# Tag Info

2

You can think about this in two different ways. One way is to look at the initial and final angular momentum. If you go from $L\cdot(0,1, 0)$ to $L\cdot(1,0,0)$ you need to remove the $Y$ component and add the $X$ component. If you just calculate the difference in the angular momentum, then you get $$\Delta L = L\cdot (1,-1,1)$$ which would immediately ...

1

In short, you are to think of the direction of the torque as pointing along the axis of the rotation it would induce in a rigid body initially at rest. But if the conception of torque as a vector out of the page seems artificial, that's because it is. Torque is not fundamentally a quantity that is a vector but a directed plane or directed area. Such an ...

1

There are probably lots of duplicates, so my apologies, but for clarity I will try a short answer, as the graphic from Wikipedia is particularly illustrative. The torque is perpendicular, ( orthogonal) to the other two vectors, so it could be the line where the hinges are located, depending on the direction of the other two forces. From Wikipedia ...

1

I believe the issue is that the sketch doesn't show all forces acting on the bar. (My guess is that this is a question concerning only rotation about the y-axis) There must be forces at point A as well. The bar seems fixed at this point so that a torque will appear here to balance the others in order to keep no rotation. The torque 159 Nm comes from force \$...

Only top voted, non community-wiki answers of a minimum length are eligible