Recent content by anuttarasammyak

In survey of "the only horizontal bar" model, I would like to know how we should choose the shape of the pivot in this figure to control orange colored x-force pair magnitude ? I should appreciate it if you could warn me the setting failure.

My bad! So we have same F.

My answer #16 which is same as the exam answer has sec, however post #13 answer has tan as coefficient to $$W\frac{L_2}{L_1}$$.

The blue length is important. The angle itself does not matter. Alternative orange rods are OK. The diagonal rod tansmit not only force along it but also shearing force to the horizontal rod connected to it. In the original model it is an inner force of the rigid body which should be cancelled...

No, not the same in a exact sense. My post #16 based on your figure deals it. Your case can be interpreted that we pull the horizontal bar tilt upward by the rope tagged in a mid point with greater force than gravity force W. Are the both two models (with or without diagonal bar) to explain...

|D|,|F| > |W|. The weight gravity generates larger y force. F is downward, however red force on screw is upward.”Only horizontal bar” model has something different from the original model. If D is parallel transport of the force the diagonal bar get from the wall, such a transport breaks the...

I mean couple not moment. Thanks to #6 in usual case Blue forces couple and red forces couple cancel. My #16 confirms it. In the calculation of forces for horizontal bar system shown, I do not find such couple cancellation. This net zero couple seems the point of the exam in OP.

Is the net couple zero in this system? The couple produced by the two forces in the x-direction is zero. However, it seems that the three forces in the y-direction do not cancel to give a zero couple.

Thanks. My thread examples were not appropriate. Net zero inhomogeneous x force act at the contact area of the remaining bar so that it produces counter torque, which should be treated by advanced mechanics.

$$F_y=w-D_y$$ it can be negative, zero or positive. Some example cases $$F_y=w > 0,\ \ D_y=0$$ $$F_y=0,\ \ D_y=w$$ $$F_y<0,\ \ D_y=w-F_y>w$$

All external forces should be considered at the points where they act on the body. D is acting on point Q in the figure you show. But D, as an external force, acts on P. Momentum balance around P $$L_2w=hF_x$$ Momentum balance around Q $$(L_2-L_1)w+L_1F_y+L_1D_y=hD_x$$ They are equivalent...

Pursuiting the calculation I showed $$\hat{L}_z=-i \hbar \frac{\partial}{\partial \phi}$$ I am not sure whether it is consistent with your picture.

We are tempted to say such and such momentum at such and such place, but QM uncertainty relation or commutation relation $$[x,p_x]=i\hbar$$ make it impossible.

Inserting $$\mathbf{p}=-i\hbar \nabla$$ in formula of L, you will prove that L has no r dependence.

We may take the problem as a study subject of statistics, sending a questionnaire to the captains: a. How old are you? b. Have you loaded 26 sheep and 10 goats? c. If yes, how old were you then? I often see this method in Medecin in order to find suspected correlation between the two events.