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If I have a situation like this enter image description here

Where my blue thing is wall and the yellow thing is the ground. If both were to be smooth then they will exert a contact force only normal to them which I have depicted as $ N_1 and N_3 $. Now let's calculate the torques :-

$\tau _1 $ = $W \times 5 ~cos 60 ~~~~~ = W \times \frac{5} {2} $

$\tau _2 $ = $ N_3 \times 10 ~ sin60 ~~~~~ = N_3\times \frac{5\sqrt{3}} {1} $

enter image description here I have taken the pivot point to be the point of contact of ground. By the right hand rule the $\tau_1 $will go into the page and $\tau_2$ will come out of the page . Now, for rotational equilibrium $$ \tau _1 - \tau _2 = 0 $$ $$ N_3 ~5\sqrt3 - W~ \frac{5}{2} = 0$$ $$ N_3 \sqrt3 = W/2 $$ $$ N_3 = \frac {W} {2 \sqrt{3}}$$ Well that just means that if $N_3 $ is $\frac{1} {2\sqrt{3} }$ of W then our ladder wouldn't rotate. SO much clear this far.

But $$ \sum F_x \neq 0 $$ because we have just $N_2$ as a horizontal force and nothing else to compensate for it. Okay, well nothing is compensating it but what it can do , all it can do is cause a motion horizontally but the ladder is pivoted at bottom and we have found if $N_3 = \frac{W} {2\sqrt{3}} $ then there would be no rotation at all about the bottom point.

But I have read that if both wall and ground were to be frictionless then the ladder would slip. What is slipping? Is it a translational motion or a rotation? As far as I can see slipping is kind of rotation. How the ladder will slip if if we have managed to do rotational equilibrium?

Thank you. Any help will be much appreciated.

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Looking at your first diagram, the ladder can not possibly be in static equilibrium. For static equilibrium both the sum of the forces and sum of the moments have to be zero. Even if the sum of the moments could be made to be zero, a free body diagram for the ladder will show that there is no horizontal force at the bottom of the ladder acting to the left opposing the horizontal normal force $N_3$ at the top of the ladder acting to the right.

Therefore 𝑁3 would cause a translational motion but due to pivoting at the bottom translation is not possible. That’s the problem. How would it slip? 𝑁3 can only cause it to rotate because the ladder is fixed at the bottom.

The ladder is not "fixed" at the bottom. You said the ground is frictionless. That means there can be no horizontal reaction at the bottom of the ladder. Without a horizontal reaction at the bottom to counter the horizontal force $N_3$ at the top, the bottom of the ladder will slip to the right.

The sum of moments (you call torques) have to equal zero AND the sum of the forces have to equal zero for equilibrium. These are two independent criteria for equilibrium and both have to be met.

Hope this helps.

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  • $\begingroup$ Therefore $N_3$ would cause a translational motion but due to pivoting at the bottom translation is not possible. That’s the problem. How would it slip? $N_3$ can only cause it to rotate because the ladder is fixed at the bottom. $\endgroup$ – user240696 Sep 18 '19 at 19:19
  • $\begingroup$ @Knight I have revised my answer to respond to your follow up comment. $\endgroup$ – Bob D Sep 18 '19 at 19:29
  • $\begingroup$ Thank you for your crystal clear explanation. Can you please tell me what will happen if both wall and ground gonna have friction? $\endgroup$ – user240696 Sep 18 '19 at 19:32
  • $\begingroup$ @Knight Do a free body diagram on the ladder. Include a static friction force parallel to the bottom surface directed to the left (to oppose the ladder slipping to the right). Include another static friction force on the wall acting vertically up (parallel to) the wall that opposes the ladder sliding down. You know that both the maximum static friction forces (forces where slipping is impending but not yet occurring) equal the normal forces (N1 and N3) times the coefficients of static friction for each surface. Lastly, do sum of moments and forces equal to zero. Good luck! $\endgroup$ – Bob D Sep 18 '19 at 20:17
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You have asserted that the torque from gravity is balanced by the torque from the normal force from the wall, but that's not right. The value you found for $N_3$ would be correct if the ground were rough enough to prevent slipping. In that case, the ladder would sit still, and the sum of all forces, and of all torques, would be zero. However, with no friction, this equilibrium is not possible, and neither the torques nor the forces sum to zero: the ladder slips down the wall, undergoing both angular and linear acceleration.

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  • $\begingroup$ Is the thing that you have said is an experimental fact? And I should accept it? $\endgroup$ – user240696 Sep 18 '19 at 18:40
  • $\begingroup$ I mean my reasoning is faulty because logical alone couldn’t account for that situation. $\endgroup$ – user240696 Sep 18 '19 at 18:40
  • $\begingroup$ I'm not sure what you're asking. Without friction, there is no force to balance N3, so the ladder cannot be in equilibrium. When you write down a formula for torques that assumes equilibrium, it is therefore not valid. $\endgroup$ – Ben51 Sep 18 '19 at 18:43
  • $\begingroup$ I’m saying that $N_3$ would cause a translational motion, because I made it such that it causes rotational equilibrium. But translation is not possible due to pivoting at the bottom. $\endgroup$ – user240696 Sep 18 '19 at 18:51
  • $\begingroup$ Again, the ladder cannot be in rotational equilibrium without a horizontal force at the bottom (aka friction). Either there is a force, in which case it's not frictionless, or there is no force, and both the top and bottom of the ladder slip. $\endgroup$ – Ben51 Sep 18 '19 at 18:53
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$N_3$ is not indeterminate, you cannot set the torque to zero, because if you assume rigidity, you can calculate both $N_1$ and $N_3$ by relating the angular acceleration to the linear acceleration. Left as homework.

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