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# Resolving forces with velocity? Watch

1. I have been working on my mechanics skills and came across this question that I am struggling with:

I understand resolving the horizontal component of the 210N force pulling the skier up the slope.

I have been able to complete questions where the object is in equilibrium and you have to find the frictional force and normal contact force; however I am not sure what to do differently now that the object is moving up the slope?
2. (Original post by x0diak)
I have been working on my mechanics skills and came across this question that I am struggling with:

I understand resolving the horizontal component of the 210N force pulling the skier up the slope.

I have been able to complete questions where the object is in equilibrium and you have to find the frictional force and normal contact force; however I am not sure what to do differently now that the object is moving up the slope?
The question is solved the same way as any "equilibrium" question.
If he is moving at constant velocity then the resultant force on him is still zero. (Newton's 1st Law.)
3. Ahh ok I understand that, however I am still not able to answer the question correctly.

The answers are: Frictional force = 38.4N and Normal contact force = 676N

If you could show me the workings to get these values that would be great
4. (Original post by x0diak)
Ahh ok I understand that, however I am still not able to answer the question correctly.

The answers are: Frictional force = 38.4N and Normal contact force = 676N

If you could show me the workings to get these values that would be great
Calculate the gravity affecting him along the horizontal and horizontal force of the component pulling him. The difference will be the frictional force, this is because these forces must balance. Basically the horizontal force of him being pulled up must be equal to the horizontal force of gravity and friction added together.

Then do the same for the vertical forces, and the forces should equal in equilibrium so the difference is the force upwards provided by the surface.

May I ask where you got this question?
5. I'll try doing that and let you know how I get on

They are from this book: http://www.amazon.co.uk/Mechanics-Ca.../dp/0521549000

It's a fantastic book if you are struggling with mechanics as it has loooooads of questions.
6. (Original post by x0diak)
I'll try doing that and let you know how I get on

They are from this book: http://www.amazon.co.uk/Mechanics-Ca.../dp/0521549000

It's a fantastic book if you are struggling with mechanics as it has loooooads of questions.
Hope it helps, I have a solution wrote down if you're still struggling.

It's Maths Mechanics book, I don't don't Mechanics, I do Statistics sadly But we do similar stuff to that question in PHY2 (AQA). I might have to invest thank you!
7. I've managed to get the frictional force to be 38.4N as I did 210cos20 - 78gsin12

However I cannot get the normal contact force to equal 676N

Would you be able to show me your solution to it?

I do statistics in AS too, I wont do Maths mechanics until A2, but this book is really helpful for AQA PHY2 as I'm doing that exam too
8. (Original post by TheAJK)
Hope it helps, I have a solution wrote down if you're still struggling.

It's Maths Mechanics book, I don't don't Mechanics, I do Statistics sadly But we do similar stuff to that question in PHY2 (AQA). I might have to invest thank you!
Would you be able to show me your solution?
9. (Original post by x0diak)
I've managed to get the frictional force to be 38.4N as I did 210cos20 - 78gsin12

However I cannot get the normal contact force to equal 676N

Would you be able to show me your solution to it?

I do statistics in AS too, I wont do Maths mechanics until A2, but this book is really helpful for AQA PHY2 as I'm doing that exam too
To get the normal contact force, you just need to resolve the forces perpendicular to the slope:

78gcos12 - 210sin20 = 675.9N or 676N to 3.s.f

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10. wow that was SOO hard. I also do statistics
Do you need to know how to do it?

For friction i did cos of 210N to get parallel to slope. and this was pulling him up. his component of weight parallel to slope was pulling him down. there is also friction puling him down. Friction+weight=pulling up

and for contact force. Sine of of 210 pulls up. Weight pulls down. Something else pushes up.
So weight=contact+(sin of 210N)

btw when i say sin 0f 210 that isnt the calc. it's 210sin20
11. (Original post by x0diak)
Would you be able to show me your solution?
Its a little messy. I apologise.
Attached Images

12. Thanks guys I get it now!
13. (Original post by TheAJK)
Its a little messy. I apologise.
Would it be ok if I post one last question that I am having difficulties with?
14. (Original post by x0diak)
Would it be ok if I post one last question that I am having difficulties with?
Sure
15. (Original post by TheAJK)
Sure
This is the question:

I have been doing F - mgsin20 = 0 for the horizontal force and then R - mgcos20 = 0 for the normal contact force as the crate is in equilibrium

The correct answers are 21.4N for the horizontal force and 62.6N for the normal contact force
16. I got 20.14 and 55.3.
I think I did the same working as you. not sure what we're doing wrong
17. (Original post by x0diak)
This is the question:

I have been doing F - mgsin20 = 0 for the horizontal force and then R - mgcos20 = 0 for the normal contact force as the crate is in equilibrium

The correct answers are 21.4N for the horizontal force and 62.6N for the normal contact force
The answers you have given are correct, the key is *horizontal* will attach picture below. Make sure you understand this stuff and then you'll be able to apply this knowledge, good luck!

(Original post by user1-4)
I got 20.14 and 55.3.
I think I did the same working as you. not sure what we're doing wrong
This is wrong I'm afraid. However it's what I thought at first glance as well. I had to read the question a few times.

I've never seen a question like this before.

Hope the picture helps, if not just message me
Attached Images

18. (Original post by TheAJK)
The answers you have given are correct, the key is *horizontal* will attach picture below. Make sure you understand this stuff and then you'll be able to apply this knowledge, good luck!

This is wrong I'm afraid. However it's what I thought at first glance as well. I had to read the question a few times.

I've never seen a question like this before.

Hope the picture helps, if not just message me
You are an absolute genius! Thank you so much!
19. (Original post by x0diak)
You are an absolute genius! Thank you so much!
I'm not, but thank you!

You're welcome

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