Newtons 3rd Law question...?

Question

This states that if an object A exerts a force on
an object B, object B will exert the same magnitude of force on A, but in the opposite direction.

So my question is; if a box is stationary and I want to push it, according to the 3rd law, the box will push back with an equal but opposite force. So if the forces cancel out, how do I get the box moving, because 1 force has to be greater than the other for there to be a non-zero resultant force.

Where have I gone wrong in this thinking?

Thanks

Answer 1

The forces will not cancel out since they are not acting on the same body.
You exert a force on the box and the box exerts and equal and opposite force back on you. The two forces are indeed acting in opposite directions with equal magnitude, however they are acting on different bodies and therefore will not cancel each other out, so there will be a net acceleration of the box (and you).

However, if you look at the system as a whole (both you and the box together), the forces will, in a way, cancel out. A force causes change in momentum. Since you exert a force on the box to change its momentum, the box exerts a force back on you which changes your momentum. Since the forces are equal, the change in momentum is equal (and in the opposite direction). So the net change in momentum of the system is zero.

Answer 2

The box and the man who's trying to push it are two different bodies. When calculating total force you take into account forces acting on the same body.

In order to be able to push the box, the man has to stand on a surface with friction (Indeed, if we try to push a box while stanfing on ice for instance then we'll probably end up with a broken nose, since, most likely, we'll land on our face).

Let's study forces on each body separately.

a) BOX There are two forces: static friction T from the surface and the force F exerted by the man. Static friction is somewhat strange. Its magnitude is not defined. What is know about static friction is its maximum magnitude Tmax = n * A, where n is the coefficient of static friction and A the normal force pressing the box on the surface (in most cases A is the weight of the box). For the magnitude of static friction T we have

0<=T<=Tmax

So when no external force is exerted on the box (i.e. it doesn't have the "tendency" to move) static friction is also zero, so Ftotal=0 and the box doesn't move.
If we exert a "small" horizontal force F on the box then T <> 0 and with an equal magnitude to F (but with opposite direction), so again Ftotal=0 and the box doesn't move.
If we increase the magnitude of F, then we have an equal increase in the magnitude of T, so again Ftotal=0. But, remember, the magnitude of T can't increase for ever. As we increase F inevitably there will come a moment when F=T=Tmax. If we keep on increasing F then Ftotal=F-Tmax>0 and the box starts moving. So no problem there!

(NOTE: as soon as the box starts moving then there's no longer static friction. Instead there is sliding friction. Since the coefficient of sliding friction is most of the times less than the coefficient of static friction, this explains why it's more difficult to pyt a body in motion that to keep it moving).

b) MAN As soon as the man pushes the box he feels the opposite force F acting on his body. In order to "keep his ground" he pushes with his feet (if the box is really heavy he may have to bend forward in order to avoid sliding) so that the friction T(man) from the ground eliminates the effects of F. So the forces that cancel out are the ones that act on the man, namely F from the box and T(man) from the ground. These have nothing to do with the force F that acts on the box!

Answer 3

I think I have missed the whole point. Let's assume I am in "outer" space and I am holding the box. Let's assume that the box has the same mass as me. When I push on the box with force F, the box pushes back at me with an equal force, but in the opposite direction. The box moves away from me at speed v/2 and I move away from the box at a speed v/2 (total relative speed v).

So the total speed of the whole system is zero. The forces result in acceleration of the box and me in opposite directions, until I let go of the box, and which time our velocities are v/2 in opposite directions. What is so difficult about that?

Answer 4

The third law assumes that no external force works on the person-box system. This would be true somewhere like space, for example, but in real life on Earth there IS another force: friction. The friction between your feet and the floor keeps you in place and allows you to move the box. You feel the box pushing back against you when IT has friction, too, but when you overcome that friction the box moves easily while you do not move.

Answer 5

ok it is true that but B will excert a force backwards in the DIRECTION of A

so like a gun and the recoil the recoil appears to be obviously less forceful than the bullet that is because of the various ways through which energy escapes when B excert its reverse force on A.

And bout your box thing yes the two eqal forces are oppsite and your right in thinking that they cancel each other out BUT remember the cancelling force which u think is working is NOT acting on the Box its acting on YOU. thats why the box moves

Answer 6

the box moves when you push it b'coz the intensity of the force applied by your body is greater than with which the box neutralises your forces effect .got it

Answer 7

The Forces you describe are acting on different body's you and the box.
you are also pushing to the Earth and vice verse

.......O
xxxx=l
xxxx..\\
positive direction
<==========
<= fby working on the box from you
=> fyb working on you from the box
<= fyE working on you from the earth
=> fEy working on Earth from you

the equation for the box
fby=m(box)*acc(box)>0

the equation for you
-fyb+fyE=m(You)*acc(You)=0

the equation for the Earth
fyE=m(Earth)*acc(Earth)=-fbY