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

In summary, the equaiton of 650*0.4+450*0.45+400=N_A*0.6 yields a value of N_A=104.2N. The magnitude of the reaction at B is 450+650=1040N. The problem statement asks for a value of P which when solved yields 311N.
  • #1
YehiaMedhat
20
3
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.
 

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  • #2
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 ?

##\ ##
 
  • #3
The picture for the problem its self
Screenshot_20230112-141024_Xodo Docs.jpg

And I did refrence the forces by the arrows and the name of the points like##N_A##
 
  • Informative
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  • #4
Okay, that explains the given ##P##.
Seen this one before, will need a moment ( :wink: ) to refresh ... my coffee
 
  • #5
YehiaMedhat said:
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 ?

##\ ##
 
  • #6
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.
 
  • #7
YehiaMedhat said:
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.
 
  • #8
haruspex said:
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##
 
  • #9
YehiaMedhat said:
Ok, that makes sense now, so it'll be##\sum{}^{} M_B=0 \rightarrow 0.45P=400-650*0.4 \rightarrow P=311N##
Yes.
 

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