vdfortd
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Homework Statement
The direction cosines of the force F are cos \varthetax= .818, cos\varthetay= .182, cos\varthetaz= -.545. The support of the beam at O will fail if the magnitude of the moment of F about O exceeds 100 kN-m. Determine the magnitude of the largest force F that can safely be applied to the beam.
In the drawing, the give the length of the beam, which would be the "r" variable, as 3 m. So r=3 m.
Also, the dots between variables stands for the dot product, and the X between variables stands for the cross product.
Homework Equations
|Mp|= D|F|
|Mp|= r X F <-- Not sure if this one applies here.
F=|F|ef with ef being the unit vector of F
The Attempt at a Solution
Our professor gave us that the direction cosines are the unit vector, so ef= .818i +.182j -.545k.
They said that the magnitude of the Moment could not exceed 100 kN-m, so I set up the equation 100=D|F|. If D is the perpendicular line from O to the force, than the length is r, which as stated above = 3 m.
So now I have:
100=3|F| or 100/3 = |F|
I know that the magnitude of F = the square root of (x2 + y2 + z2)
When I do the math I just get that F = 33.33 when the answer is supposed to be 58 kN.
Homework Statement
Use Equations 4.5 and 4.6 to determine the moment of the 20-N force about (a) the x axis, (b) the y-axis (c) the z axis.
Gives me the position (7, 4, 0)m and the force vector 20k (N)
Homework Equations
Equation 4.5 : ML= [e .(r X F)]e
Equation 4.6 : e.(r X F) = the determinate
The Attempt at a Solution
Using the equation ML= [e .(r X F)]e. I can do the r X F part. I did that and got 80i -140j +0k. When when I have to dot that with e, which is a unit vector, i don't know where to get that unit vector from.
I can't do the unit vector of (7,4,0) because it doesn't come out right.
The answers are a=80i b= -140j and c=0
Thanks for any and all help.