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    A block of mass 20kg is released from rest at the top of a rough slope. The slope is inclined to the horizontal at an angle of 30 degrees. After 6s the speed of the black is 21 m/s. As the block slides down the slope it is subject to a constant resistance of magnitude RN. Find the value of R.


    I know you work out acceleration first.

    v=u + at

    v-u/t = a

    21-0/6=3.5

    I did R- 20g cos 30 = m * 0=0
    R=20gcos30

    Then what?
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    (Original post by Deep456)
    A block of mass 20kg is released from rest at the top of a rough slope. The slope is inclined to the horizontal at an angle of 30 degrees. After 6s the speed of the black is 21 m/s. As the block slides down the slope it is subject to a constant resistance of magnitude RN. Find the value of R.


    I know you work out acceleration first.

    v=u + at

    v-u/t = a

    21-0/6=3.5

    I did R- 40g cos 30 = m * 0=0
    R=40gcos30

    Then what?
    OK. You've got the acceleration correct.

    And it looks as if you realised you need F=ma.

    You need to decide which direction is positive (it doesn't matter which but you must choose one), before forming your equation. This will tell you what sign to use for the forces and acceleration.

    Don't know where you are getting the "0" from here, as that is definitely incorrect.
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    I can follow your working for the acceleration, but after that I'm lost. Here's how I will do it:

    First you need to find the component of the weight that acts parallel to the plane of the incline. This is given by:

    

F=mg \cos \theta

    where theta is the angle of the incline. Substituting the values, and assuming the gravitational constant g = 9.81 ms^-2, we get the following:

    F = (20 kg)*(9.81 ms^-2) *cos(30 deg) = 167 N (to 3 significant figures)

    Now we know that the net force acting on the block in the direction of the block's travel (i.e. parallel to the direction of the slope) is the one that accelerates the block according to Newton's Second Law, F = ma. Hence we can find the frictional force acting from the block:

    

F_{\mbox{net}} = ma\\

F-F_{\mbox{fiction}}= ma\\

F_{\mbox{friction}} = F-ma

    Substituting the values (I use your value for a):

    F_(friction) = 167 N - (20 kg)*(3.5 ms^-2) = 97 N

    *Note, please use the accurate values, otherwise you get rounding errors in your final answer. Also I belief that your answer should obey the rules of significant figures. My working doesn't have all this, for the sake of convenience.

    Now, we also know that the frictional force is given by F_\mbox{friction}=RN[\latex], where I believe N is the normal contact force and R is the coefficient of kinetic friction (though the Greek letter [latex]\mu is more commonly used. So, the above equation can be rewritten as:

    R*N = 97 N

    Also recall that N is perpendicular to the plane of contact, so we need to now find the component of force exerted by the block on the wedge/slope thing that is perpendicular to the plane of the slope, as N is equal in magnitude (but opposite in direction) to this force. This can be found by resolving the weight like we did way above:

    

N = F_\perp = mg \sin \theta
    N = (20 kg)*(9.81 ms^-2)*sin(30 deg) = 98.1 N

    Substituting this value into the equation above for the frictional force:

    

RN = 97 N \\

R = \frac{97 N}{N}

    R = 97 N / 98.1 = 0.989 (R is dimensionless)

    Hope this helps!
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    (Original post by tridian)
    I can follow your working for the acceleration, but after that I'm lost. Here's how I will do it:

    First you need to find the component of the weight that acts parallel to the plane of the incline. This is given by:

    

F=mg \cos \theta

    where theta is the angle of the incline. Substituting the values, and assuming the gravitational constant g = 9.81 ms^-2, we get the following:

    F = (20 kg)*(9.81 ms^-2) *cos(30 deg) = 167 N (to 3 significant figures)

    Now we know that the net force acting on the block in the direction of the block's travel (i.e. parallel to the direction of the slope) is the one that accelerates the block according to Newton's Second Law, F = ma. Hence we can find the frictional force acting from the block:

    

F_{\mbox{net}} = ma\\

F-F_{\mbox{fiction}}= ma\\

F_{\mbox{friction}} = F-ma

    Substituting the values (I use your value for a):

    F_(friction) = 167 N - (20 kg)*(3.5 ms^-2) = 97 N

    *Note, please use the accurate values, otherwise you get rounding errors in your final answer. Also I belief that your answer should obey the rules of significant figures. My working doesn't have all this, for the sake of convenience.

