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A truss, its free body diagrams and also of the pin at C is shown in the figure. Ay and P, in figure II may be equated because they are acting in opposite direction on the same body(the whole truss), which is in equilibrium. Similar is the case with Ax and By. What is the explanation as to why the forces in figure II can be considered to be acting on the whole truss instead of being on the pins as in figure III ? (In the latter case, horizontal equilibrium equation involving Ax and By cannot be used as they are acting on different bodies.)

Also note that if we consider the free body diagram of pin A, then Ay and Ax would be present. How is it possible if those forces are acting on the whole body instead of being on the pin?

And if it is because they are acting on the whole body, then why will force P (which is also considered to be acting on the whole body in figure II) be not shown in the free body diagram of A?

Reference Figure

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    $\begingroup$ Regarding your last question, where is "diagram A"? Do you mean pin A? $\endgroup$ – Bob D Jul 29 at 13:45
  • $\begingroup$ Free body diagram of pin A is mentioned in the question but not shown. $\endgroup$ – Zam Jul 29 at 14:58
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What is the explanation as to why the forces in figure II can be considered to be acting on the whole truss instead of being on the pins as in figure III ? (In the latter case, horizontal equilibrium equation involving Ax and By cannot be used as they are acting on different bodies.)

The forces can be considered as acting on the whole truss because Fig II is a free body diagram of the entire truss. Therefore all external forces acting on the truss need to be shown, and these are the load P and the reactions at the pin A and roller B.

Also note that if we consider the free body diagram of pin A, then Ay and Ax would be present. How is it possible if those forces are acting on the whole body instead of being on the pin ?

Not quite sure what you mean, but if you are strictly analyzing the equilibrium of pin A, showing just $A_y$ and $A_x$ is insufficient. It is not a free body diagram of pin A. You also have to eliminate member AC and replace it with the force it applies to pin A. Without it you would obviously not have equilibrium.

And if it is because they are acting on the whole body, then why will force P (which is also considered to be acting on the whole body in figure II) be not shown in the free body diagram of A ?

Because load P is not applied to pin A. The influence of load P at pin A is accounted for if, as I said above, you eliminate member AC and replace it with the force it exerts on pin A.

The above being said, you would obviously need more info on the diagram (lengths of members, angles, etc.) in order to determine the reactions. Then, depending on exactly what is being asked, it may be easiest to use Fig II and evaluate it by setting the sum of the vertical forces equal to zero, sum of horizontal forces equal to zero, and the sum of the moments equal to zero. Regarding the last, pick a point that eliminates as many variables (unknown forces) as possible. That would be the moments about a point where unknown forces pass through the point and therefore contribute no moment.

Would the answer to the question, 'On what is Ay acting on ?', be pin A or the whole truss ?

It depends on what it is that you are looking at.

From the perspective of a FBD on pin A, $A_y$ is the vertical force that member AC exerts on pin A.

From the perspective of a FBD on the entire truss, $A_y$ is the vertical force that pin A exerts on member AC.

Per Newton's third law, the vertical force that member AC exerts on pin A is equal and opposite to the vertical force that pin A exerts on member AC.

That clears things quite a bit. But then, from the perspective of the FBD of the entire truss, Ax and Bx are forces acting on different members (different bodies) (Fig. II). So how can horizontal equilibrium equation be applied ? –

For static equilibrium both members of the truss (AC and BC) are considered to be one single rigid body. You don't analyze the influence of the various external forces on each member individually, but as a collective rigid body.

It is obvious from the external forces given that the algebraic sum of $A_x$ and $B_x$ must be zero in order that the entire truss, acting as a single rigid body, does not move in the x-direction since they are the only two external forces in the x-direction acting on the truss as a whole. (Clearly the direction of one of them has to be opposite to the direction shown).

It is equally obvious that $A_y$ in the direction as drawn must be equal in magnitude to the load $P$ in order for that the entire truss does not move in the y-direction, since these are the only two external forces acting on the truss as a whole in the y-direction. Since $P$ is presumably known, this means you can immediately determine $A_y$.

The quickest way to solve for the horizontal reactions is to take the sum of the moments about pin A. In doing so you eliminate $A_x$ and $A_y$ as variables since they act through the point A and contribute no moment about pin A. The only moments about A are due to $P$ and $B_x$. Since $P$ clearly imposes a clockwise moment about A, $B_x$ must actually be in the opposite direction to provide an equal counterclockwise moment about A so that the sum of the moments about A are zero. Given sufficient dimensions/angles you can directly calculate $B_x$ and then from there get $A_x$.

Hope this helps.

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  • $\begingroup$ Would the answer to the question, 'On what is Ay acting on ?', be pin A or the whole truss ? (It is this difference that I am referring to in my second question ) $\endgroup$ – Zam Jul 29 at 14:57
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    $\begingroup$ @Zam I updated my answer to address your follow up question. Hope it helps. $\endgroup$ – Bob D Jul 29 at 15:14
  • $\begingroup$ That clears things quite a bit. But then, from the perspective of the FBD of the entire truss, Ax and Bx are forces acting on different members (different bodies) (Fig. II). So how can horizontal equilibrium equation be applied ? $\endgroup$ – Zam Jul 29 at 16:27
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    $\begingroup$ @Zam Once again I have updated to answer to address this additional follow up question. I am assuming you know how to calculate the sum of the moments about any point on the truss. Without that, you can not determine the reactions. $\endgroup$ – Bob D Jul 29 at 17:36
  • $\begingroup$ Thanks. Actually, I wasn't in need of further steps to solve the problem. $\endgroup$ – Zam Jul 29 at 17:43

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