# Nutcracker in space, torque provided by thruster vs by reaction force

In the following image, a nutcracker is floating in space. Attached to points B and C are two thrusters (not shown). Each of them provides an equal force horizontally inwards, shown by the red arrows. As a result, torque will be experienced at the fulcrum (A), and help to crack the nut.

EDIT: it is better to think about this problem without the nut. That way, the nutcracker can "close".

In the next image, another space nutcracker is shown. Here, there is only one thruster (at C). An immovable surface is shown on the left, which is touching B.

The surface provides a reaction force against B, such that the situation is equivalent to the first image.

What I'm having trouble understanding is how this reaction force can provide a torque on the green segment around point A. My confusion arises because in my mind, I have it set that in order for the green segment to experience a torque around A, it must rotate around A. And in order for it to rotate around A, point B must move (in this case, point B must move to the right, while tracing an arc).

But because point B is fixed (the immovable object is always touching B, and the immovable object doesn't move), this can't happen.

I figure it's something to do with how motion is relative (i.e. A moving closer to B is the same as B moving closer to A), but it doesn't sit well with my mechanical intuition of what a torque is.

• Why cannot $A$ move closer to the wall? – Farcher Mar 23 '17 at 16:49
• I never said A couldn't move closer to the wall. I said that B couldn't move to the right. – spacediver Mar 23 '17 at 16:51
• If $A$ moves closer to the wall as does $C$ then the nutcracker "works" and $B$ does not move. – Farcher Mar 23 '17 at 16:53
• Right, but I'm having trouble understanding how the reaction force of the wall causes a torque to be experienced by the green segment around B, since B does not move rightwards. In the first image, B moves rightwards, and all makes sense in my head. In the second image, the situation is equivalent in terms of torque experienced at A, but also different somehow. – spacediver Mar 23 '17 at 17:00
• The green bar has two extra forces acing on it to the left due to the nut and the pivot. Those three forces produce the rotation of the green bar. – Farcher Mar 23 '17 at 17:02

The thrusters at B and C both do work to crack the nut. If the initial distance between them is $2x$ and each thruster supplies force $F$ then when they meet in the middle the work done by each is $Fx$ so the total work done is $2Fx$.
When the thruster at B is replaced by a wall, the wall also exerts force $F$ in reaction to the force $F$ applied in the opposite direction by the thruster at C. The force at the wall does not move through any distance, so it does no work. However, now points A and C move towards the wall, A by $x$ and C by $2x$. In this case the only work is done by thruster C, and is $2Fx$. So the total work done is exactly the same as before.