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    for part A)
    why is it
    T - mgsintheta + fmax = ma
    i thought it would be T - mgsintheta - fmax = ma ?
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    (Original post by ihatePE)
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    for part A)
    why is it
    T - mgsintheta + fmax = ma
    i thought it would be T - mgsintheta - fmax = ma ?
    Clearly you are resolving up the plane. Also clearly, left alone the package slides down the plane. Friction opposes motion. So friction acts up the plane. So it is positive are you are resolving upwards.
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    (Original post by Zacken)
    Clearly you are resolving up the plane. Also clearly, left alone the package slides down the plane. Friction opposes motion. So friction acts up the plane. So it is positive are you are resolving upwards.
    thanks for clearing it up
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    (Original post by ihatePE)
    thanks for clearing it up
    Sure. Do you see why that same chain of reasoning means that friction acts down the plane in (b)?
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    (Original post by Zacken)
    Sure. Do you see why that same chain of reasoning means that friction acts down the plane in (b)?
    yes i do, friction opposes motion.
    sorry quick question,
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    how would u find the area of the triangle really simply? the markscheme says 3*4 = 12
    but i went to do cosine rule and then 1/2 * b*c*sinA which is long winded for 7 marker. i dont see why they 3*4, and where they get the numbers from
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    (Original post by ihatePE)
    how would u find the area of the triangle really simply? the markscheme says 3*4 = 12
    but i went to do cosine rule and then 1/2 * b*c*sinA which is long winded for 7 marker. i dont see why they 3*4, and where they get the numbers from
    Slice the triangle down halfway from the vertex. This gives two equal right-angled triangles, each with base 3, hypotenuse 5 and so height 4. Hence has area \frac{1}{2} \times 3 \times 4 \times 2 (since there are two such triangles).

    Edit: In case this still isn't clear, the reason why each triangle has a base of 3 is because it's slices the side of the square in half, which has length 6.
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    (Original post by Zacken)
    Slice the triangle down halfway from the vertex. This gives two equal right-angled triangles, each with base 3, hypotenuse 5 and so height 4. Hence has area \frac{1}{2} \times 3 \times 4 \times 2 (since there are two such triangles).

    Edit: In case this still isn't clear, the reason why each triangle has a base of 3 is because it's slices the side of the square in half, which has length 6.
    o right pythagorous, thank u so much.
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    (Original post by ihatePE)
    thanks for clearing it up
    could you say what the answer is. I have M1 tomorrow and I'm quite stressed.
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    (Original post by AspiringUnderdog)
    could you say what the answer is. I have M1 tomorrow and I'm quite stressed.
    https://mathsbedwas.wikispaces.com/f...9729/Jun09.pdf <--- mark scheme (question 5) says 11.82N
    i have m1 tomorrow too, good luck
 
 
 
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