Question

A box weighing $2000N$ is to be slowly slid through $20m$ n a straight track having friction co-efficient $0.2$ with the box.
(a)Find the work done by the person pulling the box with a chain at an angle $\theta$ with the horizontal.
(b)Find the work when the person has chosen a value of $\theta$ which ensures him the minimum magnitude of the force.

Answer 1

$\left(a\right)$
$\begin{array}{c}N=2000-F\sin \theta \\ F\cos \theta =4N\\ F\cos \theta -\mu \left(2000-F\sin \theta \right)\\ F=\left(\frac{2000\times \mu }{\cos \theta +\mu \sin \theta }\right)\\ W=F.d.\cos \theta \\ =d\cos \theta .\frac{2000\times \mu }{\cos \theta +\mu \sin \theta }\\ =8000\frac{\cos \theta }{\cos \theta +0.25\sin \theta }\\ =\frac{8000}{1+0.2\tan \theta }\\ \left(b\right)\\ F=\frac{400}{\cos \theta +0.25\sin \theta }\\ =\frac{400}{{\left(\cos \theta +0.25\sin \theta \right)}^2}.\left(-\sin \theta +0.2\cos \theta \right)=0\end{array}$
we can minimize F or maximize
$\begin{array}{c}\cos \theta +0.25\sin \theta \\ \frac{d\left(\cos \theta +0.25\sin \theta \right)=0}{d\theta }\\ \left(-\sin \theta +0.2\cos \theta \right)=0\\ \tan \theta =0.2\\ \cos \theta =0.981\end{array}$
$\sin \theta =0.196$
So,
$W=8000\frac{\cos \theta }{\cos \theta +0.25\sin \theta }$
$=8000\frac{0.981}{0.981+0.25\left(0.2\right)}=7690J$