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Mechanics M1 Newton II acceleration?? Please help asap!! watch

    • Thread Starter

    This question is really perplexing me - help would be so much appreciated!! Please see attached...

    I managed to solve i) and ii) without any trouble; when resolving for ii) horizontally and vertically I got answers of 11.4N and 29.7N respectively for the total forces acting on the fish.

    But iii), finding the magnitude and direction of the acceleration of the fish has me stumped. This is what I've done so far anyway:

    Using Resultant force = mass x acceleration, I took the resolved vertical force of the previous question to be the total upward (resultant force) acting on the fish (i.e. 29.7N), then the mass to be 1.9kg as stated in the question.
    So therefore 29.7 =1.9 x a
    29.7 = 1.9a
    29.7/1.9 = a
    a = 15.6m^-2

    But I was meant to get an answer of 16.7m^-2 :confused:

    It would be great if someone could also let me know if my method of solving this question was completely wrong, and might be a disadvantage if I try solving other questions this way. Hopefully it's nothing major though.
    I'm also not sure how to begin finding the direction of acceleration.

    Thanks in advance!
    Attached Images

    Your approach of using:

     \displaystyle F_{\text{overall}} = ma,

    to find the acceleration seems fine but you need to apply it to the overall force on the fish, not just the vertical force component (there is still a horizontal force component on the fish!)

    You have a vertical force component and a horizontal force component.
    Can you find the overall force on the fish? (Drawing a force triangle might help and this may help with finding the direction as well).
    • Thread Starter

    I've drawn a force triangle like you suggested... sorry but I think I'm definitely still doing something silly, and it's not working out. Please see attached also.

    Using the sine rule and taking the resultant force of the third side of the triangle to be F:

    30/sin50 = F/sin90, which should mean F = 30xsin90/sin50 = 39.1622...

    Then again using F=ma gives 39.1622... / m = a
    39.1622... / 1.9 = 20.6m s^-2 (3 s.f) :confused::confused:

    Have I drawn the triangle wrong?
    Sorry I'm a bit slow to get this. Thanks for your help!
    Attached Images

    In ii), you found the horizontal force component and the vertical force component.
    Use these to find the resultant force and then the resultant acceleration.

    For the angle, I have seen the answer in the textbook of the angle of direction but I am not sure how they are measuring the direction as they state 69 degrees as the answer.
    Attached Images
    • Thread Starter

    Yay, thanks I just got the angle!!

    arctan(11.4/29.7) = 21 degrees

    Then 180- (90+21) = 69

    Now I'm just working on the other part...
    • Thread Starter

    Got it!!! Wow, I don't know how I didn't see that before!

    So the resultant force is √29.7^2+11.4^2 = 31.8127...

    Then, using F=ma, acceleration is 31.8127.../1.9 =

    Thanks again!
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Updated: February 27, 2018
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