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# M1 Edexcel -Dynamics of a particle moving in a straight line watch

1. I am doing the question in the attachment and have managed to get the answers to all three parts in the same way as the solution bank. However, for part c) the solution bank say that alternatively consider the cage.

I had 10g down, T up and R(cage) up
then
10g -(T+R) = 10x0.8.

What should I be thinking please?

Thanks
Attached Files
2. Doc1.doc (527.0 KB, 55 views)
3. (Original post by maggiehodgson)
I am doing the question in the attachment and have managed to get the answers to all three parts in the same way as the solution bank. However, for part c) the solution bank say that alternatively consider the cage.

I had 10g down, T up and R(cage) up
then
10g -(T+R) = 10x0.8.
and opposite forces).fact that
What should I be thinking please?

Thanks
You know the tension from an earlier part.
But I think you've got the direction of the normal force wrong, the normal force acts to accelerate the cage (remember Newton's equal and opposite forces, here you are using the fact that the normal force on both objects is equal).
4. (Original post by EricPiphany)
You know the tension from an earlier part.
But I think you've got the direction of the normal force wrong, the normal force acts to accelerate the cage (remember Newton's equal and opposite forces, here you are using the fact that the normal force on both objects is equal).
in part b, B's force on A was upwards so the cage's force on B should be upwards shouldn't it?
5. (Original post by maggiehodgson)
in part b, B's force on A was upwards so the cage's force on B should be upwards shouldn't it?
Yes but we are only interested here in the forces on the cage. Remember that every force has an equal and opposite 'reaction'.
6. (Original post by EricPiphany)
Yes but we are only interested here in the forces on the cage. Remember that every force has an equal and opposite 'reaction'.
No, no idea.
7. Cant see the question as I cant open the file.
8. (Original post by EricPiphany)
Yes but we are only interested here in the forces on the cage. Remember that every force has an equal and opposite 'reaction'.
If the cage is being held up by the rope then isn't the tension in the rope acting on the cage? If not, why not? Really don't get it.
9. (Original post by maggiehodgson)
I am doing the question in the attachment and have managed to get the answers to all three parts in the same way as the solution bank. However, for part c) the solution bank say that alternatively consider the cage.

I had 10g down, T up and R(cage) up
then
10g -(T+R) = 10x0.8.

What should I be thinking please?

Thanks
I would suggest either uploading the file as a PDF or taking a screenshot of the question from an online solution bank.
10. (Original post by SeanFM)
I would suggest either uploading the file as a PDF or taking a screenshot of the question from an online solution bank.
I am unable to save the document as a PDF - Word 2003 does not allow that. I have tried doing a screen shot (that's what's in the Word document) but then it won't pates into here. If you have Edexcel Modular Mathematics M1 then it's question 9 page 75. Otherwise try this attachment - it's Paint.

Thanks
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