- #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

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

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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