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    I've made a few questions and I wanted to check that the answer I have is correct
    -Find the magnitudes of the missing forces P and Q
    Name:  Physics.jpg
Views: 54
Size:  11.6 KB
    My answer:
    P= tan(20) x 120 = 43.676 N
    Q= 120/cos(20) = 127.7 N
    - A box with a weight of 1.5N is at rest on a slope inclined at 50 to the horizontal. Find the component of weight along the slope.
    My answer:
    cos(40) x 1.5N = 1.1491 N

    Thank you for your time,
    Cheese
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    I'm on my phone so can't do the math directly, but Q is equal to 120 multiplied by cos(20) instead of divided by! P would be equal to 120sin(20) since that is the horizontal component of Q and the forces cancel out. Second calculation looks good tho!
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    (Original post by CheeseIsVeg)
    I've made a few questions and I wanted to check that the answer I have is correct
    -Find the magnitudes of the missing forces P and Q
    Name:  Physics.jpg
Views: 54
Size:  11.6 KB
    My answer:
    P= tan(20) x 120 = 43.676 N
    Q= 120/cos(20) = 127.7 N
    - A box with a weight of 1.5N is at rest on a slope inclined at 50 to the horizontal. Find the component of weight along the slope.
    My answer:
    cos(40) x 1.5N = 1.1491 N

    Thank you for your time,
    Cheese

    It seems that I have a different answer to you

    Part a)
    P=Qsin20
    Qcos20=120
    Q=120/cos20 = 127.7 (3sf)
    P=43.7N
    If there's no acceleration as there are balanced forces


    Part b)
    The weight of 1.5 N acts downwards.
    Draw a right angled triangle with 1.5N acting downwards, a 50 angle at the top, 1.5cos50 opposite the 1.5N ( acting downwards). 1.5sin50 opposite the 50 degree angle acting towards the left towards the 1.5N. Look below at the triangle if it helps.

    Therefore the only force acting horizontally is 1.5sin50 to me.

    I'm not sure why we have different answers but hopefully my explanation has given food for thought.

    Name:  2016-02-20.png
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Size:  95.2 KB
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    (Original post by pinkcalculator23)
    It seems that I have a different answer to you

    Part a)
    P=Qsin20
    Qcos20=120
    Q=120/cos20 = 127.7 (3sf)
    P=43.7N
    If there's no acceleration as there are balanced forces


    Part b)
    The weight of 1.5 N acts downwards.
    Draw a right angled triangle with 1.5N acting downwards, a 50 angle at the top, 1.5cos50 opposite the 1.5N ( acting downwards). 1.5sin50 opposite the 50 degree angle acting towards the left towards the 1.5N. Look below at the triangle if it helps.

    Therefore the only force acting horizontally is 1.5sin50 to me.

    I'm not sure why we have different answers but hopefully my explanation has given food for thought.

    Name:  2016-02-20.png
Views: 50
Size:  95.2 KB

    I think that you have mistaken the angle to be at the top whereas it should be at the bottom (the incline)
    I.E:
    Name:  physik.png
Views: 39
Size:  27.7 KB
    I'm sorry, it's my fault for the bad wording, I should've added this earlier, oh well
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    (Original post by alexschmalex)
    I'm on my phone so can't do the math directly, but Q is equal to 120 multiplied by cos(20) instead of divided by! P would be equal to 120sin(20) since that is the horizontal component of Q and the forces cancel out. Second calculation looks good tho!
    Thank you for your response!
    I wasn't sure about the first one, I seemed to have made 120N the adjacent to 20 degrees with Q as the hypotenuse and P as the opposite
    I think this is why I have a different answer
    I think I need to practice these sorts of questions as I find them much harder than the one below it
    Thanks again
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    (Original post by CheeseIsVeg)
    I think that you have mistaken the angle to be at the top whereas it should be at the bottom (the incline)
    I.E:
    Name:  physik.png
Views: 39
Size:  27.7 KB
    I'm sorry, it's my fault for the bad wording, I should've added this earlier, oh well


    I think the angle should still be 50 not 40. In class we always use the same angle as that which the horizontal plane is elevated.
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    (Original post by pinkcalculator23)
    I think the angle should still be 50 not 40. In class we always use the same angle as that which the horizontal plane is elevated.
    My teacher always told us that it was 90-angle
    This angle would be between W1 and weight
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    (Original post by CheeseIsVeg)
    Thank you for your response!
    I wasn't sure about the first one, I seemed to have made 120N the adjacent to 20 degrees with Q as the hypotenuse and P as the opposite
    I think this is why I have a different answer
    I think I need to practice these sorts of questions as I find them much harder than the one below it
    Thanks again
    WAIT, ignore me! This is what happens when you try to do math right after waking up You were actually correct in the first place for the P and Q values!
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    (Original post by alexschmalex)
    WAIT, ignore me! This is what happens when you try to do math right after waking up You were actually correct in the first place for the P and Q values!
    No way!
    Are you sure?
    :I Your thinking sounded right though
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    (Original post by CheeseIsVeg)
    No way!
    Are you sure?
    :I Your thinking sounded right though
    Yep, I had it backwards in my head, mega sorry about that!
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    (Original post by alexschmalex)
    Yep, I had it backwards in my head, mega sorry about that!
    Haha don't be sorry, you've helped me so much, thank you very much!
 
 
 
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