    Now, we also know that the frictional force is given by F_\mbox{friction}=RN[\latex], where I believe N is the normal contact force and R is the coefficient of kinetic friction (though the Greek letter [latex]\mu is more commonly used. So, the above equation can be rewritten as:

    R*N = 97 N

    Also recall that N is perpendicular to the plane of contact, so we need to now find the component of force exerted by the block on the wedge/slope thing that is perpendicular to the plane of the slope, as N is equal in magnitude (but opposite in direction) to this force. This can be found by resolving the weight like we did way above:

    

N = F_\perp = mg \sin \theta
    N = (20 kg)*(9.81 ms^-2)*sin(30 deg) = 98.1 N

    Substituting this value into the equation above for the frictional force:

    

RN = 97 N \\

R = \frac{97 N}{N}

    R = 97 N / 98.1 = 0.989 (R is dimensionless)

    Hope this helps!
    My textbook has a correct answer of 28N though?
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    (Original post by ghostwalker)
    OK. You've got the acceleration correct.

    And it looks as if you realised you need F=ma.

    You need to decide which direction is positive (it doesn't matter which but you must choose one), before forming your equation. This will tell you what sign to use for the forces and acceleration.

    Don't know where you are getting the "0" from here, as that is definitely incorrect.
    I was meant to say 20g not 40g.

    This is what my textbook states so I am following that.

    = R- 20gcos30= 20 * 0=0 (Resolving perpendicular to the acceleration).
    R= 20gcos30

    which is 167N as tridian pointed out but after that I do not follow.
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    (Original post by Deep456)
    I was meant to say 20g not 40g.

    This is what my textbook states so I am following that.

    = R- 20gcos30= 20 * 0=0 (Resolving perpendicular to the acceleration).
    R= 20gcos30

    which is 167N as tridian pointed out but after that I do not follow.
    I don't know why you are resolving perpendicular to the acceleration. You are not concerned with the reaction perpendicular to the plane, nor are you concerned with the coefficient of friction.

    Resolve along the plane. You'll need Sin 30, not Cos, and the answer drops out from your first equation. No other processing is necessary.
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    (Original post by ghostwalker)
    I don't know why you are resolving perpendicular to the acceleration. You are not concerned with the reaction perpendicular to the plane, nor are you concerned with the coefficient of friction.

    Resolve along the plane. You'll need Sin 30, not Cos, and the answer drops out from your first equation. No other processing is necessary.
    Could you explain it to me? I see your calculation but I do not quite follow why it is that.
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    (Original post by Deep456)
    Could you explain it to me? I see your calculation but I do not quite follow why it is that.
    OK. I'll cut to a full solution in this circumstance.

    Spoiler:
    Show

    You've already correctly worked out that the acceleration down the plane is 3.5 ms-1.

    We'll take down the plane as positive.

    What's the component of the weight acting down the plane? 20gSin30.

    What's the frictional force acting up the plane? R Newtons.

    Resultant force = 20gSin30-R.

    Mass is 20 kg.

    So, 20gSin30-R= 20 x 3.5

    R = 22gSin30 - 20 x 3.5

    Where I am using "x" to mean "times". Taking g as 9.8, we get R = 28

    Done.

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    (Original post by ghostwalker)
    OK. I'll cut to a full solution in this circumstance.

    Spoiler:
    Show

    You've already correctly worked out that the acceleration down the plane is 3.5 ms-1.

    We'll take down the plane as positive.

    What's the component of the weight acting down the plane? 20gSin30.

    What's the frictional force acting up the plane? R Newtons.

    Resultant force = 20gSin30-R.

    Mass is 20 kg.

    So, 20gSin30-R= 20 x 3.5

    R = 22gSin30 - 20 x 3.5

    Where I am using "x" to mean "times". Taking g as 9.8, we get R = 28

    Done.

    Thanks very much. What about the coefficient of friction however?
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    (Original post by Deep456)
    Thanks very much. What about the coefficient of friction however?
    What about it?

    The question you posted makes no mention of it, nor is it required in solving what you have posted.
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    Wait, you mean RN as in R Newtons or R*N?

    If it's the former, then my first working is totally wrong, and you should follow ghostwalker's.

    If it's the latter, then I assume that R is the coefficient of friction and N is the normal contact force. Wikipedia explains it well in its article on friction. http://en.wikipedia.org/wiki/Friction.

    Take note of the diagram on the top right hand corner of the page. The frictional force is given as the product of the coefficient of kinetic friction mu and the normal contact force N.

    However, seeing that your textbook gives the correct answer in term of Newtons, I would think that R is a force and N is the unit (Newton). So I misinterpreted your question; ghostwalker's is the right solution.

    My apologies for the confusion.
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    I know this is old but is might help someone
    R(diagonal up - reaction force) = R - 20g * cos(30) = 0 as there's no acceleration in this direction
    Thus R = 20g * cos(30) = 169.74
    V=u+at
    21 = 0 + 6a
    a = 3.5
    F=ma
    F=20*3.5 = 70N
    R(diagonal down - in direction of movement) = 20g*sin(30) - RN = 70N
    RN= 20 * 9.8 * sin(30) - 70N
    =28N
 
 
 
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