Friction and Wedges

1. Nov 30, 2015

OmniNewton

1. The problem statement, all variables and given/known data
Provided the following diagram determine the reversed force of P needed to pull out the wedge, A.

2. Relevant 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)

3. 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:

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

Last edited: Nov 30, 2015
2. Nov 30, 2015

SammyS

Staff Emeritus
(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]

3. Nov 30, 2015

OmniNewton

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

4. Nov 30, 2015

SammyS

Staff Emeritus
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 " + " ?

5. Nov 30, 2015

OmniNewton

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)

6. Nov 30, 2015

SammyS

Staff Emeritus
That doesn't address the sign on 0.1NBcos15° .

7. Nov 30, 2015

OmniNewton

Sorry you are absolutely correct let me work through this with this change, thank you!

8. Nov 30, 2015

OmniNewton

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.

9. Nov 30, 2015

SammyS

Staff Emeritus
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,

10. Nov 30, 2015

OmniNewton

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.