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Size:  210.4 KB for this question I can only solve what r is by resolving f which is asked in the second part of the question. How do you do a) without knowing what f is. The mark scheme suggests that rcos15 =4g but why isn't it 4g cos15 . Also how do you know the angle between r is 15? Alternate angle?

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    (Original post by coconut64)
    Name:  1457958930241-1542586955.jpg
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Size:  210.4 KB for this question I can only solve what r is by resolving f which is asked in the second part of the question. How do you do a) without knowing what f is. The mark scheme suggests that rcos15 =4g but why isn't it 4g cos15 . Also how do you know the angle between r is 15? Alternate angle?

    Thanks.
    You can get rid of F by resolving vertically - since it is a purely horizontal force, it vanishes if you resolve vertically. If you draw a big diagram and label your angles, you should be able to find the angle between the reaction and the vertical - remember that the weight is vertical so there's no need to resolve really there.
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    (Original post by coconut64)
    Name:  1457958930241-1542586955.jpg
Views: 58
Size:  210.4 KB for this question I can only solve what r is by resolving f which is asked in the second part of the question. How do you do a) without knowing what f is. The mark scheme suggests that rcos15 =4g but why isn't it 4g cos15 . Also how do you know the angle between r is 15? Alternate angle?

    Thanks.
    you have some forces resolved horizontally and vertically and some parallel to the plane and perpendicular to the plane
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    It is not 4gcos(15) because there is a vertical force component, Fsin(15), to consider as well.

    The mark scheme has resolved vertically and horizontally to the particle, this is a bit tidier because it means there is no horizontal force component to consider at this stage, you are resolving perpendicular and parallel to the plane, which is fine, but the mark scheme has gone for the former approach.
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    (Original post by Zacken)
    You can get rid of F by resolving vertically - since it is a purely horizontal force, it vanishes if you resolve vertically. If you draw a big diagram and label your angles, you should be able to find the angle between the reaction and the vertical - remember that the weight is vertical so there's no need to resolve really there.
    How? F will also be acting upward as well. Please could you upload a diagram or working ? I don't really get it. Thanks.
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    (Original post by coconut64)
    How? F will also be acting upward as well. Please could you upload a diagram or working ? I don't really get it. Thanks.
    F is purely horizontal, horizontal and vertical are two independent components, so F does not act vertically at all.
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    (Original post by Zacken)
    F is purely horizontal, horizontal and vertical are two independent components, so F does not act vertically at all.
    Ohhh i get it ! But how do you know that the angle next to r is 15 degrees as well ?
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    (Original post by coconut64)
    Ohhh i get it ! But how do you know that the angle next to r is 15 degrees as well ?
    Draw a diagram, remember that the angle between parallel and perpendicular is 90^{\circ} and that the angle between vertical and horizontal is 90^{\circ} and you know that the angle between the perpendicular and vertical is 15^{\circ}.
 
 
 
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