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Asphyxiated
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Homework Statement
An object with weight W is dragged along a horizontal plane by a force acting along a rope attached to the object. If the rope makes an angle θ with the plane, then the magnitude of the force is given by the following equation, where μ is a constant called the coefficient of friction.
[tex] F = \frac {\mu W} {\mu sin(\theta) + cos(\theta)} [/tex]
Find the rate of change of F with respect to theta.
Homework Equations
Quotient Rule for Differentiation
[tex] \frac {d}{dx} \frac {f(x)}{g(x)} = \frac {g(x)*f'(x)-f(x)*g'(x)}{(g(x))^{2}} [/tex]
The Attempt at a Solution
So i assumed that μW is the same as something like 2x so I took its derivative to be μ and I think that's all I need to explain so...
[tex] \frac {dF}{d\theta} = \frac { (\mu sin(\theta) + cos (\theta))(\mu) - (\mu W) (\mu cos (\theta)-sin(\theta))} { (\mu sin(\theta) + cos (\theta))^{2}} [/tex]
[tex] \frac {dF}{d\theta} = \frac {\mu^{2} sin(\theta) + \mu cos (\theta) - (\mu^{2}Wcos(\theta)-\mu W sin(\theta))} { (\mu sin(\theta) + cos (\theta))^{2}} [/tex]
[tex] \frac {dF}{d\theta} = \frac {\mu^{2} sin(\theta) + \mu cos (\theta) - \mu^{2}Wcos(\theta)+\mu W sin(\theta)} { (\mu sin(\theta) + cos (\theta))^{2}} [/tex]
Thats what I get, which is wrong according to the program we use for my class but it will never tell me what the right answer is so I am kind of at a loss here until monday unless someone can point out where I went wrong. I also re-did it while assuming μW to be a total constant with a derivative of just 1 and the only thing that changes is then:
[tex] \frac {dF}{d\theta} = \frac {\mu sin(\theta) + cos (\theta) - \mu^{2}Wcos(\theta)+\mu W sin(\theta)} { (\mu sin(\theta) + cos (\theta))^{2}} [/tex]
Which is also wrong... thanks for any help in advance!