Finding the force to pull out a wedge with friction

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OmniNewton
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Homework Statement


Provided the following diagram determine the reversed force of P needed to pull out the wedge, A.
6a9b6b6b3b4780d93b80dd6917a408fc.png


Homework Equations


Given the coefficient of static friction between A and C and between B and D is 0.2 and between A and B static friction is 0.1. Weights of wedges are neglected.
Force of friction = static coefficient * normal (When impending motion is occurring)

The Attempt at a Solution


Let NC be the normal at C
ND be the normal at D
NB be the normal between wedge A and wedge B
FC be the friction force between wedge A and wall C
FB be the friction force between wedge A and wedge B
FD be the friction force between wedge B and wall D

For FBD of wedge A:

X-direction:
0 = 0.2 NC - NBsin15 - 0.1NBcos15 - P
simplifying 0= 0.2 NC - 0.3554NB - P (Equation 1)
Y-direction:
Nc - NBcos15 + 0.1NBcos15=0
simplifying Nc=0.94 NB (Equation 2)

For FBD of wedge B:[/B]

X-direction:
NB= ND/(sin15) (Equation 3)

Y-direction:

NBcos15 - 3000 + 0.2ND = 0 (Equation 4)

substituting equation 3 into 4 and solving ND = 763lb

From equation 3 then NB = 2947.9 lb

From Equation 2 NC = 2771lb

From Equation 1 P = 493.5 lb

My answer is considerably off the correct answer which is 106 lb
 
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OmniNewton said:

Homework Statement


Provided the following diagram determine the reversed force of P needed to pull out the wedge, A.
6a9b6b6b3b4780d93b80dd6917a408fc.png


Homework Equations


Given the coefficient of static friction between A and C and between B and D is 0.2 and between A and B static friction is 0.1. Weights of wedges are neglected.
Force of friction = static coefficient * normal (When impending motion is occurring)

The Attempt at a Solution


Let NC be the normal at C
ND be the normal at D
NB be the normal between wedge A and wedge B
FC be the friction force between wedge A and wall C
FB be the friction force between wedge A and wedge B
FD be the friction force between wedge B and wall D

For FBD of wedge A:

X-direction:
0 = 0.2 NC - NBsin15 - 0.1NBcos15
simplifying 0.2 NC - 0.3554NB - P (Equation 1)
Y-direction:
Nc - NBcos15 + 0.1NBcos15=0
simplifying Nc=0.94 NB (Equation 2)

For FBD of wedge B:[/B]

X-direction:
NB= ND/(sin15) (Equation 3)

Y-direction:

NBcos15 - 3000 + 0.2ND = 0 (Equation 4)

substituting equation 3 into 4 and solving ND = 763lb[/B]

From equation 3 then NB = 2947.9 lb

From Equation 2 NC = 2771lb

From Equation 1 P = 493.5 lb

My answer is considerably off the correct answer which is 106 lb
(Try to not use Bold type face except where necessary.)

Isn't P supposed to be pulling the wedge out? That would mean that friction forces acting on wedge A are exerted toward the right.[/B]
 
SammyS said:
(Try to not use Bold type face except where necessary.)

Isn't P supposed to be pulling the wedge out? That would mean that friction forces acting on wedge A are exerted toward the right.

This is why the question says the reverse force of P so -P. I accommodated for this in my equations. Sorry about the bold my bad
 
OmniNewton said:
This is why the question says the reverse force of P so -P. I accommodated for this in my equations. Sorry about the bold my bad
OmniNewton said:
X-direction:
0 = 0.2 NC - NBsin15 - 0.1NBcos15
simplifying 0.2 NC - 0.3554NB - P (Equation 1)

Equation 1 is not an equation. There's no " = " in it.

In the line before that there is no P, and shouldn't 0.1NBcos15° be to the right, so have " + " ?
 
SammyS said:
Equation 1 is not an equation. There's no " = " in it.

In the line before that there is no P, and shouldn't 0.1NBcos15° be to the right, so have " + " ?
Sorry I forgot to add that variable when typing the corrections have been made,
0 = 0.2 NC - NBsin15 - 0.1NBcos15 - P
simplifying 0= 0.2 NC - 0.3554NB - P (Equation 1)
 
OmniNewton said:
Sorry I forgot to add that variable when typing the corrections have been made,
0 = 0.2 NC - NBsin15 - 0.1NBcos15 - P
simplifying 0= 0.2 NC - 0.3554NB - P (Equation 1)
That doesn't address the sign on 0.1NBcos15° .
 
SammyS said:
That doesn't address the sign on 0.1NBcos15° .
Sorry you are absolutely correct let me work through this with this change, thank you!
 
Excellent, Thank you so much for your time it worked out! Sorry for wasting your time on such a trivial mistake. I was looking at this for hours and didn't notice it.
 
OmniNewton said:
Excellent, Thank you so much for your time it worked out! Sorry for wasting your time on such a trivial mistake. I was looking at this for hours and didn't notice it.
No problem.

Lately I've been helping some people who understand almost nothing and at times seem to act clueless on purpose. It's great to be able to make a few suggestions and have them recognized for what they are,
 
SammyS said:
No problem.

Lately I've been helping some people who understand almost nothing and at times seem to act clueless on purpose. It's great to be able to make a few suggestions and have them recognized for what they are,

Yeah I really appreciate that I think it is important to work from ones mistakes and learn from them. I appreciate you not just giving me the answer but rather guiding me.
 
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