# Statics/Two-Dimensional Equilibrium

There are six equations expressing the equilibrium of a rigid body in 3 dimensions.

Sum of Forces: ${\displaystyle \sum _{}^{}F_{x}=0}$, ${\displaystyle \sum _{}^{}F_{y}=0}$, ${\displaystyle \sum _{}^{}F_{z}=0}$

Sum of Moments: ${\displaystyle \sum _{}^{}M_{x}=0}$, ${\displaystyle \sum _{}^{}M_{y}=0}$, ${\displaystyle \sum _{}^{}M_{z}=0}$

In two dimensions one direction of force and two directions of moments can be ignored. When forces exist only in the x and y directions, there cannot be a moment in any direction except z. The equations of concern when forces only exist in the x and y directions are shown below.

Sum of Forces: ${\displaystyle \sum _{}^{}F_{x}=0}$, ${\displaystyle \sum _{}^{}F_{y}=0}$

Sum of Moments: ${\displaystyle \sum _{}^{}M_{z}=0}$

To solve two dimensional statics problems:

• draw a free body diagram
• Write equations for force equilibrium
• Write equations for moment equilibrium
• Once you have the same number of equations as unknowns the problem can be solved, you may have to strategically pick points to write moment equations etc.

## Couple

F1 and F2 are equal

A couple exerts the same moment at every point as demonstrated in this section. After looking at the picture to the right the following equations can be written. Counter clockwise moments are considered positive and clockwise negative. F1 = F2

${\displaystyle \sum M_{A}=F_{1}*a-F_{2}*(b+a)=F_{1}*a-F_{1}*(b+a)=-F_{1}*b=-F*b}$

${\displaystyle \sum M_{B}=-F_{1}*(b+c)+F_{2}*(c)=-F*b}$

${\displaystyle \sum M_{C}=-F_{1}*d-F_{2}*e=-F*(d+e)=-F*b}$

The moment about all points is the force multiplied by the distance between the forces. If you found the moment about D, or any other point you would continue to find the same moment. All points have the same moment, even points that aren't in the x-y plane.

## Example 1

### Question

The picture to the right shows the forces acting on a parked car. If the weight of the car acts exactly halfway between the two wheels and the weight is 1000 lbs how much force is exerted on the rear wheel? What about the front wheel?

Writing the force equations

${\displaystyle \sum F_{x}=0}$

There are no forces in the x direction

${\displaystyle \sum F_{y}=0=R_{f}+R_{B}-W=R_{f}+R_{B}-1000lbs}$

Writing moment equation about front wheel

${\displaystyle \sum M_{f}=0=-5ft*W+10*R_{B}=-5000lbft+10*R_{B}}$

${\displaystyle \ R_{B}=500lb}$

subbing ${\displaystyle R_{B}}$ back into the ${\displaystyle sumF_{y}}$

${\displaystyle \sum F_{y}=0=R_{f}+R_{B}-W=R_{f}+500lbs-1000lbs}$

${\displaystyle \ R_{f}=500lbs}$

Please note that ${\displaystyle R_{f}}$ and ${\displaystyle R_{B}}$ are distributed over two wheels. Each front wheel supports half of ${\displaystyle R_{f}}$ and each back wheel supports half of ${\displaystyle R_{B}}$.

## Example 2

Question:

A uniform ring of mass ${\displaystyle m\,\!}$ and radius ${\displaystyle r\,\!}$ carries an eccentric mass ${\displaystyle m_{0}\,\!}$ at a radius ${\displaystyle b\,\!}$ and is in equilibrium position on an incline, which makes an angle ${\displaystyle \alpha \,\!}$ with the horizontal. If the contacting surfaces are rough enough to prevent slipping, write the expression for the angle ${\displaystyle \theta \,\!}$ which defines the equilibrium position.
Answer: ${\displaystyle \theta =sin^{-1}[{\frac {r}{b}}(1+{\frac {m}{m_{0}}})sin\alpha ]\,\!}$