- #1
OmniNewton
- 105
- 5
Homework Statement
Provided the following diagram determine the reversed force of P needed to pull out the wedge, A.
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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