Elevator normal force

Question

In the above case, the normal force exerted by the elevator on the box would be $= 10\mathrm{kg}\ g + 10\mathrm{kg}\ 5\ \mathrm{m/s^2}$, assuming $g = 10\ \mathrm{m/s^2}$, the elevator would exert a normal force of $150\ \mathrm N$ 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 $mg = 10\ \mathrm{kg}\ g = 100\ \mathrm N$ on the elevator, or, since the elevator is pushing on the box with $150\ \mathrm N$, it will react by pushing down on the elevator with $150\ \mathrm N$

Answer 1

When you do a force balance on a body, you include only the forces that are acting on that body, not forces that the body exerts on other bodies. The force that the box exerts on the elevator should not included in the force balance on the box. Similarly, the force that the elevator exerts on the box should not be included in the force balance on the elevator.

Answer 2

This is a clever question, because you're thinking for yourself, rather than just letting yourself be spoon-fed formulas and answers. In short, you are correct that there is an equal and opposite force, but you seem to be forgetting another force.

One technique that I've found useful is to draw a dotted line around anything you're considering. Make sure it's a closed loop, so it doesn't have any gaps. Then, to analyze the motion of the thing inside that loop, you have to look at all of the forces passing through the loop. In particular, it doesn't matter if there are forces inside of the loop that don't pass through the loop. If there is an equal-and-opposite pair of forces inside, it doesn't change anything.

Take the simple example first. Draw a dotted line around the box, passing right between the floor of the elevator and the box. Gravity acts through that line, and the normal force from the floor acts through that line. So you're right that $150\mathrm{N}$ is applied, and the box accelerates upward at $5\mathrm{m}/\mathrm{s}$. You're also right that the force applied by the floor on the box is balanced out by an equal and opposite force from the box on the floor. But at the moment, we're just looking at the forces on the box, and they tell us that there's a net acceleration of the box.

Okay, so now let's look at the forces on the elevator, including the box. So you have to draw a dotted line around the entire elevator. And here's the value in actually drawing such a line: you now notice that the cable supporting the elevator passes through your dotted line. That means you need to take the force it exerts into account as well. There is motion because that force is large enough to overcome the $150\mathrm{N}$ exerted on the floor by the box, as well as the gravitational force on the elevator itself. For example, say the elevator itself has a mass of $1000\mathrm{kg}$. The total mass of the elevator and box is $1010\mathrm{kg}$, and its effective acceleration is $15\mathrm{m}/\mathrm{s}^2$. So the total force exerted by the cable must be $15150\mathrm{N}$. No contradiction here. This is sufficient to analyze the motion of the system.

But let's go a little further, which might help clarify things. We can also look at the motion of just the elevator. I didn't mention it above, but you can draw another dotted line inside any other one, to remove things within this inner line. So now draw a dotted line inside of the elevator, passing right between the floor and the box. (This is the same as the first line that we drew just around the box.) We're looking at the motion of just the elevator, ignoring the box, so we need all of the forces acting through either the inner or outer dotted lines. Run the numbers again, and we have (using negative for down and positive for up)

1. $-10000\mathrm{N}$ (the downward force of gravity on the elevator itself)
2. $-150\mathrm{N}$ (the downward force of the box on the floor, as you pointed out)
3. $+15150\mathrm{N}$ (the upward force of the cable, as we saw above)

Add them all up, and you find that there's a total force of $+5000\mathrm{N}$ on the elevator. Since its mass is $1000\mathrm{kg}$, it must be accelerating upward at $5\mathrm{m}/\mathrm{s}^2$, which is the same as what you gave for the acceleration of the box. Everything works out nice and consistently, and that $150\mathrm{N}$ force you correctly pointed out is included (as item 2 in the list above).

Answer 3

Two forces (action and reaction) would have cancelled each other if they would act on the same body but such is not the case here.
Here in your question gravity is pulling the mass down but the body is accelerating upward. (The body lying in the gravitational field is equivalent to body accelerating in space.)So in order to calculate the normal reaction on the elevator we use F_net=mg_net. And, for body accelerating upwards g_eff=g+a. Body accelerating upwards means, the force applied upwards = the force of gravity + the force required to produce acceleration on the body.The force applied upwards(on elevator) is equal to the normal reaction(for body) on the surface of elevator.This involves action and reaction on two different bodies which cannot cancel each other.

Answer 4

Action and reaction pairs act on different objects and won't cancel out each other (unless you deliberately add them together under suitable situations, e.g., in defining total force acting on a system.) So you are right that the box will exert an equal and opposite force ACTING on the elevator. But the force acting by the elevator on the box is still $150$ N. They won't cancel each other.

Answer 5

The box on elevator force is equal and opposite to the elevator on box force - Newton's third law. However note that these forces are acting on different bodies.

The box has two forces acting on it - the force due to the elevator upwards (150 N) and the gravitational attractive force due to the Earth downwards (100 N).
So the net force on the box is 150 - 100 = 50 N upwards and this is the force which accelerates the box upwards.

Answer 6

There is a problem with the axis that you have considered. If the elevator is going up with $5 ms^{-2}$. Then if you take the UPWARD portion of the $y$ axis. So, taking $g = 10 ms^{-2}$ . $g$ is there fore negative and the resultant acceleration would be $a = (5 - g)ms^{-2}= -5ms^{-2}$ . If you consider the $y$ Axis DOWNWARD then because of the change of the sides, that is you take reverse direction to be positive the resultant accelaration would be $5 ms^{-2}$.

And for the confusion of the conservative forces cancelling each other you have to remember $100 N $ force is the property of the Earth - Box system. Whereas $50 N$ is caused by the acceleration of $5ms^{-2}$ that the lift makes upward. They are no way action-reaction pair force. Note that only action-reaction pair force cancel each other by newtons third law .So, by newtons law they don't cancel Each other.