How can I solve problem 2/120 without knowing the vectors for the cross product?

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In summary, the student is having trouble solving for Mz in a problem that is most easily solved if you express each vector in unit vector notation. He is unclear on how to write vector r in unit vector notation. For vector P, he first needs to write it in terms of two unit vectors, one along the normal n and one along the z-axis. Once he has done this, he can write n in terms of unit vectors along the x and y axes. Finally, he needs to write r in terms of unit vector notation.
  • #1
charlies1902
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  • #2
This problem is most easily done if you express each vector in unit vector notation, then take the cross product.

Can you write vector r in unit vector notation?

For vector P you first need to write it in terms two unit vectors, one along the normal n and one along the z-axis. Once you have done this, write n in terms of unit vectors along the x and y axes and you're done.
 
  • #3
kuruman said:
This problem is most easily done if you express each vector in unit vector notation, then take the cross product.

Can you write vector r in unit vector notation?

For vector P you first need to write it in terms two unit vectors, one along the normal n and one along the z-axis. Once you have done this, write n in terms of unit vectors along the x and y axes and you're done.

doesn't having them both in unit vector notation means you'll get a diff magntitude when taking the cross product than if you were to take the cross product without the unit vector?
 
  • #4
Perhaps you don't understand unit vector notation. If vector A has components Ax= 3 units and Ay= 4 units, we would write it in unit vector notation as
[itex]\vec{A}=3\widehat{x}+4\widehat{y}[/itex] units,
where [itex]\hat{x}[/itex] stands for "in the x-direction" and [itex]\hat{y}[/itex] stands for "in the y-direction". So the above equation in plain English translates as "Vector A is the same as going three units in the x-direction and then going 4 units in the y-direction." Note that the magnitude of vector A is not one but five units. You get 5 by squaring whatever multiplies i-hat, adding to it the square of whatever multiplies j-hat and then taking the square root of this sum (Pythagorean theorem.)
 
  • #5
pyroknife said:
doesn't having them both in unit vector notation means you'll get a diff magntitude when taking the cross product than if you were to take the cross product without the unit vector?


Of course not.
 
  • #6
I'm confused now. I always thought unit vector notation was that the vector has a magnitude of 1. so for your vector wouldn't unit vector notation be 3/5i+4/5j??
 
  • #7
pyroknife said:
I'm confused now. I always thought unit vector notation was that the vector has a magnitude of 1. so for your vector wouldn't unit vector notation be 3/5i+4/5j??
It would. I note that (3/5)i+(4/5)j is a unit vector (a vector of magnitude 1) that points along the direction of A. Observe that vector A, as I have written it in unit vector notation, is the magnitude of A times a unit vector in the direction of A, i.e. A=5[(3/5)i+(4/5)j] = 3i+4j units. In this problem, for r, you have to write down a vector that has magnitude 900 mm and looks like r = (so many mm)i+(so many other mm)j.
 
  • #8
I got the right answer (208k) by shifting the x&y axis by 20 degrees thus making the force perpendicular with the y axis. Is that an accurate way to do it?
 

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