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enter image description hereSuppose we have a bar with non-zero mass inside a hemispherical bucket fixed to the ground. Suppose, too, that there is friction between the bar and the inner spherical surface of the bucket. Part of the bar is outside the bucket. Why, when the bar is about to tip over, is the normal force of the bucket on the lower edge of the bar pointed in the direction of the bar?

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  • $\begingroup$ please add some pictures describing the scenario. it's not very clear the way it currently is. $\endgroup$ Feb 6 at 18:06
  • $\begingroup$ Ok, @Dev Not Taken. It's done! $\endgroup$ Feb 6 at 18:23
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    $\begingroup$ Normal forces are by definition perpendicular to the surface they originate from - that's what "normal" means. Any normal force from the bucket must be perpendicular to the surface of the bucket. Any component of the force applied by the bucket that points in a non-normal direction is not a normal force. The direction of the bar here certainly doesn't look perpendicular to the bucket surface - a force pointing in that direction is not a normal force. $\endgroup$ Feb 6 at 18:30
  • $\begingroup$ The force which constrains the bar's tip to the bucket's contour will be normal to the surface of the bucket, not along the bar. One can however, resolve a component of the normal force along the bar's direction. Further since the bar makes two points of contact it will experience a normal for not only at the bar's tip, but also at the point of pivot at the top of the bucket. $\endgroup$ Feb 6 at 18:46
  • $\begingroup$ So please take a look at the following solution to question 22 in the link below. How can I explain where is the normal, is that the $N_1$ one? Is the solution wrong? physoly.tech/static/files/MockFMA-2021_sol.pdf $\endgroup$ Feb 6 at 19:00

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When the bar is about to tip over, it is impossible for the normal force to act in the direction of the bar.

Since the final situation is not very clearly specified, I will take 2 cases.

  1. The rod left side to the hemisphere is not too long.

case 1

In this case, the lower end of the rod traces the path of the hemisphere and at the end, when the rod is horizontal, normal force would be perpendicular to the rod.

cas1

Normal force at the other end would be 0 as there is no component of gravity acting at that direction at the moment.

  1. The rod left side to the hemisphere is too long.

case2

In this case, the contact point remains constant and the whole rod rotates about that point.

cas2 Here also, it is impossible for the normal force to act along the rod.

By both the cases, we can clearly see that the normal force can never be directed along the rod.

However it is to note that in the first case, a infinitely small time before the rod becomes horizontal, an infinitely small normal force acts in the direction of the rod since we can almost consider the angle between rod and hemisphere as 90 and there would be a component of $mg$ acting in that direction.

Contact If contact force was meant instead of normal force, then in all cases, the friction force at point A acts along the rod except when the rod is horizontal at which point friction force = 0.

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In this kind of problem, there are often two kinds of force. I will use a block on a table as an example.

A block and a table are both rigid objects. When gravity pushes the block downward, the block tries to accelerate downward. That would dig into the table. The table exerts an upward force on the block just strong enough to prevent the block from digging in. This force is perpendicular to the surface between the two. This is sometimes called a normal force because it is normal to (perpendicular to) the surfaces. It is sometimes called a reaction force because if the block exerts a force on the table, the table exerts an equal and opposite reaction force on the block.

Then there is friction. Friction acts sideways along the surface.

You can see the difference by considering a block on ice. There would still be a force perpendicular to the ice surface because ice is rigid. But there would be no sideways force because ice is friction free.


In your problem, the surface where the rod rests on the bowl is small and not clearly defined. It is hard to separate the normal force from friction by the direction of the surface.

So imagine the bowl is made of ice. What would be the direction of the forces at the points of contact?

You would expect the rod to slip. The length of the rod, the diameter of the bowl, and the angle of the rod would affect which way and how far.

  • A short rod nearly vertical would slip into the bowl. The bottom would slide right.
  • A longer rod nearly horizontal with a little hanging outside the bowl would slip the other way. The bottom would slip left.
  • You can find a combination of length, radius, and angle where the rod doesn't slip either way.

In both of these examples, the middle of the rod would slip along the edge of the bowl. At the edge of the bowl, friction is along the rod.

The other component is normal to the rod. This is the component that tries to make the rod dig into the bowl.

If friction is present, both forces would affect your problem. If not, the total force of the rod on the edge of the bowl is normal to the rod.

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  • $\begingroup$ so please take a look at the following solution to question 22 in the link below. How can I explain where is the normal, is that the N1 one? Is the solution wrong? physoly.tech/static/files/MockFMA-2021_sol.pdf – $\endgroup$ Feb 6 at 20:09

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