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:

\[\text{ Weight = 2000 N, s = 20 m }, \mu = 0 . 2\]

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

(a) From the figure,

\[R + P \sin \theta - 2000 = 0 . . . (i)\]
\[P \cos \theta - 0 . 2 R = 0 . . . (ii)\]

From (i) and (ii),

\[P \cos \theta - 0 . 2 \left( 2000 - P \sin \theta \right) = 0\]
\[P \left( \cos \theta + 0 . 2 \sin \theta \right) = 400\]
\[P = \frac{400}{\cos \theta + 0 . 2 \sin \theta} . . . (iii)\]

 

So, work done by the person,

\[W = PS \cos \theta\]
\[ = \frac{8000 \cos \theta}{\cos \theta + 0 . 2 \sin \theta}\]
\[ = \frac{8000}{1 + 0 . 2 \tan \theta}\]
\[ = \frac{40000}{5 + \tan \theta} . . . (iv)\]

(b) For minimum magnitude of force from equation (iii),

\[\frac{d}{dk} \left( \cos \theta + 0 . 2 \sin \theta \right) = 0\]
\[ \Rightarrow \tan \theta = 0 . 2\]

Putting the value in equation (iv),

 

\[W = \frac{40000}{\left( 5 + \tan \theta \right)}\]
\[ = \frac{40000}{5 + 0 . 2} \approx 7692 J\]