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    15.

    A horizontal force of 2N is just sufficient to prevent a block of mass 1kg from sliding down a rough plane inclined at arcsin7/25 to the horizontal. Find the coefficient of friction between the block and the plane, and the acceleration with which the block will move when the forice is removed.

    Thank you, ive got 3 really hard questions, and i spent 3 days trying, a bit each day, and i couldnt do it, so im posting them all, this is the first, if anyone could provide an explanation step by step etc, thanks :P ive got the correct answer, so can compare thanks
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    If you just want confirmation of the correct answer, why do you want a step by step solution? We don't tend to go through every step questions. If you have a particular bit you don't understand then we'll help - but we won't do your homework for you :p:
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    It's not homework, i have the answer because its in the back of the book ,

    ohhh the reason i wanted a step by step, is because theres a place where im going wrong... but the fact is i dont know where it is

    if that makes sense....

    thats why if i would compare mine to yours, its not homework but i need to be able to do this, i jus wish i understood
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    Are you aware of the relation between F and coeff. of friction and normal reaction?
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    Well i would suggest posting your working so far if you want people to show you your mistake. People on this board won't do the full solution for you.

    Also, was it really necessary to post the 3 questions in three different threads?
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    I know the way i asked the qeustions seemed like i wanted you all to help 1 by 1 etc,

    when i get more time ill post all that i have done ,and ask the questions which is im unsure off

    yeah im aware of the coefficient of friction etc,


    if i posted it all in one thread with working...would be all too confusoing? for me anyway
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    for this question, i know

    F = L ( taking l as lamptha) R,

    arcsin 7/25, = inv sin 25, = 16.3 rounded up...

    i dont know how to find out friction....

    for R, is it just, mgcostheta, = 1*9.81*cos16.3???
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    No, I'm afraid mg\cos\theta is not the normal reaction force: It would be if it and weight were the only forces perpendicular to the slope. However, the question states that the 2N force acts horizontally, so there will be a component of this acting into the slope: in the same direction as weight.

    If you have drawn a diagram you should be able to see that there are 3 forces to consider perpendicular to the slope: Weight, the 2N force which I'll call F and the Normal reaction. Now, perpendicular to the slope the system is in equilibrium, so The Normal Reaction force balances a component of wieght and a component of F. You've resolved weight correctly, so find that component of F and set up the equation perpendicular to the slope.

    You then find out friction using F=\mu R There F is the frictional force (make sure you know which way it will act) \mu is the coeffecient of friction and R is the normal reaction force. Unfortunatly, you know neither friction nor the co-efficient of friciton. So, consider the forces parallel to the slope. Again, the system is in equilibrium parallel to the slope, so, a component of weight is equal to a component of F and Friction. A second equation to find Friction, then come back to F=\mu R to get \mu. Easy

    Then, when you remove the force, the coeffecient of friciton is always constant, but you have to find the normal reaction force again, use F=\mu R to find the frictional force and say net force downslope = Downslope component of weight - friction. Finally, Net downslope force =ma so divide through by m (which happens to be 1) to get the downslope acceleration.

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    What you say makes sense

    So, the

    Normal = weight + F,

    normal = 9.81 + F...., and F = 2? so normal = 11.81?
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    (Original post by lilpenguin)
    15.

    A horizontal force of 2N is just sufficient to prevent a block of mass 1kg from sliding down a rough plane inclined at arcsin7/25 to the horizontal. Find the coefficient of friction between the block and the plane, and the acceleration with which the block will move when the forice is removed.

    Thank you, ive got 3 really hard questions, and i spent 3 days trying, a bit each day, and i couldnt do it, so im posting them all, this is the first, if anyone could provide an explanation step by step etc, thanks :P ive got the correct answer, so can compare thanks
    gN [≈9.81N) is the force of the block downwards, because m is 1. Work everything out relative to the angle that the plane is at - convert the 2N horizontal force into components parallel and perpendicular to the plane, the parallel one being the important one. Likewise, convert g into vector components parallel and perpendicular to the plane.

    When you've done that, you'll have 2 opposing forces acting parallel to the plane. There ought to be a net downward force

    F = μR, where μ is the coefficient of friction. R is the perpendicular force you've worked out when finding the component vectors of g. Subtracting the component of the 2N force from the component of the g force will give you a value for F. Now you have F and R, μ is your [Blue] Öyster [Cult].

    When you've done that, the final part is dead easy. You merely need to calculate parallel component of g - F [force of friction, which = μR]. F = ma, but as m is 1 (the examiner is kind here), F = a, so this value is the acceleration. :tsr2:
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    thanks... that kind of makes sense...but i may aswell give up now, 3 days...
    i odnt understand all the words... english isnt first language... so it just doesnt make sense to me... i can only work with the numbers....

    normal = weight + 2n force acting... as its equilibrium...

    weight = 9.81....2n acting horizonal...so R = 11.81?
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    (Original post by lilpenguin)
    normal = 9.81 + F...., and F = 2? so normal = 11.81?
    Afraid not again. F does not act directly into the slope. Nor does it act parallel. Only a component of F acts into the slope, in the same way a component of weight does.

    Therefore, R=mg\cos\theta + F\sin\theta

    R= 9.81 \times \frac{24}{25} + 2 \times \frac{7}{25}

    R=9.9776N OK?
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    omg you are great henerz! :P i understand it with numbers much more, i get really bad confused otherwise..

    i understand greatly the 9.81* 24/25,

    but where does the 2 * 7/25 come from?? thats what i cant get,

    + F sin theta, is so 2 * sin 7/25,
    why?
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    Ok, consider just the component F acting into the slope: F \sin\theta

    F=2 and \theta = \arcsin \frac{7}{25}

    \sin \theta= \sin( \arcsin (\frac{7}{25}))

    therefore, \sin \theta = \frac {7}{25}

    So, F \sin \theta = 2 \times \frac{7}{25}
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    What level is this question? i guess it must be university physics or engineering of some sort.
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    thanks for all the help i dont understand that,.... sin= o/h... arcsin = inv sin 7/25....

    wtf
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    its not university physics... and i dont think that 3 hours trying to do 18 questions is helping now im just restless and its startign to p*** me off
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    (Original post by Andrew P)
    What level is this question? i guess it must be university physics or engineering of some sort.
    You're joking, right? This is AS-level Maths or Physics!
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    (Original post by Andrew P)
    What level is this question? i guess it must be university physics or engineering of some sort.
    :rofl:
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    LOL. My mistake. Its justs that i'm doing IB HL Physics and we havent done arcsin or coefficients of friction. For the moment anyway...
 
 
 
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