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enter image description here

Simplified the question. Now if anybody can kindly explain the FBD for direction of normal force.

PS: do not need the solution to problem of roller curb at a step but how the Normal force at A is pointed toward the point of application of force and not toward the center of roller

Sage
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A rigid body is rigid because it pushes back when you push on it. This push back is the reaction force. The body pushes back just hard enough to keep you from pushing into the body.

For a flat surface like your first picture, the reaction force is perpendicular to the surface. Any sideways force is friction. Friction prevents or opposes two surfaces sliding past each other. This is different from penetrating into each other.

Typically, you figure out how strong a reaction force is by how strong other forces are. Gravity pulls the roller down with force $mg$. The reaction force is just strong enough to prevent this. The roller has no downward acceleration. The total vertical force is $0$. The magnitude of the reaction force must also be $mg$.

The reaction force is the same if the surface is made of ice. The roller can slide sideways freely. But it can't penetrate into the ice.

Even thought there is no perpendicular direction at the step, you can apply these ideas to figure out the direction of the reaction force.

When the roller contacts the step, which direction would cause it to penetrate into the step and which would allow the roller to move without distortion. The roller can rotate around the point of contact, but not decrease the distance from the center of the roller to the step.

There are tension and gravity forces that try to push the roller. You can resolve it into a component in the forbidden direction and perpendicular to that direction. The reaction force opposes motion in the forbidden direction.

mmesser314
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  • If the step was made of ice, ie no friction, would the roller traverse the step? – Farcher Sep 23 '23 at 14:51
  • @Farcher - Friction is perpendicular to the forbidden direction. It affects slipping as the roller passes over the step. It affects torques that make the roller rotate around the point of contact with the step. Without friction, it might be possible for the roller to rotate in place without traversing the step. It depends on the geometry and forces. – mmesser314 Sep 23 '23 at 16:16
  • "It depends on the geometry and forces" this is what i think may answer. any references for the statement. – Sage Sep 23 '23 at 17:12
  • Moreover would it make a difference if force is applied at different point as in blue diagram. a/c to me it should as moment arm would decrease in 2nd case (blue diagram). – Sage Sep 23 '23 at 17:20
  • also i am not into the solution of the problem but rather interpretation of reaction force at point A pointing towards the pull "F" from rope as shown in explanation 1. How is this interpretation made. any reference in text would help. Quoting https://physics.stackexchange.com/questions/688445/normal-force-direction "The normal force is always normal (i.e. perpendicular) to both surfaces" How is this intreptation possible in any case if above stack answer is correct. since multiple online sources have hinted at explanation 1 being real case of normal force direction but no further explanation. – Sage Sep 23 '23 at 17:25
  • @mmesser314 kindly have a relook at the question. can i generalize if >=2 forces are acting then normal force is direction towards the point of intersection of the forces and th8us if point of application is at center of roller, then normal force is towards the center iof roller which happens to be also the point of aplicatioon of force. – Sage Sep 24 '23 at 04:53
  • @Farcher kindly have a relook at the question – Sage Sep 24 '23 at 04:53
  • This may help you think address the torque - Toppling of a cylinder on a block – mmesser314 Sep 24 '23 at 14:03
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I think have resolved my own query. For anybody else for a static system we can generalize if >=2 forces are acting then normal force is direction towards the point of intersection of the forces and thus if point of application is at center of roller, then normal force is towards the center of roller which happens to be also the point of application of force. as depicted. enter image description here as if >=2 forces are acting in static equilibrium then forces have to be concurrent or parallel. in each of the case assume the static equilibrium state with only contact at A.

Sage
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  • Almost there! Forces being parallel to one another is not in itself a condition for static equilibrium as they could exert a torque but forces being concurrent is. In general what you have called a normal force is the resultant of a force acting perpendicular to the surface (normal force) and a frictional force so the red forces in your diagrams are not necessarily perpendicular to the surface of the step. – Farcher Sep 24 '23 at 06:38
  • ty sir, conceptualization of reaction vs normal force is more appropriate question now if i am not wrong. but that will do for now. just a side note - it may be that normal force is special case for 2 force system, i.e., it always opposite to any force which may be labeled as action force. Thus, for 2 surfaces (2 bodies) interacting will always be along the center of bodies (line of action of forces) which would make it to be normal (perpendicular to both surfaces) . – Sage Sep 24 '23 at 06:57