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

[charlies1902]

Homework Statement

http://books.google.com/books?id=0g...m=1&sqi=2&ved=0CBgQ6AEwAA#v=onepage&q&f=false

problem 2/120 is the question I'm having a hard time with. I know how to do everything except I just can't EYE this one.


I would like to solve for Mz by taking the cross product rxF. but I don't know how to get those vectors for this problem.

 

[kuruman]

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.

 

[pyroknife]

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?

 

[kuruman]

Perhaps you don't understand unit vector notation. If vector A has components A_x= 3 units and A_y= 4 units, we would write it in unit vector notation as
$\vec{A}=3\widehat{x}+4\widehat{y}$ units,
where $\hat{x}$ stands for "in the x-direction" and $\hat{y}$ 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.)

 

[pabloenigma]

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.

 

[pyroknife]

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??

 

[kuruman]

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.

 

[charlies1902]

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?