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Math Help - moving particle with friction

  1. #1
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    Feb 2007
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    moving particle with friction

    Hello,

    I can’t see where I am going wrong in answering this question.

    The question is “ a particle with mass 9kg is sliding down a smooth inclined slope with an acceleration of 4.9ms. Find the angle of the incline”.

    the smooth slope means no friction.

    +x is down the slope, (which i think is all i need), so i reslove this this direction.

    After resolving along F I get F = 9gsin x

    Therefore I use F=ma which gives me

    9gsin x = (9g)(4.9)

    solving for x gives me

    x = arcsin (9g X 4.9) / 9g

    the 9g’s cancel giving me

    x = arcsin 4.9 which can’t be solved.

    Any help on where I went wrong?
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  2. #2
    MHF Contributor
    Joined
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    Quote Originally Posted by pablohacker View Post
    Hello,

    I can’t see where I am going wrong in answering this question.

    The question is “ a particle with mass 9kg is sliding down a smooth inclined slope with an acceleration of 4.9ms. Find the angle of the incline”.

    the smooth slope means no friction.

    +x is down the slope, (which i think is all i need), so i reslove this this direction.

    After resolving along F I get F = 9gsin x

    Therefore I use F=ma which gives me

    9gsin x = (9g)(4.9)

    solving for x gives me

    x = arcsin (9g X 4.9) / 9g

    the 9g’s cancel giving me

    x = arcsin 4.9 which can’t be solved.

    Any help on where I went wrong?
    Your (9g)sinX = (9g)(4.9) is not right.
    It should be (9g)sinX = 9(4.9)

    F = ma = mass*acceleration.
    9g is a force, not a mass.

    So,
    g*sinX = 4.9
    Since g = 9.8 m/sec/sec, thenn
    sinX = 4.9/9.8 = 1/2
    X = arcsin(1/2) = 30 degrees ------------answer.
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  3. #3
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    earboth's Avatar
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    Quote Originally Posted by pablohacker View Post
    Hello,

    I can’t see where I am going wrong in answering this question.

    The question is “ a particle with mass 9kg is sliding down a smooth inclined slope with an acceleration of 4.9ms. Find the angle of the incline”.

    the smooth slope means no friction.

    +x is down the slope, (which i think is all i need), so i reslove this this direction.

    After resolving along F I get F = 9gsin x

    Therefore I use F=ma which gives me

    9gsin x = (9g)(4.9)

    solving for x gives me

    x = arcsin (9g X 4.9) / 9g

    the 9g’s cancel giving me

    x = arcsin 4.9 which can’t be solved.

    Any help on where I went wrong?

    Hello,

    if the particle is accelerated at a=4.9 \frac{m}{s^2} then you can calculate the accelerating force:

    F=m \cdot a \Longrightarrow F=9 kg\cdot 4.9 \frac{m}{s^2}=44.1N

    The weight of the particle is directed to the centre of the eart and can be calculated by:

    W=m \cdot g \Longrightarrow W=9 kg \cdot 9.81 \frac{m}{s^2} \approx 88.29 N

    Now \sin(\alpha)=\frac{F}{W} \Longrightarrow \sin(\alpha)=\frac{44.1N}{88.29N} \approx 0.49949... \approx 0.5

    Therefore α = 30°

    EB
    Attached Thumbnails Attached Thumbnails moving particle with friction-schief_ebeneohnreibg.gif  
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  4. #4
    Newbie
    Joined
    Feb 2007
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    thank you allot, i see its just that i put the mass as 9g when its measured in killograms not newtond. thank you for pointing out my silly mistake. (now that helps with the rest of the homwork.)
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