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

    3
    ReputationRep:
    (Original post by uberteknik)
    You have the correct answer so you have to say equilibrium cannot be achieved with the stated variables. The only way of balancing is to either increase the length of the plank, or increase the mass (b), or decrease the distance of mass(a) from the pivot, or decrease mass (a).
    no, its a show that, and it does work. i've actually reworded the question... but I was trying to calculate x, so that I could 1.5 - x = 30....... here is the original.... part (ii)
    Attached Images
     
    • Study Helper
    Offline

    21
    ReputationRep:
    Study Helper
    (Original post by SexyNerd)
    i dont get the correct answer.... use clock wise as positive from pivot .. b is on right side of pivot, a is on the left....

    so

    5g(x) - 50g(.20) = 0 (equal to 0 because its in equilibrium)

    5g(x) = 10g

    x = 10/5

    x = 2

    which is not correct, can you see where my problem is?
    Have they given you a mass for the plank? If yes, the answer is nit the simple gcse type.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by uberteknik)
    Have they given you a mass for the plank? If yes, the answer is nit the simple gcse type.
    yes they did, i've posted the question....
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by uberteknik)
    Have they given you a mass for the plank? If yes, the answer is nit the simple gcse type.
    any success?
    • Study Helper
    Offline

    21
    ReputationRep:
    Study Helper
    Oh Ok. Just seen that.

    I will be back shortly but before that, you now have to take into account the turning force produced by the beam itself which will be solved by assuming the mass of the plank is evenly distributed along the entire beam and you have to integrate all of the moments produced by that distributed mass.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by uberteknik)
    Oh Ok. Just seen that.

    I will be back shortly but before that, you now have to take into account the turning force produced by the beam itself which will be solved by assuming the mass of the plank is evenly distributed along the entire beam and you have to integrate all of the moments produced by that distributed mass.
    i've not done that before, can you show me when you return?
    Offline

    9
    The weight of the plank, as its mass is uniform, acts downwards at its centre. (At the centre of gravity.)
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by Stonebridge)
    The weight of the plank, as it's mass is uniform, acts downwards at it's centre. (At the centre of gravity.)
    oh yes, of course!
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by uberteknik)
    Oh Ok. Just seen that.

    I will be back shortly but before that, you now have to take into account the turning force produced by the beam itself which will be solved by assuming the mass of the plank is evenly distributed along the entire beam and you have to integrate all of the moments produced by that distributed mass.
    (Original post by Stonebridge)
    .......
    can you explain part (ii) to me please?
    Attached Images
     
    • Study Helper
    Offline

    21
    ReputationRep:
    Study Helper
    Torque is a force vector producing a rotation about a pivot.

    This is the same as a moment (force x distance from pivot) but this time with the angle made between the force vector and the lever arm now taken into account.

    measured in Newton meters

    where r = radius of the force vector from the pivot (axis) of rotation.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by uberteknik)
    Torque is a force vector producing a rotation about a pivot.

    This is the same as a moment (force x distance from pivot) but this time with the angle made between the force vector and the lever arm now taken into account.

    measured in Newton meters

    where r = radius of the force vector from the pivot (axis) of rotation.
    so how would you apply this to the above?
    • Study Helper
    Offline

    21
    ReputationRep:
    Study Helper
    (Original post by SexyNerd)
    so how would you apply this to the above?
    Same as you would for a moment. In this case the forces at opposite ends of the lever (distance to the pivot = radius of the wheel) are acting in the same 'rotational' direction so sum.

    The force vector is acting at a tangent to the radius. i.e. 90o​.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by uberteknik)
    Same as you would for a moment. In this case the forces at opposite ends of the lever (distance to the pivot = radius of the wheel) are acting in the same direction so sum.

    The force vector is acting at a tangent to the radius. i.e. 90o​.
    so the answer should be the wheel is in equilibrium.
    • Study Helper
    Offline

    21
    ReputationRep:
    Study Helper
    (Original post by SexyNerd)
    so the answer should be the wheel is in equilibrium.
    Equilibrium means no influences cause it to move. i.e. the Net forces sum to zero.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by uberteknik)
    Equilibrium means no influences cause it to move. i.e. the Net forces sum to zero.
    so it doesn't move?
    • Study Helper
    Offline

    21
    ReputationRep:
    Study Helper
    (Original post by SexyNerd)
    so it doest move?
    The forces at opposite ends of the radius are acting to rotate the wheel in the same direction. There are no other forces opposing this.

    Those forces will therefore cause the wheel to accelerate.

    So yes, the wheel will accelerate and no, the forces are not in equilibrium.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by uberteknik)
    The forces at opposite ends of the radius are acting to rotate the wheel in the same direction. There are no other forces opposing this.

    Those forces will therefore cause the wheel to accelerate.

    So yes, the wheel will accelerate and no, the forces are not in equilibrium.
    thanks.....
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by uberteknik)
    The forces at opposite ends of the radius are acting to rotate the wheel in the same direction. There are no other forces opposing this.

    Those forces will therefore cause the wheel to accelerate.

    So yes, the wheel will accelerate and no, the forces are not in equilibrium.
    part i,

    v = u +at

    so should v = u + (8)(4)

    v = 32???
    Attached Images
     
    • Study Helper
    Offline

    21
    ReputationRep:
    Study Helper
    (Original post by SexyNerd)
    part i,

    v = u +at

    so should v = u + (8)(4)

    v = 32???
    No, because the acceleration is not constant over the 4 seconds. i.e. it starts at 0 and rises to 8m/s2 during that 4 seconds.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by uberteknik)
    No, because the acceleration is not constant over the 4 seconds. i.e. it starts at 0 and rises to 8m/s2 during that 4 seconds.
    thought so, thanks...
 
 
 
  • 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
    Brussels sprouts
  • 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.