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    Hi.
    For question 7(ii) b, is the particle now at the bottom of the slope?
    https://211c25b87002c8b34761c926d793...20M1%20OCR.pdf
    thanks in advance.
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    (Original post by SamuelN98)
    Hi.
    For question 7(ii) b, is the particle now at the bottom of the slope?
    https://211c25b87002c8b34761c926d793...20M1%20OCR.pdf
    thanks in advance.
    The particle will be at its furthest point up the slope where it reaches instantaneous rest. The
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    (Original post by SamuelN98)
    Hi.
    For question 7(ii) b, is the particle now at the bottom of the slope?
    https://211c25b87002c8b34761c926d793...20M1%20OCR.pdf
    thanks in advance.
    No the particle is on the slope - the question says the particle comes to rest before reaching the top of the slope.

    For this question, it's important to realise that the particle is at rest so all forces acting on the particle must be balanced. So any force exterted by the plane on the particle (the contact force) must be equal and opposite to the particles weight.

    This contact force will have a normal reaction component and a friction component that acts up the slope since the particle will be on the point of slipping down the slope.

    This type of question can be confusing so let us know if you're still unsure.
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    (Original post by notnek)
    No the particle is on the slope - the question says the particle comes to rest before reaching the top of the slope.

    For this question, it's important to realise that the particle is at rest so all forces acting on the particle must be balanced. So any force exterted by the plane on the particle (the contact force) must be equal and opposite to the particles weight.

    This contact force will have a normal reaction component and a friction component that acts up the slope since the particle will be on the point of slipping down the slope.

    This type of question can be confusing so let us know if you're still unsure.
    I think i understand it, i have to find the resultant of the reaction force and friction right?
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    (Original post by SamuelN98)
    I think i understand it, i have to find the resultant of the reaction force and friction right?
    Yes you can do this.

    This may be more than is expected for the 2 marks in b) but it may be more beneficial if you go through the process of resolving friction and the normal reaction like you said.

    For b) the particle will have a tendency to slide down the plane so friction will be acting up the slope. Also, friction will not be limiting.
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    (Original post by notnek)
    Yes you can do this.

    This may be more than is expected for the 2 marks in b) but it may be more beneficial if you go through the process of resolving friction and the normal reaction like you said.

    For b) the particle will have a tendency to slide down the plane so friction will be acting up the slope. Also, friction will not be limiting.
    I was able to get the correct answer using this method is there a quicker way?
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    (Original post by SamuelN98)
    I was able to get the correct answer using this method is there a quicker way?
    For a particle on a sloped plane, the contact force from the plane has two components : the normal reaction and the friction force.

    For a particle at rest on a slope, the friction force is equal to the component of the weight parallel to the slope but in the opposite direction. And the normal reaction is equal to the component of the weight perpendicular to the slope but in the opposite direction.

    So overall the contact force must be equal to the weight but in the opposite direction. So the contact force is mg and will have angle 90 - 60 = 30 degrees to the slope.
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    (Original post by notnek)
    For a particle on a sloped plane, the contact force from the plane has two components : the normal reaction and the friction force.

    For a particle at rest on a slope, the friction force is equal to the component of the weight parallel to the slope but in the opposite direction. And the normal reaction is equal to the component of the weight perpendicular to the slope but in the opposite direction.

    So overall the contact force must be equal to the weight but in the opposite direction. So the contact force is mg and will have angle 90 - 60 = 30 degrees to the slope.
    Thanks you very much for explaining this, made it much clearer, i see why its only 2 marks now.
 
 
 
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