Could anyone explain the case of a ladder leaning on a smooth floor and against a rough wall? It's not clear how there is a vertical frictional force at the wall despite having zero normal reaction there

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    $\begingroup$ If the ladder is leaning against a rough wall, there will be a normal reaction at that point. $\endgroup$ – Sam Feb 19 at 6:04
  • $\begingroup$ but how is the horizontal equilibrium satisfied for the ladder then since there are no other horizontal forces(because the floor is smooth) $\endgroup$ – Swarnab_D Feb 19 at 6:15
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    $\begingroup$ Floor friction is required for equilibrium $\endgroup$ – Bob D Feb 19 at 8:17

As discussed below, a normal reaction force by the wall is required for rotational static equilibrium. Given a normal reaction force is needed, it follows that an equal and opposite floor friction force is needed for horizontal equilibrium.

Let’s say the reaction force normal to the wall was zero. If the force normal to the wall were zero then the static friction force at the wall would also have to be zero. In which case the wall would exert no force on the ladder and it can be removed . Clearly without the wall the ladder would fall because there would be a net counterclockwise torque about the point of contact with the floor due to the weight of the ladder and load.

Although floor friction is needed for equilibrium wall friction is not. In fact, if there is both wall and floor friction the problem becomes statically indeterminate, meaning the equations for static equilibrium are insufficient for the number of unknowns. Additional equations of deformation are required.

Hope this helps.

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