# Related Rate

1. Jul 1, 2008

### cmajor47

1. The problem statement, all variables and given/known data
The force F (in pounds) acting at an angle θ with the horizontal that is needed to drag a crate weighing W pounds along a horizontal surface at a constant velocity is given by
F= (μW)/(cosθ+μsinθ)
Where μ is a constant called the coefficient of sliding friction between the crate and the surface. Suppose that the crate weighs 150 lbs and that μ=0.3

Find dF/dθ when θ=30°. Express the answer in units of pounds/degree.

2. Relevant equations
F= (μW)/(cosθ+μsinθ)

3. The attempt at a solution
F= 45/(cosθ+.3sinθ)
dF/dθ=[(cosθ+.3sinθ)(d/dθ 45)]-[45(d/dθ cosθ + .3d/dθ sinθ)] / (cosθ+.3sinθ)2
dF/dθ=[-45(-sinθ+.3cosθ)]/(cosθ+.3sinθ)2
dF/dθ=(45sinθ-13.5cosθ)/(cosθ+.3sinθ)2

My problem is, when I plug in θ=30° isn't my answer in lbs not lbs/degree ?

2. Jul 1, 2008

### HallsofIvy

Staff Emeritus
The derivative formulas used for the trig functions assume that the angle is given in radians not degrees. And radians, since they are not really angles at all, do not have units. What you really have is "pounds per radian"- and there are 180 degrees per $\pi$ radians. Multiply by the "unit fraction" $\pi/180$ to convert that to "pounds per degree".