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.
 

Attachments

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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
Likes BvU
  • #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.
 

1. What is the significance of finding the value of ##N_A## and P in a frame?

Finding the value of ##N_A## and P in a frame is important because it allows us to determine the number of atoms present in a unit cell and the probability of finding a particular atom in that unit cell. This information is crucial in understanding the structure and properties of a material.

2. How is the value of ##N_A## and P calculated in a frame?

The value of ##N_A## can be calculated by dividing the total number of atoms in the unit cell by the number of atoms in one unit cell. The value of P can be calculated by dividing the number of atoms of a specific type by the total number of atoms in the unit cell.

3. What factors can affect the value of ##N_A## and P in a frame?

The value of ##N_A## and P can be affected by the crystal structure of the material, the size of the unit cell, and the types of atoms present in the unit cell. Additionally, defects or impurities in the crystal structure can also impact these values.

4. How can the value of ##N_A## and P be experimentally determined?

The value of ##N_A## and P can be experimentally determined using techniques such as X-ray diffraction, which can provide information about the crystal structure and the number of atoms in a unit cell. Other methods such as electron microscopy and spectroscopy can also be used to determine these values.

5. What is the relationship between the value of ##N_A## and P in a frame and the overall properties of a material?

The value of ##N_A## and P in a frame can greatly influence the properties of a material. For example, a higher P value indicates a higher probability of finding a particular atom in a unit cell, which can affect the material's electrical, thermal, and mechanical properties. Additionally, the value of ##N_A## can also impact the density and stability of the material.

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