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If you suddenly increase the force, is the reaction force equal to the force? Press a force on the object, and then suddenly increase the force. In this process, is the reaction force equal to the force? I don't think they are equal. Because in the process of increasing the force, the reaction force will not increase immediately. Only in the inertial system is the force equal to the reaction force.

For example, if you put a 1 kg object on the rubber block, the force of the object on the rubber block must be less than 9.8 N at first, and then when the object is still, the force of the object on the rubber block is 9.8 N.

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Since you are using the Newtonian mechanics tag here, then we will stay in that regime.

According to Newton's third law, all forces have an equal but opposite reaction force. There is no additional condition of "after enough time has passed". In other words if you have two bodies interacting by some force, the force of body 1 on body 2 is always equal and opposite to the force body 2 exerts on body 1.

You might be getting mixed up with how bodies aren't exactly rigid. So that if you start pushing on an object, then it takes a finite time for the entire body to start moving, as the part of the body closer to the point of application will start moving before the parts farther away, and the body will initially compress. But this doesn't mean the reaction force is not equal to the applied force, as this action-reaction pair is only concerned with what is happening right at the application point, not with what the entire body is doing.

For example, if you put a 1 kg object on the rubber block, the force of the object on the rubber block must be less than 9.8 N at first, and then when the object is still, the force of the object on the rubber block is 9.8 N.

The weight and the rubber block will interact via an action-reaction force pair. There is no time delay to this. I think you are confused about what a reaction force is. This pair isn't the weight of the 1 kg and the reaction force of the block. That's not a valid action-reaction force pair. The two relevant pairs are 1)"gravity from Earth on 1k weight and gravity from 1kg weight on Earth" and 2)" Force from 1kg weight on rubber block and force from rubber block on 1kg weight". Just because the rubber block isn't pushed on by a force of 9.8N at all times doesn't mean the reaction force is delayed. The weight of the 1kg weight isn't linked to the force experienced by the rubber block through Newton's third law.

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  • $\begingroup$ @enbinzheng see my edit $\endgroup$ – BioPhysicist Dec 23 '19 at 11:50
  • $\begingroup$ For example, if you put a 1kg object on the rubber block, the force of the object on the rubber block must be less than 1kg at first, and then when the object is still, the force of the object on the rubber block is 1kg. $\endgroup$ – enbin zheng Dec 23 '19 at 12:17
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    $\begingroup$ @enbinzheng When that happens, the force of the rubber block on the object is also less than 1 kg at first. The force being applied to the rubber block still has an equal and opposite reaction on the object. $\endgroup$ – JMac Dec 23 '19 at 12:38
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I don't think they are equal. Because in the process of increasing the force, the reaction force will not increase immediately. Only in the inertial system is the force equal to the reaction force.

For example, if you put a 1 kg object on the rubber block, the force of the object on the rubber block must be less than 9.8 N at first, and then when the object is still, the force of the object on the rubber block is 9.8 N

The reaction force is equal and opposite at all times as required by Newton’s 3rd law. If that were not the case then momentum would not be conserved. Let’s analyze the example more carefully.

There are two forces acting on the object. There is the 9.8 N downwards force of gravity (the weight) and the variable upwards contact force.

The third law pair of the 9.8 N weight is an upwards gravitational force on the earth. This is also a constant 9.8 N, so it is equal and opposite at all times.

The third law pair of the upwards contact force on the object is a downward contact force on the rubber block. These forces vary over time, but at each moment they are equal and opposite to each other. When the object first touches the block the forces are much smaller than 9.8 N, but as the block rapidly decelerates they quickly become larger than 9.8 N before reducing back down.

The contact forces are independent of the gravitational force except in the special case when the system is in equilibrium. It seems in your scenario that you are trying to analyze the contact force and the gravitational force as a 3rd law pair, which is incorrect. Third law pairs always act on different objects and are the same “type” of force.

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