2.1. Force, Moment of a Force, Distributed Actions#

2.1.1. Concentrated Force#

A (concentrated) force is a vector quantity with physical dimensions,

\[[\text{force}] = \frac{\text{[mass]}\text{[length]}}{\text{[time]}^2}\]

which can be measured using a dynamometer, and whose effect can alter the equilibrium or motion conditions of a physical system.

In addition to the typical information of a vector quantity - magnitude, direction, and sense - contained in the force vector \(\vec{F}\), it is often necessary to know the point of application or the line of application of the force.

2.1.2. Moment of a Concentrated Force#

The moment of a force \(\vec{F}\) applied at point \(P\), or with a line of application passing through \(P\), relative to point \(H\) is defined as the vector product,

\[\vec{M}_H = (P - H) \times \vec{F}\]

2.1.3. System of Forces, Resultant of Actions, and Equivalent Loads#

Given a system of \(N\) forces \(\left\{ \vec{F}_n \right\}_{n=1:N}\), applied at points \(P_n\), we define:

  • resultant of the system of forces: the sum of the forces,

    \[\vec{R} = \sum_{n=1}^{N} \vec{F}_n \ ,\]
  • resultant of the moments with respect to a point \(H\): the sum of the moments

    \[\vec{M}_H = \sum_{n=1}^{N} (P_n - H) \times \vec{F}_n \ ,\]
  • an equivalent load: a system of forces that has the same resultant of forces and moments; for a system of forces, an equivalent load can be defined as a single force, the resultant of the forces \(\vec{R}\) applied at point \(Q\) derived from the equivalence of moments

    \[\begin{split}\begin{aligned} \vec{R} & = \sum_{n=1}^{N} \vec{F}_n \\ (Q - H) \times \vec{R} & = \sum_{n=1}^{N} (P_n - H) \times \vec{F}_n \\ \end{aligned}\end{split}\]

2.1.4. Couple of Forces#

A couple of forces is an equivalent load to two forces of equal magnitude and opposite sense, \(\vec{F}_2 = - \vec{F}_1\), applied at two points \(P_1\), \(P_2\) not aligned along the line of application of the forces to have non-zero effects.

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The resultant of the forces is zero,

\[\vec{R} = \vec{F}_1 + \vec{F}_2 = \vec{F}_1 - \vec{F}_1 = \vec{0} \ ,\]

while the resultant of the moments does not depend on the moment pole,

\[\begin{split}\begin{aligned} \vec{M}_H & = (P_1 - H) \times \vec{F}_1 + (P_2 - H) \times \vec{F}_2 = \\ & = (P_1 - H) \times \vec{F}_1 - (P_2 - H) \times \vec{F}_1 = \\ & = (P_1 - P_2) \times \vec{F}_1 =: \vec{C} \ . \end{aligned}\end{split}\]

2.1.5. Force Fields#

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2.1.6. Distributed Actions#

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