Question

A box weighing 2000 N is to be slowly slid through 20 m on a straight track with friction coefficient 0⋅2 with the box. (a) Find the work done by the person pulling the box with a chain at an angle θ with the horizontal. (b) Find the work when the person has chosen a value of θ, which ensures him the minimum magnitude of the force.

Answer 1

Given:
$\mathrm{Weight}=2000\mathrm{N},\mathrm{s}=20\mathrm{m},\mu =0.2$

The free-body diagram for the box is shown below:




(a) From the figure,
$$\begin{gathered} R+P\sin \mathrm{\theta }-2000=0...\left(i\right) \\ P\cos \mathrm{\theta }-0.2R=0...\left(ii\right) \end{gathered}$$

From (i) and (ii),
$$\begin{gathered} P\cos \mathrm{\theta }-0.2\left(2000-P\sin \mathrm{\theta }\right)=0 \\ P\left(\cos \mathrm{\theta }+0.2\sin \mathrm{\theta }\right)=400 \\ P=\frac{400}{\cos \mathrm{\theta }+0.2\sin \mathrm{\theta }}...\left(iii\right) \end{gathered}$$

So, work done by the person,

$$\begin{gathered} W\mathit{=}PS\cos \mathrm{\theta } \\ =\frac{8000\cos \mathrm{\theta }}{\cos \mathrm{\theta }+0.2\sin \mathrm{\theta }} \\ =\frac{8000}{1+0.2\tan \mathrm{\theta }} \\ =\frac{40000}{5+\tan \mathrm{\theta }}...\left(iv\right) \end{gathered}$$

(b) For minimum magnitude of force from equation (iii),
$$\begin{gathered} \frac{d}{dk}\left(\cos \mathrm{\theta }+0.2\sin \mathrm{\theta }\right)=0 \\ \Rightarrow \tan \mathrm{\theta }=0.2 \end{gathered}$$
Putting the value in equation (iv),
$$\begin{gathered} W=\frac{40000}{\left(5+\tan \mathrm{\theta }\right)} \\ =\frac{40000}{5+0.2}\approx 7692\mathrm{J} \end{gathered}$$