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    You've crossed it out so I'm assuming you got it.
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    (Original post by Clarity Incognito)
    You've crossed it out so I'm assuming you got it.
    yeah, thanks anyway for your message mate
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    Problem 6
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    "Two students are lowering a trunk down a rough slope inclined at 25 degrees to the horizontal using a single rope which is parallel to the slope, as shown in the diagram. Using mg for the weight of the trunk, T for tension in the rope, F for frictional force between the trunk and the ground and R for the normal reaction of the ground on the trunk.

    The two students have a combined mass of 110kg and can just prevent the trunk from sliding down a slope by pulling with a force which is, in magnitude 20% of their combined weight/

    (i) show that the tension in rope is about 216N"

    T +F - Mgcos25 = 0......F = 20/100*110 = 22N

    T +22 - 976N = 0, T = 954N, TOTALLY wrong, help please
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    (Original post by boromir9111)
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    "Two students are lowering a trunk down a rough slope inclined at 25 degrees to the horizontal using a single rope which is parallel to the slope, as shown in the diagram. Using mg for the weight of the trunk, T for tension in the rope, F for frictional force between the trunk and the ground and R for the normal reaction of the ground on the trunk.

    The two students have a combined mass of 110kg and can just prevent the trunk from sliding down a slope by pulling with a force which is, in magnitude 20% of their combined weight/

    (i) show that the tension in rope is about 216N"

    T +F - Mgcos25 = 0......F = 20/100*110 = 22N

    T +22 - 976N = 0, T = 954N, TOTALLY wrong, help please
    For this part of the question, you only need to read the last paragraph. Find the weight of the combined students by Weight = Mass x 9.8 and then multiply it by 0.2 to find 20%.
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    (Original post by Ramjams)
    For this part of the question, you only need to read the last paragraph. Find the weight of the combined students by Weight = Mass x 9.8 and then multiply it by 0.2 to find 20%.
    And done, very silly mistake made. Thanks for your help, much appreciated
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    Problem 7

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    http://www.mei.org.uk/files/papers/m106ju_hwi4.pdf 7(v)......i get you have to solve parallel to the slope and therefore you T + F = mg.....but i don't know how to figure it out if it's sin 40 or cos 40......i would really like a good explanation there please
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    (Original post by boromir9111)
    Problem 7

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    http://www.mei.org.uk/files/papers/m106ju_hwi4.pdf 7(v)......i get you have to solve parallel to the slope and therefore you T + F = mg.....but i don't know how to figure it out if it's sin 40 or cos 40......i would really like a good explanation there please
    Consider it to be a right-angled triangle, with the line which the weight acts down being one side, the other two being the slope and horizontal surface. The angle formed by the weight side and the surface is 90 degrees, and you know that the other angle in this triangle is 40 degrees, so the final angle is 180-90-40=50 degrees. This means that it is cos50 or sin40.

    Here's an image that might help...though it's a bit hard to see...

    Spoiler:
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    Since you are resolving perpendicular to the slope, the angle between the force and the slope is 50, hence you use cos50. Alternatively, you could use sin(90-50) or sin40 to get the same result.
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    (Original post by Ramjams)
    Consider it to be a right-angled triangle, with the line which the weight acts down being one side, the other two being the slope and horizontal surface. The angle formed by the weight side and the surface is 90 degrees, and you know that the other angle in this triangle is 40 degrees, so the final angle is 180-90-40=50 degrees. This means that it is cos50 or sin40.

    Here's an image that might help...though it's a bit hard to see...

    Spoiler:
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    Since you are resolving perpendicular to the slope, the angle between the force and the slope is 50, hence you use cos50. Alternatively, you could use sin(90-50) or sin40 to get the same result.
    Perpindicular to slope are we solving?? i thought it's parallel?? perpindicular is a straight to the slope so, vertical is always sin isn't it?? how did you get cos in there???
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    Ah yeah sorry, I mean't parallel :P

    You can use cos or sin, it just depends on which angle you are using. As a rule, if the angle you are using is between the force and axis which you are resolving along, then use cos

    if the angle you are using is between the force and angle you aren't resolving along, then use sin.

    In this example, the angle for cos50 is the angle between the slope and the weight.
    The angle for sin40 is between the weight and the line perpendicular to the slope.

    Or, you could just say that cos(x) = sin(90-x) and vice versa.
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    (Original post by Ramjams)
    Ah yeah sorry, I mean't parallel :P

    You can use cos or sin, it just depends on which angle you are using. As a rule, if the angle you are using is between the force and axis which you are resolving along, then use cos

    if the angle you are using is between the force and angle you aren't resolving along, then use sin.

    In this example, the angle for cos50 is the angle between the slope and the weight.
    The angle for sin40 is between the weight and the line perpendicular to the slope.

    Or, you could just say that cos(x) = sin(90-x) and vice versa.
    The part you highlighted, do you mind showing that to me as a image so i can vividly see it???
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    (Original post by boromir9111)
    The part you highlighted, do you mind showing that to me as a image so i can vividly see it???
    Of course. Here are a few scanned in images.

    For the first one, say that we are resolving forces parallel to the slope.



    Here, as the angle is between the slope and the weight, we use cos, and hence it is mgcos50. Alternatively, we could've used mgsin40 as cosx=sin90-x.

    This second image is the same example, but we are given a different angle.



