Solve the following :

Question:

Suppose the friction coefficient between the ground and the ladder of the previous problem is $0.540 .$ Find the maximum weight of a mechanic who could go up and do the work from the same position of the ladder.

Solution:

Here,$\mu=0.54$

Translatory Equilibrium

$N_{1}=m g+16 g{ }_{-(i)}$

$N_{2}=f f=\mu N_{1}-$ (ii)

Rotational Equilibrium about point 'O'

$N_{2}\left(10 \sin 53^{\circ}\right)=m g(8 \cos 53)+16 g \cos 53^{\circ}$-(iii)

Solving (i), (ii) and (iii)

$m=44 \mathrm{~kg}$

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now