00:01
Consider the following diagram.
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We have a box with its edge placed at the origin of a cartesian system.
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In this diagram, we have a vector p going from point a to point b, which are two opposing diagonals of our box.
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In our first question, question a, we want to determine the cartesian coordinates of vector p.
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So generally speaking, any vector can be described by the product of its magnitude times its unit vector, say p -hat.
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The magnitude of vector p was given to us in our problem statement and is equal to 500 newtons.
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So now, what we need to do is find p -hat, which is the unit vector of vector p.
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So p -hat will correspond to the vector going from point a to point b divided by its magnitude.
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So let's start by finding vector ab.
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So vector ab starts at point a and has a length of 0 .42 in the negative i direction, a length of 0 .81 in the negative y direction, and a length of 0 .54 in the positive z direction, and this is in meters.
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Now let's calculate the magnitude of ab, and conveniently for us, the magnitude of this vector is equal to 1, which means that vector p -hat is simply equal to vector ab.
02:02
So the cartesian coordinates of vector p will be equal to 500 newtons times minus 0 .42 i minus 0 .81 j plus 0 .54 k, which will give us the following vector, minus 210 i minus 405 j plus 270 k.
02:51
Now in the next question, we want to find the moment of force of vector p around point c.
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So the moment of force, tau, is given by the force to the point c in this case.
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So vector r cross our force vector, vector p.
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So we've already determined vector p, now let's determine r.
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So r is the vector going from point c to point a.
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So this will only have a component in the x direction, which will be equal to 0 .42 i.
04:00
So now the only thing we need to do is to calculate, oh in meters, excuse me, this is in newtons, so the only thing left to do now is to calculate the cross product r with vector p.
04:25
So in the cross product, unit vectors that are parallel give us a cross product equal to 0, which means that i cross i here will be equal to 0.
04:38
So the only contributing factors are the components in j and k in this case.
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So we will obtain minus 0 .42 times 405 times i cross j plus 0 .42 times 270 times i cross k.
05:18
And the units will be in newton times meter...