# What is Statics: Definition and 891 Discussions

Statics is the branch of mechanics that is concerned with the analysis of (force and torque, or "moment") acting on physical systems that do not experience an acceleration (a=0), but rather, are in static equilibrium with their environment. The application of Newton's second law to a system gives:

F

=
m

a

.

{\displaystyle {\textbf {F}}=m{\textbf {a}}\,.}
Where bold font indicates a vector that has magnitude and direction.

F

{\displaystyle {\textbf {F}}}
is the total of the forces acting on the system,

m

{\displaystyle m}
is the mass of the system and

a

{\displaystyle {\textbf {a}}}
is the acceleration of the system. The summation of forces will give the direction and the magnitude of the acceleration and will be inversely proportional to the mass. The assumption of static equilibrium of

a

{\displaystyle {\textbf {a}}}

F

=
0

.

{\displaystyle {\textbf {F}}=0\,.}
The summation of forces, one of which might be unknown, allows that unknown to be found. So when in static equilibrium, the acceleration of the system is zero and the system is either at rest, or its center of mass moves at constant velocity. Likewise the application of the assumption of zero acceleration to the summation of moments acting on the system leads to:

M

=
I
α
=
0

.

{\displaystyle {\textbf {M}}=I\alpha =0\,.}
Here,

M

{\displaystyle {\textbf {M}}}
is the summation of all moments acting on the system,

I

{\displaystyle I}
is the moment of inertia of the mass and

α

{\displaystyle \alpha }
= 0 the angular acceleration of the system, which when assumed to be zero leads to:

M

=
0

.

{\displaystyle {\textbf {M}}=0\,.}
The summation of moments, one of which might be unknown, allows that unknown to be found.
These two equations together, can be applied to solve for as many as two loads (forces and moments) acting on the system.
From Newton's first law, this implies that the net force and net torque on every part of the system is zero. The net forces equaling zero is known as the first condition for equilibrium, and the net torque equaling zero is known as the second condition for equilibrium. See statically indeterminate.
A physicist who does research in statics is called a statician.

View More On Wikipedia.org
1. ### Engineering Frame Statics Problem

So for this problem I have already solved for the Y forces: Dy = 171.43 N, Cy = 228.57 N, and By = -428.57 N. For the X forces I split up the frame and took the moment of DE. Me = 1.5(300) + 3.5(300) - 5(Dx), Dx = 300 For CD Dx = -Cx fo I got Cx = 300, as when you forces are two member Dx =...
2. ### B Basic Angle Explanation for Statics

My textbook introduces this angle concept really early on and I still don't understand it. It just shows that a normal to a line and some other random angle shown is the same. I don't see any transversal angles or anything. Where did they get the secondary line to form theta for the normal line?
3. ### B The law of total probability with extra conditioning

Hello, I am studying probability and came across this theorem, it's the law of total probability with extra conditioning, I tried to work out a proof but couldn't ,does anyone know the proof for this : thanks!
4. ### Solving Fluid Statics Problem: Accounting for Atmospheric Pressure

Hi! For this fluid statics problem, One of the answers is: However, why did they assume the pressure at the top was zero? I thought the pressure at the top would be 1 atm? So tried to take atmospheric pressure into account putting 1 atm at the top, Do you please know how to get P_0A/2 so...

49. ### Statics and Structural Analysis Question

I am able to draw shear and bending moment diagrams. We were never taught how to even start with a problem like this.
50. ### A cart with two cylindrical wheels connected by a rod

Firstly I only consider one of the wheels. This wheel consists of a big wheel (black) with mass M and radius R and inside it a circular region with a negative mass (-m) and radius R/2. (I assume they have same mass density but with opposite signs. I do this because I don't know where the center...