Timeline for Why doesn't limiting friction contradict Newton's Third Law?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 5, 2017 at 15:27 | vote | accept | Pancake_Senpai | ||
| Jul 5, 2017 at 15:26 | comment | added | Pancake_Senpai | Thank you both for all your help Javier and @Steeven. I finally "get" the third law. I would accept both your answers if I could but I guess I'll just flip a coin and see who gets the big green tick. | |
| Jul 5, 2017 at 15:23 | comment | added | Pancake_Senpai | It's clicked. I understand it now. The object's weight does NOT act on the surface in any way, and neither does the $30sin(30)$ force. They both act on the object to pull/push it down. This force is transmitted by the ball to the table, so the table pushes back on the object with force R. Meanwhile, the $30cos(30)$ force pushes horizontally on the object. This has NO impact on the ground. However, when the object moves right it pushes the surface to the right (friction), so the surface responds by pushing the object to the left. | |
| Jul 5, 2017 at 15:05 | comment | added | Bill N | The 3rd Law tells you how the forces on two different objects compare with each other due to the singular interaction of those two objects. The ball touches the ground, therefore, the ground exerts a force on the ball AND the ball exerts an equal and opposite force on the ground. There might be other forces from other interactions. Those other forces affect the acceleration, based on 2nd Law. The 3rd Law tells us nothing about the acceleration or about other forces that might act on an object. 3rd Law doesn't apply to the net force, only to the singular direct interaction. | |
| Jul 5, 2017 at 14:54 | comment | added | Javier | The third law doesn't say that all forces must be balanced; it says that if object A exerts a force on object B, the object B exerts an equal an opposite force on object A. Having two objects is essential. In your example, the vertical force on the object is balanced because of the second law, not because of the third law. Since there's no assumption that the object doesn't move horizontally, there is no reason the horizontal forces should be balanced. See physics.stackexchange.com/questions/45653/… | |
| Jul 5, 2017 at 14:39 | comment | added | Pancake_Senpai | R is counterbalancing the net vertical force, yet Fr isn't counterbalancing the net horizontal force. Why does this occur according to the THIRD law? I understand the second law, but its the third law I'm struggling with. | |
| Jul 5, 2017 at 14:12 | comment | added | Javier | Again, the reaction force is equal to the total downwards force on the object because of the second law, not because of the third, and it's only because we're assuming the object doesn't move vertically. There is no reason to expect the horizontal forces will be balanced. | |
| Jul 5, 2017 at 14:11 | comment | added | Javier | Indeed you have to assume that. The ground has a maximum force it can withstand before collapsing, and if the pushing force is smaller than that, the object won't move vertically. R is a force exerted by the ground on the object, so its pair is a force exerted by the object on the ground, and it has the same magnitude. It doesn't appear in your diagram because you only drew the forces acting on the object. | |
| Jul 5, 2017 at 14:03 | comment | added | Pancake_Senpai | Doesn't that require the assumption that the object is not going to move vertically? If you were asked to calculate the direction of motion of the object, you would have to assume that there is no vertical motion to apply that logic. Also, don't the 2nd and 3rd laws contradict in this case? What is the reaction pair force with R? If it is the weight of the object only, then by increasing R is not balanced, which disobeys the 3rd law. If it is the net vertical force as transmitted through the object, then why is the reaction pair of Fr not the net horizontal force? | |
| Jul 5, 2017 at 13:59 | comment | added | Javier | $R$ balances the vertical forces because, since the object is not moving vertically, its vertical acceleration is zero. Therefore, the sum of all vertical components of forces is zero. | |
| Jul 5, 2017 at 13:15 | comment | added | Pancake_Senpai | I understand from your answer that the 30N force is not acting between the particle and the surface, hence the frictional force is not matching it (because the frictional force's force pair is the equal in magnitude but opposite in direction frictional force, also between the particle and the ground). However, why then does R balance out all downwards vertical forces. The normal reaction is between the particle and the ground, so why should the 30N force effect it? | |
| Jul 5, 2017 at 13:05 | history | answered | Javier | CC BY-SA 3.0 |