How do I determine the magnitude of torque from a cross product?

  • Thread starter Thread starter smatt_31
  • Start date Start date
  • Tags Tags
    Magnitude Torque
smatt_31
Messages
6
Reaction score
0
[SOLVED] the magnitude of torque

Homework Statement


Determine the magnitude of the torque about A by evaluating lABxFl (the magnitude of vector AB cross vector F) where AB = -15i + 12j and F= -200lbs (cos theta)i - 200lbs (sin theta)j

Homework Equations


cross product and maybe the magnitude of torque

The Attempt at a Solution


so i did the cross product and ended up with
k = (-15(-200 sin theta))-(12(-200cos theta)) whick i simplified
to 3000(sin theta) + 2400(cos theta)

so k = 3000(sin theta) + 2400(cos theta)

so am I done or should I take the sqrt of the values = to k ??
the problem i have with this is theta is undefined and i would end up with a more complicated expression.
so what do i do
 
Physics news on Phys.org
smatt_31 said:

Homework Statement


Determine the magnitude of the torque about A by evaluating lABxFl (the magnitude of vector AB cross vector F) where AB = -15i + 12j and F= -200lbs (cos theta)i - 200lbs (sin theta)j

Homework Equations


cross product and maybe the magnitude of torque

The Attempt at a Solution


so i did the cross product and ended up with
k = (-15(-200 sin theta))-(12(-200cos theta)) whick i simplified
to 3000(sin theta) + 2400(cos theta)

so k = 3000(sin theta) + 2400(cos theta)

so am I done or should I take the sqrt of the values = to k ??
the problem i have with this is theta is undefined and i would end up with a more complicated expression.
so what do i do
That is not "k= ". What you have is the coefficient of k (unit z-vector). The magnitude is the value of that coefficient. If you wanted to be real technical, the "magnitude" is the square root of the square of that: its absolute value and so just what you have as long as it is positive.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top