti-clockwise moments?

Answer: the clockwise and anti-clockwise moments are _ _ _ _ _ (1 word)

Reply 7

rizamadiyar

Original post by SeanFM

equal?

Reply 8

Kevin De Bruyne

Original post by rizamadiyar

equal?

Correct.

Now the anti-clockwise moment is equal to ? * ?

And the clockwise moment is equal to ??? + ???

And since they are equal, x = ?

Reply 9

rizamadiyar

Original post by SeanFM

Correct.

Now the anti-clockwise moment is equal to ? * ?

And the clockwise moment is equal to ??? + ???

And since they are equal, x = ?

500*20=200*x+200*40, x=10 cm, but it says that the answer is 30 cm

Reply 10

rizamadiyar

Original post by rizamadiyar

500*20=200*x+200*40, x=10 cm, but it says that the answer is 30 cm

I guess it has smth to do with the weight of the beam=800 N

Reply 11

Kevin De Bruyne

Original post by rizamadiyar

I guess it has smth to do with the weight of the beam=800 N

That was my initial thought, however the beam is uniform and the pivot is in the centre. Hence the distance of the force due to the weight is 0 from the centre, and so it has no impact if you take moments about where the weight is (which we know is the centre)

I suspect that the answer is just incorrect, I do not see anything else that can be done,

Reply 12

rizamadiyar

Original post by SeanFM

I guess so as well, what about the next one? 8000(moment to the right off the pivot)=200(50-x)+10*(what? 30?)

Reply 13

RogerOxon

Original post by SeanFM

I suspect that the answer is just incorrect, I do not see anything else that can be done,

I agree, x=10cm.

Original post by rizamadiyar

I guess so as well, what about the next one? 8000(moment to the right off the pivot)=200(50-x)+10*(what? 30?)

The clockwise moment of the 800N load is 8000Ncm. The anti-clockwise moment of the 200N load is not what you've written - remember, we're taking moments about the pivot.

The uniform beam is 60cm, so we model its weight as a single force at the centre - how far from the pivot is that?

(edited 7 years ago)

Reply 14

rizamadiyar

Original post by RogerOxon

I agree, x=10cm.

The clockwise moment of the 800N load is 8000Ncm. The anti-clockwise moment of the 200N load is not what you're wirtten - remember, we're taking moments about the pivot.

The uniform beam is 60cm, so we model its weight as a single force at the centre - how far from the pivot is that?