# Resulting force: how exercises are calculated and solved

The **resulting ****force** is the sum of all the forces acting on the same body. When a body or object is subjected to the action of several forces simultaneously, an effect occurs. Driving forces can be replaced by a single force that produces the same effect. This single force is the resultant force also known as net force and is represented by the symbol *F *_{R }**.**

The effect of *F ** _{R} *will depend on their size and direction. Physical quantities that have direction and meaning are vector quantities.

As the forces acting on a body quantity vector, the resultant force *F ** _{R} *is a vector sum of all forces and can be graphically represented with an arrow indicating the direction and sense.

With the resulting force, the problem of a body affected by multiple forces is simplified, reducing it to a single acting force.

**Formula**

The mathematical representation of the resulting force is a vector sum of the forces.

*F *_{R}** = ∑ F** (1)

**∑ F = F **

_{1 }

*+ F*

_{2 }

*+ F*

_{3}*+…*(2)

**F**_{N}*F ** _{R}* = Resultant Force

**∑ F =** Sum of forces

*N* = number of forces

The resulting force can also be represented with Newton’s second law equation.

*F *_{R}* = m. to* (3)

*m* = body mass

** a =** body acceleration

If equation (1) is substituted into equation (3), the following equations will be obtained:

**∑ F**

*= m.*(4)

**to***F *_{1 }*+ F *_{2 }*+ F *_{3}* +… F _{N =} m. to* (5)

Mathematical expressions (4) and (5) provide information about the state of the body, obtaining the acceleration vector ** a** .

**How is the net force calculated?**

The resulting force is obtained by applying Newton’s Second Law, which states the following:

*The net force acting on a body is equal to the product of its mass by the acceleration it acquires* . (Equation (3))

The acceleration of the body will have the direction of the net force applied. If all the forces acting on the body are known, it would be enough to add it vectorally to obtain the net force. Likewise, if the net force is known, it would be possible to divide it by the body mass to obtain its acceleration.

If the resulting force is zero, the body will be at rest or at constant velocity. If the body acts on a single force, the resulting force is equal to the force *F *_{R}** = F** .

When several forces act on the same body, the vector components of the force must be taken into account and whether these forces are parallel or not.

For example, if we slide a book horizontally across a table, forces in the horizontal direction are the only ones that provide acceleration to the body. The net vertical force on the book is nil.

If the force applied to the book has an inclination relative to the horizontal plane of the table, the force will be written according to the vertical and horizontal components.

**Resulting ****from parallel forces**

Parallel forces acting on a body are those acting in the same direction. They can be of two types in the same direction or in the opposite direction.

When the forces acting on a body have the same direction and the same or the opposite direction, the resulting force is obtained by performing the algebraic sum of the numerical values of the forces.

Force resulting from two parallel forces.

**Unparalleled Forces**

When non-parallel forces are applied to a body, the resulting forces will have rectangular and vertical components. The mathematical expression to calculate the net force is:

*F _{R }^{2} =* (

*∑ F*)

_{x}^{2}+ (

*∑ F*)

_{y}^{2}(6)

tan *θ _{x}* =

*∑ F*(7)

_{y}/ ∑ F_{x}*∑ F _{x} y ∑ F _{x} =* Algebraic sum of the

*x*and

*y*components of the applied forces

*θ _{x}* = angle forming the resultant force

*F**to the axis*

_{R}*x*

Note that the resulting force of expression (6) is not highlighted in bold and is because it expresses only the numerical value. The direction is determined by the angle *θ _{x}* .

Expression (6) is valid for forces acting on the same plane. When forces act in space , the *z* force component is taken into account if you are working with rectangular components.

**solved exercises**

Parallel forces from the same direction are added and the parallel force from the opposite direction is subtracted.

*F _{R}* = 63 N + 50 N – 35 N = 78N

The resulting force has a magnitude of 78N with horizontal direction.

**2. Calculate the resultant force of a body under the influence of two forces F _{1}_{ and }F _{2} . Force F _{1} has a magnitude of 70N and is being applied horizontally. Force F _{2} has a magnitude of 40N and is being applied at an angle of 30° to the horizontal plane.**

To solve this free-body exercise diagram with the coordinate axes *x* and *e* is drawn.

All the *x* and *y* components of the forces acting on the body are determined. The force *F ** _{1}* has only one horizontal component about the

*x*axis . The force

*F**has two components*

_{2}*F*that are obtained from the sine and cosine functions of the 30° angle.

_{2x}and F_{2y}*F _{1x}* =

*F*70N

_{1}=*F _{2x}* =

*F*cos 30 ° = 40 N.cos 30 ° = 34.64N

_{2}*F _{1y}* = 0

*F _{2y}* =

*F*without 30 ° = 40 without 30 ° = 20N

_{2}*∑ F _{x} =* 70N + 34.64N = 104.64N

*∑ F _{y} =* 20N + 0 = 20N

After determining the net forces on the *x* axis and *e* proceeds to obtain the numerical value of the net force.

*F _{R }^{2} =* (

*∑ F*)

_{x}^{2}+ (

*∑ F*)

_{y}^{2}

The resulting force is the square root of the square sum of the force components.

*F _{R} =* √ (104.64N)

^{2}+ (20N)

^{2}

*F _{R} =* 106.53N

The angle between the resultant force *F ** _{R}* is obtained from the following expression:

*θ _{x}* = tan

^{-1}(

*∑ F*)

_{y}/ ∑ F_{x}*θ _{x} =* tan

^{-1}(20N

*/*104.64N) = 10.82 °

The resultant force *F ** _{R}* has a magnitude 106,53N and has a given direction by the angle of 10.82 ° which forms with the horizontal.