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AQA A2 Mathematics MM2B Mechanics 2 - Monday 22nd June 2015 [Exam Discussion Thread] Watch

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    (Original post by Haza100)
    Im so confused as to how to work this out, supposedly the answers are Ta = 24.2N and Tb = 4.6N.Attachment 433177

    If anyone can help out I would be massively grateful - I've wasted an hour trying to figure it out haha.
    Apologies for the poor picture quality/lighting and my handwriting! Hope this helps

    EDIT: Just gonna clear some things up that are slightly less visible in the pic: the angle between the tension and the vertical is the same for the top part and bottom part, and the radius of the circle is root 0.48 (0.693) due to use of pythagoras for the string length and the height. the exact value of root 0.48 is (2*root3)/5 which is where the 72root3/5 comes from

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    (Original post by Haza100)
    Im so confused as to how to work this out, supposedly the answers are Ta = 24.2N and Tb = 4.6N.Attachment 433177

    If anyone can help out I would be massively grateful - I've wasted an hour trying to figure it out haha.
    Does this help?

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Size:  167.9 KB


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    First, you have to use Pythagoras to find the radius, but you have to divide the distance between A and B by two.
    Then, use v = r*omega to find the velocity.
    Next, find the angle between the string and the horizontal by using trig.
    Finally, use Tsin theta = (mv^2)/r to find the tension. The tension in both strings are the same. I got T = 49.9N

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    No this book:
    http://www.amazon.co.uk/Advancing-Ma.../dp/0435513370
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    I've just realised that I should resolve vertically and horizontally. I'll try to do that again.

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    (Original post by benny1152)
    Apologies for the poor picture quality/lighting and my handwriting! Hope this helps

    EDIT: Just gonna clear some things up that are slightly less visible in the pic: the angle between the tension and the vertical is the same for the top part and bottom part, and the radius of the circle is root 0.48 (0.693) due to use of pythagoras for the string length and the height. the exact value of root 0.48 is (2*root3)/5 which is where the 72root3/5 comes from

    Name:  WIN_20150621_130400.JPG
Views: 104
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    (Original post by CD223)
    Does this help?

    Name:  ImageUploadedByStudent Room1434888847.027347.jpg
Views: 145
Size:  167.9 KB


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    Awesome, thanks so much both of you. Got it now, I wasn't taking into account the extra downwards force exerted by Tb. Really cleared things up, thanks!
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    (Original post by Haza100)
    Awesome, thanks so much both of you. Got it now, I wasn't taking into account the extra downwards force exerted by Tb. Really cleared things up, thanks!
    No worries! Ironic if a similar question appears tomorrow.


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    Would we ever be asked a ladder question where the wall is rough?
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    (Original post by Tibooster)
    Would we ever be asked a ladder question where the wall is rough?
    Perfectly possible yeah. It's in the CGP guide


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    (Original post by Tibooster)
    Would we ever be asked a ladder question where the wall is rough?
    (Original post by CD223)
    Perfectly possible yeah. It's in the CGP guide


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    I guess you would now just have to take into account a vertically upward force?
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    (Original post by Reverse Swing)
    I guess you would now just have to take into account a vertically upward force?
    Yup. A rough wall and a rough floor would be a bit harsh though as when you take moments you'll have another unknown. They'd have to provide enough information though, so I wouldn't panic.


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    What does it mean if you do moments about a point (ACW-CW) and you get a value like 400N
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    (Original post by Stepidermis)
    What does it mean if you do moments about a point (ACW-CW) and you get a value like 400N
    Do you mean taking moments about a point with ACW positive?

    I'm confused as to whether you mean there's a resultant moment or you're just working out a force which happens to be 400N?


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    (Original post by CD223)
    Do you mean taking moments about a point with ACW positive?

    I'm confused as to whether you mean there's a resultant moment or you're just working out a force which happens to be 400N?


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    yeah what does there being a resultant moment mean? that the forces at that point should negate this force?
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    (Original post by Stepidermis)
    yeah what does there being a resultant moment mean? that the forces at that point should negate this force?
    A resultant moment means there's a turning effect about a pivot. It's like turning a door handle. The resultant moment is clockwise so there's a turning effect about the pivot and the door mechanism opens the door.


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    (Original post by CD223)
    A resultant moment means there's a turning effect about a pivot. It's like turning a door handle. The resultant moment is clockwise so there's a turning effect about the pivot and the door mechanism opens the door.


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    You know when a car is travelling up a hill at which is inclined at an angle to the vertical, does the resistance force act down the hill or up the hill?
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    (Original post by ProbablyPassing)
    Down the hill, the resistance acts in an opposite direction to the movement in this scenario.
    Thank you! I doing June 2010 and I don't know how the the component of the weight parallel to the hill is greater or less than the resistance force.
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    (Original post by Tiwa)
    Thank you! I doing June 2010 and I don't know how the the component of the weight parallel to the hill is greater or less than the resistance force.
    In those cases where the body moves up the hill, the component of the weight parallel to the hill is in the same direction as the resistive force.

    This means at maximum/constant speed the driving force is equal to the component of the weight down the slope added to the resistive force. This is because at maximum/constant speed there's no resultant.


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    (Original post by CD223)
    In those cases where the body moves up the hill, the component of the weight parallel to the hill is in the same direction as the resistive force.

    This means at maximum/constant speed the driving force is equal to the component of the weight down the slope added to the resistive force. This is because at maximum/constant speed there's no resultant.


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    Thank you!
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    (Original post by Tiwa)
    Thank you!
    No worries!

    Gonna try another unofficial MS depending on how the exam goes - could be a tough one given last year's was fairly kind
 
 
 
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