Finding Value of ##N_A## and P in a Frame

[YehiaMedhat]

Homework Statement

The rigid body ABC is in equilibrium in the position shown. If $W=650N, P=450N, M=400N.m, and x=0.35m$, then: (Neglect the the thickness of the frame):
The samllest force P for equilibrium in the position shown is (N):

Relevant Equations

$$\sum_{}^{} F_y =0$$
$$\sum_{}^{} F_x =0$$
$$\sum_{}^{} Moments =0$$

The first thing I did is to get the value of $N_A$ by the equaiton of $\sum{}^{} M_B=0 \rightarrow 650*0.4+450*0.45+400=N_A*0.6 \rightarrow N_A = 104.2N$ This is the first.
The magnitude of the reaction at B: $\sum{}^{} F_y=0 \rightarrow B_y=450+650,\sum{}^{} F_x=0 \rightarrow B_x=N_A=104.2N$
This was until I came to the question whose required is to get the minimum value of P to keep the body in equilibrium, my question is: will I use the value of N_A nevertheless that I just got from the previous P value?? isn't this wierd? so, what should I do?
Below is the FBD of the frame in the problem.

 

[BvU]

Hi,

Can you explain the diagram ? Or do I have to sleuth to find out where A, B and C are located ?

And can you explain why it says $P = 450$ N when the problem statement asks for a value of $p$ ?

Finally, what is the reference point for this $M = 400$ N.m ?

$\$

 

[YehiaMedhat]

The picture for the problem its self

And I did refrence the forces by the arrows and the name of the points like$N_A$

 

[BvU]

Okay, that explains the given $P$.
Seen this one before, will need a moment ( ) to refresh ... my coffee

 

[BvU]

so, what should I do?

At the minimum $P$ the contraption will almost start to rotate anti-clockwise around B. So what $N_A$ is associated with that minimum ?

$\$

 

[YehiaMedhat]

Zero, isn't it?
Maybe I'm asking because the model answer for this problem used the $N_A$ we got earlier in the calculations, so that made me feel that there's sth wrong.

 

[haruspex]

Zero, isn't it?
Maybe I'm asking because the model answer for this problem used the $N_A$ we got earlier in the calculations, so that made me feel that there's sth wrong.

The part of the diagram where $N_A$ makes contact has been drawn on, so it is hard to see the details. But yes, I would have interpreted it as just a buffer, so as P is reduced it should go to zero.

 

[YehiaMedhat]

The part of the diagram where $N_A$ makes contact has been drawn on, so it is hard to see the details. But yes, I would have interpreted it as just a buffer, so as P is reduced it should go to zero.

Ok, that makes sense now, so it'll be$\sum{}^{} M_B=0 \rightarrow 0.45P=400-650*0.4 \rightarrow P=311N$

 

[haruspex]

Ok, that makes sense now, so it'll be$\sum{}^{} M_B=0 \rightarrow 0.45P=400-650*0.4 \rightarrow P=311N$

Yes.