# Force equilibrium

Part of the Statics course offered by the Division of Applied Mechanics, School of Engineering and the Engineering and Technology Portal

## Lecture

Static equilibrium defines the state in which the sum of the forces, and torque, on each particle of the system is zero. A particle in mechanical equilibrium is undergoing neither linear nor rotational acceleration; however it could be translating or rotating at a constant velocity.

### Irrotational Equilibrium

Newton's Second Law for a static system (Equilibrium):

${\displaystyle \sum {\vec {F}}\ =M*{\vec {a}}=0}$


The sum of the forces is equal to the mass times the acceleration. The 2nd Law tells us that if the object or system is motionless, the acceleration is equal to zero. Therefore the sum of the vector forces must be equal to zero.

Example:

Consider a table with four legs and a 200 kg object resting motionless in the center of the table. What force acts upon the bottom of each table leg ${\displaystyle {\vec {F}}_{L}}$?

The force upward on the legs is found by balancing these forces in the equation given by Newton's Law. Gravity is acting downward on the 200 kg object and the 100 kg table. Therefore, we may substitute the acceleration due to gravity on Earth ${\displaystyle {\vec {g}}}$ which is ${\displaystyle 9.81{\frac {m}{s^{2}}}}$, for ${\displaystyle {\vec {a}}}$.

Substitute in known values...

${\displaystyle \sum {\vec {F}}_{y}=M*{\vec {a}}=0=-(200kg*9.81{\frac {m}{s^{2}}})-(100kg*9.81{\frac {m}{s^{2}}})+(F_{L}*4)}$

And simplifying...

${\displaystyle {\vec {F}}_{L}={\frac {200kg*9.81{\frac {m}{s^{2}}}+100kg*9.81{\frac {m}{s^{2}}}}{4}}}$

The unit ${\displaystyle {\frac {kg.m}{s^{2}}}}$ is equivalent to the unit of force called a Newton N. Thus when multiplying through, the kilograms cancel out, and we may solve for ${\displaystyle {\vec {F}}_{L}}$...

${\displaystyle {\vec {F}}_{L}={\frac {1962N+981N}{4}}=735.75N}$

This means that each leg exerts a downward force of 735.75 Newtons on the floor, and the floor simultaneously exerts the same force upward on each table leg.

Notice that it is critical that a consistent sign convention be followed throughout the entire analysis effort. The sign convention is typically chosen in more complex problems to ease the total amount of algebra necessary to analyze the equations in the ${\displaystyle x}$, ${\displaystyle y}$ and ${\displaystyle z}$ axis. We chose positive values to mean upward forces and negative values to mean downward forces, but we could have also used the opposite, provided we were consistent.

### Rotational Equilibrium

Lesson: In Statics the Sum of the Torques (Moments) is Equal to Zero.

The sum of all rotational forces, or torques, denoted by the capital Greek letter tau (${\displaystyle \mathrm {T} }$), is also zero. Commonly used units for torque are Foot•pounds (ft•lb) and Newton•meters (N•m).

Newton's Second Law (applied to torques):

${\displaystyle \sum \tau =\sum ({\vec {\omega }}*I)=({\vec {\omega }}_{1}*I_{1})+({\vec {\omega }}_{2}*I_{2})+({\vec {\omega }}_{3}*I_{3})+...+({\vec {\omega }}_{n}*I_{n})}$


The sum of the torques is equal to the rotational mass or moment of inertia (I) times the angular acceleration, denoted by the lower case Greek letter omega (${\displaystyle \omega }$). The 2nd Law tells us that if the object or system is motionless, the angular acceleration is equal to zero. Therefore the sum of the vector torques must be equal to zero.

Torques may also be calculated as forces times distances:

${\displaystyle \sum {\vec {\tau }}=\sum ({\vec {F}}*d)={\vec {F}}_{1}*d_{1}+{\vec {F}}_{2}*d_{2}+{\vec {F}}_{3}*d_{3}+...+{\vec {F}}_{n}*d_{n}}$


Example:

Consider a massless lever with two weights attached and a single massless support:

Weight of Object 1, ${\displaystyle {\vec {F}}_{1}}$ = 10 lb
Distance of Object 1 from fulcrum = 10 ft
Distance of support from fulcrum = 7 ft
Weight of Object 2, ${\displaystyle {\vec {F}}_{2}}$ = 80 lb
Distance of Object 2 from fulcrum = 4 ft

What forces act upon the massless lever?

First let's write the static torque equation for the system. Forces 1 and 2 are actually the weights of the two objects...

${\displaystyle \sum \tau =\sum ({\vec {F}}*d)={\vec {F}}_{1}*d_{1}+{\vec {F}}_{2}*d_{2}+{\vec {F}}_{S}*d_{S}=0}$

Substitute in known values...

${\displaystyle \sum \tau =\sum ({\vec {F}}*d)=(-10lb)*(10ft)+(-80lb)*(4ft)+{\vec {F}}_{S}*(7ft)=0}$

Simplify...

${\displaystyle \sum \tau =\sum ({\vec {F}}*d)=(-100ft.lb.)+{\vec {F}}_{S}*(7ft)+(-320ft.lb.)=0}$

And solve for ${\displaystyle {\vec {F}}_{S}}$ ...

${\displaystyle {\vec {F}}_{S}*(7ft)\ =(420ft.lb.)}$ and therefore ${\displaystyle {\vec {F}}_{S}=(420ft.lb.)/(7ft)\ =60lb.}$

Thus, a force of 60 pounds acts upward on the lever from the support in order to balance the torques on the lever. Since the total weight on the lever is 90 pounds and 60 of those pounds are countered by the support, the remaining 30 pounds must act upward on the lever at the fulcrum in the lower, left corner.

Note that we didn't need to use the fulcrum as the point from which all distances are measured, we could have chosen any point along the lever. However, as with forces, a consistent sign convention must be used. In this case, positive may be used for clockwise (CW) torques measured from one point of view, with negative torques used for counterclockwise (CCW) torques measured from the same POV. The opposite system could also be used, so long as we were consistent.

Also note that the table example used previously stated that the object was in the center of the table. If not, then balancing the torques (in two directions) would result in more force supported by some table legs and less by others. If the object was directly over one of the legs, for example, then it's entire weight would be supported by that leg, in addition to one-fourth of the table's weight. The other legs would then only support one-fourth of the table's weight each.

Activities: