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    Need some help with this question:

    A block of mass 4 kg is on a rough plane inclined at 30° to the horizontal with a horizontal force P acting on it.

    The coefficient of friction between the block and the plane is 0.4. Find the range of possible values of P if the block is to remain stationary.
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    Can you post some of your attempts of the problem?

    (Original post by awkwardesiguy)
    Need some help with this question:

    A block of mass 4 kg is on a rough plane inclined at 30° to the horizontal with a horizontal force P acting on it.

    The coefficient of friction between the block and the plane is 0.4. Find the range of possible values of P if the block is to remain stationary.
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    Name:  ImageUploadedByStudent Room1420670431.300016.jpg
Views: 79
Size:  145.0 KBSorry, it's quite messy


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    (Original post by Phichi)
    Its against TSR rules and custom to just give the solution to a question. We try to help them by reviewing their working and pointing them in the right direction.
    I asked the question haha! It's my attempt at solving it, and it's definitely wrong
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    (Original post by awkwardesiguy)
    I asked the question haha! It's my attempt at solving it, and it's definitely wrong
    Terribly sorry, it's been a long day, didn't see the name I'll look through it.

    Edit: Before reading through it all, remember, when finding R, you need to remember both the force P and weight have a component perpendicular to the plane. You've included just P.
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    (Original post by Phichi)
    Terribly sorry, it's been a long day, didn't see the name I'll look through it.

    Edit: Before reading through it all, remember, when finding R, you need to remember both the force P and weight have a component perpendicular to the plane. You've included just P.
    I subtracted the weight force from the reaction force, which I presumed would total to zero, only leaving the P force as the resultant force. I think that's probably where I've gone wrong, but I have no idea how to go about correcting it.


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    (Original post by awkwardesiguy)
    I subtracted the weight force from the reaction force, which I presumed would total to zero, only leaving the P force as the resultant force. I think that's probably where I've gone wrong, but I have no idea how to go about correcting it.


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    Just resolve perpendicular like so:

     R = mgcos\theta + Psin\theta where \theta = 30

    Remember, to solve this, you need to imagine what the frictional force would be doing, under certain values of P. If P is large, friction will be going down the plane to stop the block from sliding up the plane, and if P is small, friction will be directed up the plane, to stop the block sliding down. You need to consider these two circumstances, to find the upper and lower bounds for P.

    Resolving parallel to the plane, in the direction up the plane:

    If friction is directed up the plane:

     F_r + Pcos\theta = mgsin\theta

    If friction is directed down the plane:

     Pcos\theta = mgsin\theta + F_r

    Where F_r = \mu R
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    (Original post by Phichi)
    Just resolve perpendicular like so:

     R = mgcos\theta + Psin\theta where \theta = 30

    Remember, to solve this, you need to imagine what the frictional force would be doing, under certain values of P. If P is large, friction will be going down the plane to stop the block from sliding up the plane, and if P is small, friction will be directed up the plane, to stop the block sliding down. You need to consider these two circumstances, to find the upper and lower bounds for P.

    Resolving parallel to the plane, in the direction up the plane:

    If friction is directed up the plane:

     F_r + Pcos\theta = mgsin\theta

    If friction is directed down the plane:

     Pcos\theta = mgsin\theta + F_r

    Where F_r = \mu R
    Yeah, I was able to solve it using that! Thank you so much! It's really helped me understand this topic and answer questions on it.


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