Question

(1 point) 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 t with the plane, then the magnitude of the force is F=μWμsint+cost, where μ is a constant called the coefficient of friction. Let W=45 lb and μ=0.6. (a) Find the rate of change of F with respect to t.

Answer 1

Answer:

$\frac{27(sin t - 0.6 cost)}{(0.6 sint +cost)^2}$

Explanation:

From the text of the complete problem found on internet, the expression of the force is:

$F=\frac{\mu W}{\mu sin t + cos t}$

The rate of change of F with respect to t can be find by calculating the derivative of F with respect to t.

First of all, let's recall the derivative of sine and cosine:

$\frac{d}{dt} sin t = cos t\\\frac{d}{dt} cos t = - sin t$

We have to calculate the derivative of a composite function, of the form

$\frac{k}{g(f(x))}$

Its derivative is given by

$-\frac{k}{g^2(f(x))} f'(x)$

By applying this formula to our expression, we find:

$F'=-\frac{\mu W}{(\mu sin t + cos t)^2} (\mu cos t - sin t) = \frac{\mu W(sin t - \mu cost)}{(\mu sint +cost)^2}$

And by substituting the given data:

W = 45 lb

$\mu = 0.6$

we find:

$\frac{27(sin t -0.6 cost)}{(0.6 sint +cost)^2}$