    Here, we know the angle between the weight and the perpendicular line of the slope. However, we are resolving parallel to the slope, hence we need to use mgsin40 or mgcos(90-40) to find the component of the weight along the slope.

    Now, consider this final example which, in my opinion, makes things easier to understand.



    If we were resolving vertically, we could use cos30, or we could use sin60.

    If we were resolving horizontally, we could use cos60 or sin30.

    I hope this makes it easier to understand; it's actually very difficult to describe and it can be tricky to get your head around.

    EDIT: By the way, the first two images are from your question, but I forgot to write the 40 degrees as the angle formed between the slope and horizontal.
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    problem 8

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    http://www.mei.org.uk/files/papers/m105ju_kdht28.pdf 4(iii) i don't know where 2g comes from and why no angle is used at all.... please explain
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    (Original post by boromir9111)
    problem 8

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    http://www.mei.org.uk/files/papers/m105ju_kdht28.pdf 4(iii) i don't know where 2g comes from and why no angle is used at all.... please explain
    Markschemes are always confusing for these questions.

    Anyway, in part (ii) you calculate tension, which you've probably done, but if not you have to resolve parallel to the slope, which gives the equation:

    10+Tcos30-4gSin60=0

    Hence, rearrange to get T=27.7(3 s.f.)

    Now, for (iii) you have to resolve perpendicular to the slope.
    This results in the equation:

    R+Tsin30-4gSin30=0

    You may think that this doesn't look similar to the markscheme equation, but if you remember that Sin30 = 0.5, then you can simplify it down to:

    R+\frac{T}{2}-2g=0

    Then, sub in T and rearrange to get the equation for R.

    Hope this helps
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    (Original post by Ramjams)
    Markschemes are always confusing for these questions.

    Anyway, in part (ii) you calculate tension, which you've probably done, but if not you have to resolve parallel to the slope, which gives the equation:

    10+Tcos30-4gSin60=0

    Hence, rearrange to get T=27.7(3 s.f.)

    Now, for (iii) you have to resolve perpendicular to the slope.
    This results in the equation:

    R+Tsin30-4gSin30=0

    You may think that this doesn't look similar to the markscheme equation, but if you remember that Sin30 = 0.5, then you can simplify it down to:

    R+\frac{T}{2}-2g=0

    Then, sub in T and rearrange to get the equation for R.

    Hope this helps
    Okay thanks for that so, say the angle is 50 instead, so parallel would be sin50 and perpindicular would be sin40 right?? this is always the case isn't it?? when solving parallel, it should be cos but it's sine instead and vice versa, right???

    edit, this is the case only when it's on slope right??
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    (Original post by boromir9111)
    Okay thanks for that so, say the angle is 50 instead, so parallel would be sin50 and perpindicular would be sin40 right?? this is always the case isn't it?? when solving parallel, it should be cos but it's sine instead and vice versa, right???

    edit, this is the case only when it's on slope right??
    I would say to you that being able to resolve forces is paramount in M1, go back over how to resolve forces using triangles until you can resolve forces in any situation. The idea is that you want to resolve a force, break it up into two other forces therefore you don't want to draw lines perpendicular to the force, i.e. you have got to be careful where you draw your right angle. Ramjam's attempt above is respectable but like he says, it is quite hard to explain through text. The book will be your best bet.
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    problem 9

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    http://www.mei.org.uk/files/papers/m108ju_vik3.pdf question 7 (iv)

    horizontally 10 + Tcqcos25 - Tpccos45 = 0

    vertically Tpc sin45 + Tcqsin25 - 8g = 0

    that i got to, sin45 = cos45 so i made a substitution for them

    10 + TqcCos25 - 8g + TqcSin25 but got no idea how to solve this......?
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    anyone?
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    (Original post by boromir9111)
    problem 9

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    http://www.mei.org.uk/files/papers/m108ju_vik3.pdf question 7 (iv)

    horizontally 10 + Tcqcos25 - Tpccos45 = 0

    vertically Tpc sin45 + Tcqsin25 - 8g = 0

    that i got to, sin45 = cos45 so i made a substitution for them

    10 + TqcCos25 - 8g + TqcSin25 but got no idea how to solve this......?
    You've resolved correctly but what's up with the extra Tc on every force? just leave the forces as P and Q or it looks like you have extra variables to solve when in fact you have two. This is a standard simultaneous equation, two variables, two equations.
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    (Original post by Clarity Incognito)
    You've resolved correctly but what's up with the extra Tc on every force? just leave the forces as P and Q or it looks like you have extra variables to solve when in fact you have two. This is a standard simultaneous equation, two variables, two equations.
    Thanks for your message mate and yeah extra c is not needed lol

    yeah i get the simultaneous equation part but one has cos and the other has sine, so how do i cancel it??? sin45 = cos45....can't i make a substitution?? that don't work even if i do cause then i have sin25 and cos25 and got no idea how to cancel them down??
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    (Original post by boromir9111)
    Thanks for your message mate

    yeah i get the simultaneous equation part but one has cos and the other has sine, so how do i cancel it??? sin45 = cos45....can't i make a substitution??
    You're looking for P and Q. sin45 and cos45 have values that you can plug in. So do sin25 and cos25

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     \sin(45) = \cos(45) = \dfrac{1}{\sqrt{2}}
 
 
 
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