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A 3m long ladder weighing 20 kg leans on a frictionless wall. Its feet rest on the floor 1 m from the wall as shown in Fig.7.27. Find the reaction forces of the wall and the floor.

Answer» <html><body><p></p>Solution :<img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NCERT_GUJ_PHY_XI_P1_C07_SLV_009_S01.png" width="80%"/> <br/> The ladder AB is 3 m long, its foot A is at distance `AC = 1 m` from the wall. From Pythagoras theorem, `BC = 2 <a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a>(2) m`. The forces on the ladder are its weight W acting at its centre of gravity D, reaction forces `F_(1) and F_(2)` of the wall and the floor respectively. <a href="https://interviewquestions.tuteehub.com/tag/force-22342" style="font-weight:bold;" target="_blank" title="Click to know more about FORCE">FORCE</a> `F_(1)` is perpendicular to the wall, <a href="https://interviewquestions.tuteehub.com/tag/since-644476" style="font-weight:bold;" target="_blank" title="Click to know more about SINCE">SINCE</a> the wall is frictionless. Force `F_(2)`is resolved into two components, the normal reaction N and the force of friction F. Note that F <a href="https://interviewquestions.tuteehub.com/tag/prevents-605529" style="font-weight:bold;" target="_blank" title="Click to know more about PREVENTS">PREVENTS</a> the ladder from sliding away from the wall and is therefore directed toward the wall. <br/> For translational equilibrium, taking the forces in the vertical direction, <br/> `N – W = 0` (i)<br/> Taking the forces in the horizontal direction, `F – F_(1) = 0` (ii) <br/> For <a href="https://interviewquestions.tuteehub.com/tag/rotational-625601" style="font-weight:bold;" target="_blank" title="Click to know more about ROTATIONAL">ROTATIONAL</a> equilibrium, taking the moments of the forces about A, <br/> `2sqrt(2)F_(1)-(1//2)W=0` (iii) <br/> Now `W = 20 g = 20 × 9.8 N = 196.0 N` <br/> From (i) `N = 196.0 N` <br/> From (iii) `F_(1)=W//4sqrt(2)=196.0//4sqrt(2)=34.6N` <br/> From (ii) `F=F_(1)=34.6N` <br/> `F_(2)=sqrt(F^(2)+N^(2))=199.0N` <br/> The force `F_(2)` makes an angle `α` with the horizontal, <br/> `tanalpha=N//F=4sqrt(2),alpha=tan^(-1)(4//sqrt(2))~~80^(@)`</body></html>


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