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

    15
    ReputationRep:
    The total mass of two children is 60kg. They are sitting on a seesaw one at each end.
    Find their separate masses.
    -Child 1 is 1.6 metres from the centre and child 2 is 1.4 metres from the centre?

    Any hints? please help
    • Community Assistant
    • Welcome Squad
    Offline

    20
    ReputationRep:
    (Original post by joyoustele)
    The total mass of two children is 60kg. They are sitting on a seesaw
    Find their separate masses.
    -Child 1 is 1.6 metres from the centre and child 2 is 1.4 metres from the centre?

    Any hints? please help
    Draw a diagram.
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by RDKGames)
    Draw a diagram.
    I did.
    -Should I do trial and error to find their separate masses?
    As their combined moment should be equal to 0?
    Offline

    8
    ReputationRep:
    Take moments about the pivot point. That will give you an equation that relates m1 and m2. The question also tells you that m1 + m2 = 60. Solve simultaneously or by substitution.
    • Community Assistant
    • Welcome Squad
    Offline

    20
    ReputationRep:
    (Original post by joyoustele)
    I did.
    -Should I do trial and error to find their separate masses?
    As their combined moment should be =0
    Trial and error is a waste of time if there is a solid systematic way of getting the answers.

    Say their masses are m_1 \mathrm{kg} and m_2 \mathrm{kg} respectively.

    You know that m_1+m_2=60

    Mark on the reaction force in the middle of the seesaw, what would it be?

    Then use one kid as a pivot, and calculate moments from there and work out the mass of the other kid.

    There isn't much to the context from what you've said so I just assume the seesaw is perfectly horizontal with the system in equilibrium.
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by RDKGames)
    Trial and error is a waste of time if there is a solid systematic way of getting the answers.

    Say their masses are m_1 \mathrm{kg} and m_2 \mathrm{kg} respectively.

    You know that m_1+m_2=60

    Mark on the reaction force in the middle of the seesaw, what would it be?

    Then use one kid as a pivot, and calculate moments from there and work out the mass of the other kid.

    There isn't much to the context from what you've said so I just assume the seesaw is perfectly horizontal with the system in equilibrium.
    m_1 g1.6 -(60-m_1)g1.4=60g - is this the right equation?
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by old_engineer)
    Take moments about the pivot point. That will give you an equation that relates m1 and m2. The question also tells you that m1 + m2 = 60. Solve simultaneously or by substitution.

     m_1 g1.6 -(60-m_1)g1.4=60g is this right?
    Offline

    8
    ReputationRep:
    (Original post by joyoustele)
     m_1 g1.6 -(60-m_1)g1.4=60g is this right?
    No. Where did the right hand side come from?
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by old_engineer)
    No. Where did the right hand side come from?
    Reaction force for the total mass
    • Thread Starter
    Offline

    15
    ReputationRep:
    I haven't been taught moments yet, im trying to learn it before we start
    Offline

    8
    ReputationRep:
    (Original post by joyoustele)
    Reaction force for the total mass
    OK, moments gives you 1.6(m1)g = 1.4(m2)g. Clockwise moment equals anti-clockwise moment, as the system is in equilibrium. Now proceed from there with the substitution you already worked out from m1 + m2 = 60.
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by old_engineer)
    OK, moments gives you 1.6(m1)g = 1.4(m2)g. Clockwise moment equals anti-clockwise moment, as the system is in equilibrium. Now proceed from there with the substitution you already worked out from m1 + m2 = 60.
    Thank you very much
    I get m1 is 35kg and m2=25kg
    Offline

    8
    ReputationRep:
    (Original post by joyoustele)
    Thank you very much
    I get m1 is 35kg and m2=25kg
    That's not right. You need to check those numbers in the moments equation then check your arithmetic.
    Offline

    18
    ReputationRep:
    (Original post by joyoustele)
    The total mass of two children is 60kg. They are sitting on a seesaw one at each end.
    Find their separate masses.
    -Child 1 is 1.6 metres from the centre and child 2 is 1.4 metres from the centre?

    Any hints? please help
    Set the mass of child 1 to be x and the mass of child 2 to be 60-x. The moment is equal to the force times the distance from the pivot, so find the moments for each child. The total moment should be 0, and the two moments are in opposite directions, so set them equal to each other!

    Don't try to do it with two variables. If you can easily eliminate one (as I did with the 60-x bit), do so.
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by old_engineer)
    That's not right. You need to check those numbers in the moments equation then check your arithmetic.
    oops I meant 28Kg for M1 and 32 for M2

    m1 should be 84g/3g =28
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by TheMindGarage)
    Set the mass of child 1 to be x and the mass of child 2 to be 60-x. The moment is equal to the force times the distance from the pivot, so find the moments for each child. The total moment should be 0, and the two moments are in opposite directions, so set them equal to each other!

    Don't try to do it with two variables. If you can easily eliminate one (as I did with the 60-x bit), do so.
    that's the first method I tried, I couldn't solve it. Maybe I went wrong in my calculations.
    Offline

    8
    ReputationRep:
    (Original post by joyoustele)
    oops I meant 28Kg for M1 and 32 for M2

    m1 should be 84g/3g =28
    Yes that's it. You should always be looking for quick consistency checks you can apply to your answers.
    Offline

    18
    ReputationRep:
    (Original post by joyoustele)
    that's the first method I tried, I couldn't solve it. Maybe I went wrong in my calculations.
    1.6x = 1.4(60-x)
    1.6x = 84 - 1.4x
    3x = 84
    x = 28

    Should be 28kg and 32kg, as you managed to get earlier.
 
 
 
  • 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
    What newspaper do you read/prefer?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • 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.