Vectors (1 Viewer)

carrotsss

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7. Find the unit vector of each direction vector, and then multiply them both by their respective magnitudes, then add them and you’ve found the total force (and obv you know how to find magnitude from there)

Other q: just use vector projection, and then for the perpendicular one find a line perpendicular to L (i.e. (4,-3)) and find the projection to that - I think if I remember this q correctly the solution uses Pythagoras but imo this is easier
 

CC_Cecilia

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sorry for the bad picture
I'm only year 10 now, so I'm not quite sure about that.
 

user18181818

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7. Find the unit vector of each direction vector, and then multiply them both by their respective magnitudes, then add them and you’ve found the total force (and obv you know how to find magnitude from there)

Other q: just use vector projection, and then for the perpendicular one find a line perpendicular to L (i.e. (4,-3)) and find the projection to that - I think if I remember this q correctly the solution uses Pythagoras but imo this is easier
why do we find unit vectors? can u show me ur working pls sorry i'm dumb
 

carrotsss

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why do we find unit vectors? can u show me ur working pls sorry i'm dumb
don’t have time to do full working but basically if we just multiply them by their magnitude then the force vector will have a greater magnitude than intended the one we were told to give it, which obviously isn’t right.
 

CC_Cecilia

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o
wait i don't thnk that is right
yeah. There is a mistake in the calculations because I rushed. Actually F1 is equal to 2 forces and one is (0,4) with a mag of 40 N. The other is (3,0) with a mag of 30N.
And then we do the same thing at F2.
 

Luukas.2

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don’t have time to do full working but basically if we just multiply them by their magnitude then the force vector will have a greater magnitude than intended the one we were told to give it, which obviously isn’t right.
The direction vectors given are not unit vectors as they each have a magnitude that isn't 1. The vector 3i + 4j has magnitude 5, for example. If you multiplied the given direction vector by 50, you would get a vector for the force that has magnitude 250 N.

So, to get a magnitude = 50 N vector in that direction, change the direction vector into its unit vector (3i + 4j) / 5, and then multiply by 50 to get the required vector 30i + 40j with the required magnitude.

Alternatively, since the magnitude is already 5, multiply by 10 get a vector of magnitude 50 N.

Similarly, the other vector has magnitude 13 and you want a vector in that direction with a magnitude of 65 N, so the required vector is
(65 / 13) * (12i - 5j) = 60i - 25j.

The sum of the vectors is 90i + 15j and this is the net force applied.
 

carrotsss

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The direction vectors given are not unit vectors as they each have a magnitude that isn't 1. The vector 3i + 4j has magnitude 5, for example. If you multiplied the given direction vector by 50, you would get a vector for the force that has magnitude 250 N.

So, to get a magnitude = 50 N vector in that direction, change the direction vector into its unit vector (3i + 4j) / 5, and then multiply by 50 to get the required vector 30i + 40j with the required magnitude.

Alternatively, since the magnitude is already 5, multiply by 10 get a vector of magnitude 50 N.

Similarly, the other vector has magnitude 13 and you want a vector in that direction with a magnitude of 65 N, so the required vector is
(65 / 13) * (12i - 5j) = 60i - 25j.

The sum of the vectors is 90i + 15j and this is the net force applied.
yeah this is what I was trying to say except I was too lazy

i have no clue how you have the time to do this 3 days before english though
 

Luukas.2

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Of course I'm not in High School, it's 11 o'clock at night... and it's a Sunday... what kind of a weirdo would I have to be to be in a high school now?
 

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