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

    2
    ReputationRep:
    Hi all, was hoping someone could explain where I went wrong on this moments question on an OCR paper. The correct answer is 110N, I've attached both the question and my working

    Thanks!

    Attachment 536507536509Attachment 536507536509
    Attached Images
      
    Offline

    1
    ReputationRep:
    The weight is 45g.
    You can work out the angle theta by using the 0.15m length they've given you at the bottom to form a right angled triangle with the 0.3m.
    Then take moments about P.
    Offline

    2
    ReputationRep:
    (Original post by RBoss)
    Hi all, was hoping someone could explain where I went wrong on this moments question on an OCR paper. The correct answer is 110N, I've attached both the question and my working

    Thanks!

    Attachment 536507536509Attachment 536507536509
    So you've made a simple error which has complicated this.

    First, looking at the definition of a moment:
    "Force multiplied by the PERPENDICULAR distance from the pivot"

    If you look at the diagram, the force F is perpendicular the the slab of length 0.600m, so we dont have to resolve any angle.

    Simple, equal 0.600F to the clockwise moment and solve for F which will give you the answer of 110N.

    Hope this helped
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by voltz)
    So you've made a simple error which has complicated this.

    First, looking at the definition of a moment:
    "Force multiplied by the PERPENDICULAR distance from the pivot"

    If you look at the diagram, the force F is perpendicular the the slab of length 0.600m, so we dont have to resolve any angle.

    Simple, equal 0.600F to the clockwise moment and solve for F which will give you the answer of 110N.

    Hope this helped
    I'm definitely overlooking something very obvious xD

    I only formed the second equation cos theta = weight/F (using vertical resultant force = 0) in an attempt to eliminate the cos theta part from the moments equation, as the weight is not acting perpendicularly to the slab/slope. Surely only the normal component of the weight to the slope needs to be considered in the moments equation?
    Offline

    20
    ReputationRep:
    (Original post by RBoss)
    I'm definitely overlooking something very obvious xD

    I only formed the second equation cos theta = weight/F (using vertical resultant force = 0) in an attempt to eliminate the cos theta part from the moments equation, as the weight is not acting perpendicularly to the slab/slope. Surely only the normal component of the weight to the slope needs to be considered in the moments equation?
    note: lengths on the exam paper diagram aren't to scale which could cause confusion.

    You seem to have overlooked that the vertical component of F should be W/2... though you wrote it on line 1. Not sure if you've confused yourself by trying to make the vertical forces always equal W - but you don't want to do that when you're taking moments, There's a vertical component at the pivot but you can worry about that separately if you need to.

    W=45*g = 441 N
    cos ϴ = 1/2

    fwiw I resolved F vertically and said

    cos ϴ = F/Fvert
    Fvert cos ϴ = F
    W/2 cos ϴ = F

    F=W/4
    =110 N (3sf)

    (there was no need to resolve horizontally for this question)

    ---
    if would also work if you prefer to resolve parallel and perpendicular to the slab
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Joinedup)
    note: lengths on the exam paper diagram aren't to scale which could cause confusion.

    You seem to have overlooked that the vertical component of F should be W/2... though you wrote it on line 1. Not sure if you've confused yourself by trying to make the vertical forces always equal W - but you don't want to do that when you're taking moments, There's a vertical component at the pivot but you can worry about that separately if you need to.

    W=45*g = 441 N
    cos ϴ = 1/2

    fwiw I resolved F vertically and said

    cos ϴ = F/Fvert
    Fvert cos ϴ = F
    W/2 cos ϴ = F

    F=W/4
    =110 N (3sf)

    (there was no need to resolve horizontally for this question)

    ---
    if would also work if you prefer to resolve parallel and perpendicular to the slab
    Aha, yes. I was looking too much at the parallel/perpendicular components, without considering Fv = W/2, as the gradient of the slope is the same, and F has twice the vertical distance of W... *facepalm*

    That's a major bit of thinking space recovered though - thanks a lot for the clear explanation
 
 
 
  • 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
    Have you ever been hacked?
  • 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

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