3D Equilibrium and Constraints

Introduction

A rigid body must be properly constrained to ensure static equilibrium. There are two different kinds of constraints: redundant constraints and improper constraints. Redundant constraints are when there are too many constraints, making the system tough to solve and a statically indeterminate problem. Improper constraints are when there aren’t enough constraints, causing the system to move. A stable system requires that the reaction forces are not parallel to another or share a common axis.

3D Supports

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Notation

F → = ( x i ̂ + y j ̂ + z k ̂ ) \overrightarrow{F} = (x\widehat{i} + y\widehat{j} + z\widehat{k})

Procedure

  1. Draw a FBD

  2. Solve Equilibrium Equations

Cross Products to Remember

i ̂ × i ̂ = 1 \widehat{i} \times \widehat{i} = 1\

j ̂ × i ̂ = − k ̂ \widehat{j} \times \widehat{i} = - \widehat{k}

k ̂ × i ̂ = j ̂ \widehat{k} \times \widehat{i} = \widehat{j}

i ̂ × j ̂ = k ̂ \widehat{i} \times \widehat{j\ } = \widehat{k}

j ̂ × j ̂ = 1 \widehat{j} \times \widehat{j} = 1

k ̂ × j ̂ = − i ̂ \widehat{k} \times \widehat{j} = - \widehat{i}

i ̂ × k ̂ = − j ̂ \widehat{i} \times \widehat{k} = - \widehat{j}

j ̂ × k ̂ = i ̂ \widehat{j} \times \widehat{k} = \widehat{i}

k ̂ × k ̂ = 1 \ \widehat{k} \times \widehat{k} = 1

Example Problem: The boom is used to support the 75-lb flowerpot. Determine the tension developed in wires AB and AC.

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Equilibrium Equations

F → = F ( r → r ) \overrightarrow{F} = F(\frac{\overrightarrow{r}}{r})

T A B = 2 7 T A B i ̂ − 6 7 T A B j ̂ + 3 7 T A B k ̂ T_{AB} = \frac{2}{7}T_{AB}\widehat{i} - \frac{6}{7}T_{AB}\widehat{j} + \frac{3}{7}T_{AB}\widehat{k}

T A C = − 3 7 T A C i ̂ − 6 7 T A C j ̂ + 3 7 T A C k ̂ T_{AC} = - \frac{3}{7}T_{AC}\widehat{i} - \frac{6}{7}T_{AC}\widehat{j} + \frac{3}{7}T_{AC}\widehat{k}

↺ Σ M O = 0 = r → × ( T A B + T A C + W ) ; 0 = ( 18 7 T A B + 18 7 T A C − 450 ) i ̂ + ( − 12 7 T A B + 12 7 T A C ) k ̂ \circlearrowleft\Sigma M_{O}^{} = 0 = \overrightarrow{r} \times (T_{AB} + T_{AC} + W);0 = (\frac{18}{7}T_{AB} + \frac{18}{7}T_{AC} - 450)\widehat{i} + ( - \frac{12}{7}T_{AB} + \frac{12}{7}T_{AC})\widehat{k}

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  1. ↺ Σ M x = 0 = 18 7 T A B + 18 7 T A C − 450 = 0 \circlearrowleft\Sigma M_{x} = 0 = \frac{18}{7}T_{AB} + \frac{18}{7}T_{AC} - 450 = 0

  2. ↺ Σ M z = 0 = − 12 7 T A B + 12 7 T A C \circlearrowleft\Sigma M_{z} = 0 = - \frac{12}{7}T_{AB} + \frac{12}{7}T_{AC}

Solving equation 1 and 2, T A B = T A C = 87.5 l b T_{AB} = T_{AC} = 87.5lb