connection from one set of axes to the other. Someone may say, " That looks just like what we did with vectors. And indeed, that is exactly what we are intending to do, Then he may say, " Well, isnt torque just a vector? It does turn out to be a vector, but we do not know that right away without making an analysis. So in the following steps we shall make the analysis, We shall not discuss every step in detail, since we only want to illustrate how it works. The torques calculated by Joe ry=xFy- yF (20.1) We digress at this point to note that in such cases as this one may get the wrong sign for some quantity if the coordinates are not handled in the right way. Why not write Tuz= zFy-yFz? The problem arises from the fact that a coordinate system may be either "right-handed"or"left-handed. Having chosen(arbitrarily)a sign for, say Try, then the correct expressions for the other two quantities may always be found by interchanging the letters xyz in either order 'y'sx'Fy'-y'l Ty'r=yF2-2'F z'Fr'-x'F2 Now we suppose that one coordinate system is rotated by a fixed angle 0, such that the z-and z'-axes are the same. (This angle 0 has nothing to do with rotating objects or what is going on inside the coordinate system. It is merely the relation ship between the axes used by one man and the axes used by the other, and is supposedly constant. ) Thus the coordinates of the two systems are related b y'= cos 0-x sin 8 (20.3) Likewise, because force is a vector it transforms into the new system in the same yay as do x, y, and z, since a thing is a vector if and only if the various components transform in the same wa Fr cos 0+ F sin Fy'= Fy cos 0- Fx sin e x', y, and zthe expressions(20.3), and for Fr, Fy, Fx those given by(20.4), all into(20. 2). So, we have a rather long string of terms for Try and (rather sur ingly at first) it turns out that it comes right down to xFy-yFr, which we re nize to be the y 0)(Fy Fr sin 8) e)(Fr cos 0+ Fy sin 0) xF,(cos+sin2)-yF-(sing cos26 +xFxCsin 0 cos 0 sin g cos 0) +yF,(sin 0 cos e