Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    2
    ReputationRep:
    I have been working on my mechanics skills and came across this question that I am struggling with:

    Name:  Question.jpg
Views: 101
Size:  74.9 KB

    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? :confused:
    Offline

    9
    (Original post by x0diak)
    I have been working on my mechanics skills and came across this question that I am struggling with:

    Name:  Question.jpg
Views: 101
Size:  74.9 KB

    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? :confused:
    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.)
    • Thread Starter
    Offline

    2
    ReputationRep:
    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
    Offline

    2
    ReputationRep:
    (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?
    • Thread Starter
    Offline

    2
    ReputationRep:
    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.
    Offline

    2
    ReputationRep:
    (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!
    • Thread Starter
    Offline

    2
    ReputationRep:
    I've managed to get the frictional force to be 38.4N as I did 210cos20 - 78gsin12 :banana:

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

    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
    • Thread Starter
    Offline

    2
    ReputationRep:
    (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?
    Offline

    0
    ReputationRep:
    (Original post by x0diak)
    I've managed to get the frictional force to be 38.4N as I did 210cos20 - 78gsin12 :banana:

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

    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


    Posted from TSR Mobile
    Offline

    1
    ReputationRep:
    wow that was SOO hard. I also do statistics
    I have worked out the answers now (or methods to your answers).
    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
    Offline

    2
    ReputationRep:
    (Original post by x0diak)
    Would you be able to show me your solution?
    Its a little messy. I apologise.
    Attached Images
     
    • Thread Starter
    Offline

    2
    ReputationRep:
    Thanks guys I get it now! :2euk48l:
    • Thread Starter
    Offline

    2
    ReputationRep:
    (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? :eek3:
    Offline

    2
    ReputationRep:
    (Original post by x0diak)
    Would it be ok if I post one last question that I am having difficulties with? :eek3:
    Sure
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by TheAJK)
    Sure
    This is the question:
    Name:  Question.jpg
Views: 60
Size:  50.8 KB

    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 :confused:

    The correct answers are 21.4N for the horizontal force and 62.6N for the normal contact force
    Offline

    1
    ReputationRep:
    I got 20.14 and 55.3.
    I think I did the same working as you. not sure what we're doing wrong
    Offline

    2
    ReputationRep:
    (Original post by x0diak)
    This is the question:
    Name:  Question.jpg
Views: 60
Size:  50.8 KB

    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 :confused:

    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
     
    • Thread Starter
    Offline

    2
    ReputationRep:
    (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! :bban:
    Offline

    2
    ReputationRep:
    (Original post by x0diak)
    You are an absolute genius! Thank you so much! :bban:
    I'm not, but thank you!

    You're welcome
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Would you like to hibernate through the winter months?
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.