Elevator action-reaction pair/normal forces

[EddiePhys]

In the above case, the normal force exerted by the elevator on the box would be = 10kg * g + 10kg*5m/s^2, assuming g = 10m/s^2, the elevator would exert a normal force of 150N on the box. But wouldn't the box exert an equal and opposite force on the elevator resulting in no motion? I know this is a really stupid question but I've not been able to understand where I'm going wrong.

Another related question: the box exert a force of mg = 10kg*g = 100N on the elevator, or, since the elevator is pushing on the box with 150N, it will react by pushing down on the elevator with 150N

 

[jbriggs444]

the elevator would exert a normal force of 150N on the box. But wouldn't the box exert an equal and opposite force on the elevator resulting in no motion?

The only forces that affect the motion of the box are forces on the box. The only forces that affect the motion of the elevator are forces on the elevator.

The force of the box on the elevator is not a force on the box. It is a force on the elevator.

 

[EddiePhys]

The only forces that affect the motion of the box are forces on the box. The only forces that affect the motion of the elevator are forces on the elevator.

The force of the box on the elevator is not a force on the box. It is a force on the elevator.

I never stated otherwise.

 

[jbriggs444]

I never stated otherwise.

Then why would you expect the downward 150 N force of the box on the elevator to prevent the box from moving?

 

[EddiePhys]

Then why would you expect the downward 150 N force of the box on the elevator to prevent the box from moving?

No, what I'm saying is that if the elevator is exerting a force of 150N on the box, then by Newton's Third Law, the box too should exert an equal and opposite force on the elevator, preventing motion. And the other question I was confused about was whether the box would exert an equal and opposite force on the elevator =150N or whether it would exert of force of mg = 100N and the reasons/explanation behind it.

 

[jbriggs444]

No, what I'm saying is that if the elevator is exerting a force of 150N on the box, then by Newton's Third Law, the box too should exert an equal and opposite force on the elevator, preventing motion.

Equal and opposite force, yes. But preventing motion... What other forces act on the elevator? Must they total to 150 N upward?

 

[EddiePhys]

Equal and opposite force, yes. But preventing motion... What other forces act on the elevator? Must they total to 150 N upward?

Just assuming that the elevator exerts 150N force upwards

 

[jbriggs444]

Just assuming that the elevator exerts 150N force upwards

If you are considering the motion of the elevator then the force that it exerts upwards is irrelevant. See post #2 above.

 

[A.T.]

But wouldn't the box exert an equal and opposite force on the elevator resulting in no motion? I know this is a really stupid question but I've not been able to understand where I'm going wrong.

https://www.lhup.edu/~dsimanek/physics/horsecart.htm

 

[SammyS]

[ ATTACH=full]99361[/ATTACH]
In the above case, the normal force exerted by the elevator on the box would be = 10kg * g + 10kg*5m/s^2, assuming g = 10m/s^2, the elevator would exert a normal force of 150N on the box. But wouldn't the box exert an equal and opposite force on the elevator resulting in no motion? I know this is a really stupid question but I've not been able to understand where I'm going wrong.
...

There are other forces being exerted on the elevator.

The NET force exerted on the elevator must be such that it accelerates upward, in this case at 5 m/s² .

 

[Chestermiller]

The force balance on a body should only include the forces that other bodies exert on it, not the force that it exerts on other bodies. Didn't they teach you how to do a force balance in your course?

 

[EddiePhys]

The force balance on a body should only include the forces that other bodies exert on it, not the force that it exerts on other bodies. Didn't they teach you how to do a force balance in your course?

If the elevator exerts 150N on the box, why wouldn't the box exert 150N on the elevator? Everywhere I'm looking, it seems like the box is only exerting mg = 100N on the elevator.

 

[Dale]

In the above case, the normal force exerted by the elevator on the box would be = 10kg * g + 10kg*5m/s^2, assuming g = 10m/s^2, the elevator would exert a normal force of 150N on the box. But wouldn't the box exert an equal and opposite force on the elevator resulting in no motion?

For Newtons 2nd law the acceleration of the box is determined by the sum of the forces on the box. Other forces acting on other objects are irrelevant. The forces acting on the elevator are acting on the elevator and not on the box. Therefore they are not relevant for the acceleration of the box.

 

[Michael C]

If the elevator exerts 150N on the box, why wouldn't the box exert 150N on the elevator? Everywhere I'm looking, it seems like the box is only exerting mg = 100N on the elevator.

If the elevator exerts 150N upwards on the box then the box does indeed exert 150N downwards on the elevator.

 

[EddiePhys]

Alright. Thanks tons guys, I finally got it.

 

[Chestermiller]

If the elevator exerts 150N on the box, why wouldn't the box exert 150N on the elevator? Everywhere I'm looking, it seems like the box is only exerting mg = 100N on the elevator.

Then you're looking in the wrong places. When the elevator is not accelerating, the box is exerting 100 N on the elevator, but when the elevator is accelerating, the box is exerting 150 N on the elevator. The action-reaction pair always match.

 

[OldYat47]

I'll jump in with my simple-minded answer. The box weighs (10 kg * G). The acceleration is upward at (5 m/s^2). So the floor of the elevator is pushing up on the box with a force of [10 kg * (G + 5 m/s^2)]. If we put a scale on the floor of the elevator under the box it will show a weight of 10 kg * (G + 5 m/s^2).

Now let's look at the cable at the top of the elevator. It is under tension, that tension is [(10 kg + mass of the elevator) * (G + 5 m/s^2)]. When the elevator is not moving the tension on the cable is (10 kg + mass of the elevator) * G.

 

[A.T.]

When the elevator is not moving the tension on the cable is (10 kg + mass of the elevator) * G.

Or when it's moving at constant velocity.

 

[lychette]

draw a free body diagram for each object...there are 2 forces acting on the box...gravity (weight of the block) and the force of the floor of the lift.
there are 2 forces acting on the floor of the lift...gravity (ie weight) and the force of the block on the floor.
Stop waffling !