Dynamics: Forces in mechanism problem

[Andy Salter]

Homework Statement

Viewing a solution on the internet to this question, the free body diagram is drawn as:

The forces here are then used to solve for $F_D$ and this is given as the answer to "the force which the bar exerts on the link at D". I'm just confused as to why there is no vertical component of the force at D, ie $F_Dy$. Then the total force at D is $$F_D = F_{Dx}\hat {\mathbf i} + F_{Dy}\hat {\mathbf j}$$

Cheers

 

[Dr.D]

So, did you have a question? Did you try to work the problem?

 

[PKM]

Think of the situation carefully... The whole mass of the bar is sustained by the rod $AB$ when the rod passes the horizontal position, and the torque at $B$ pushes it up at this instant; so there is no effect of the weight $mg$ upon the rod $CD$ in this position, right? Where from then, the end $D$ would experience $F_{D_y}$?

 

[Andy Salter]

So, did you have a question? Did you try to work the problem?

Yes:
$$\textbf{I'm just confused as to why there is no vertical component of the force at D}$$

The whole mass of the bar is sustained by the rod A B AB when the rod passes the horizontal position

Why's that?

I would have thought it would be equally distributed? What about tangential acceleration? Why would this not be felt at both B and D?

Cheers

 

[Dr.D]

What about tangential acceleration?

Be careful not to confuse force and acceleration; they are two different concepts.

In this problem, there are three moving bodies. You should draw a free body diagram (FBD) for each one. Actually drawing the FBDs cannot be over emphasized; it is critical!

 

[Andy Salter]

You should draw a free body diagram (FBD) for each one. Actually drawing the FBDs cannot be over emphasized; it is critical!

My FBD looked something like this:

 

[PKM]

I would have thought it would be equally distributed?

No, it won't be equally distributed. At this instant, the torque $M$ on the shaft $AB$ solely does the job of lifting the bar. The upper shaft is made to move with the bar by the torque $M$, not that the shaft $CD$ moves the bar. That's why the whole weight of the bar & the shaft $CD$ (considered negligible here) is sustained by $M$.

 

[Dr.D]

Acceleration terms (m*a terms) do not belong in an FBD. You will get the wrong results every time.

 

[haruspex]

To think of it another way, what would happen if you took away CD?

 

[Andy Salter]

No, it won't be equally distributed. At this instant, the torque $M$ on the shaft $AB$ solely does the job of lifting the bar. The upper shaft is made to move with the bar by the torque $M$, not that the shaft $CD$ moves the bar. That's why the whole weight of the bar & the shaft $CD$ (considered negligible here) is sustained by $M$.

Ahh right, so only only if you had torques exerted on both links would the mass be sustained by both?

Acceleration terms (m*a terms) do not belong in an FBD. You will get the wrong results every time.

Yeah I usually would have done those as dotted lines to show they are resultant forces, i drew that very quickly. Your pedantism has been very helpful /s

 

[haruspex]

Ahh right, so only only if you had torques exerted on both links would the mass be sustained by both?

Yes.

 

[PKM]

Ahh right, so only only if you had torques exerted on both links would the mass be sustained by both?

Of course. If there is a force acting on any shaft, a corresponding torque must be present there. You have noted that there is no torque at $D$ at this instant (but there is a horizontal force on $CD$ due to the normal acceleration of the bar). Why does it not create a torque?