created at April 10, 2021

Lesson 3 - Force and Torque

As you already know, the subject of mechanics is mechanical interaction between bodies. This interaction needs to be evaluated and measured somehow, since it has some intensity, area of application, direction. For this purpose the concept of "Force" is introduced. Force is a measure of power interaction between the bodies. This interaction occurs when bodies come in contact with each other.

Force is represented with a vector. This means, it has a magnitude (length), direction, line of impact and application point.

fig. 1.3.1 Graphical representation of a Force

Note that all units during this course will be presented in SI system.

Force is measured in Newtons [N]. One Newton is roughly on tenth of a kilogram. Thus one kilogram contains 10 Newtons approximately. Actually, such a comparison is not quite correct, since Newton is a measure of force, and kilogram is a measure of mass. However, you can keep this relation in mind since it gives you a hint of how big one Newton actually is. So, if you put 1 kilo bottle on the table, it will impact the table with 10 Newton force approximately (9.81 N to be exact).

Force is a sliding vector. It means that its application point can be translated along its impact line. Result will be the same.

fig. 1.3.2 Translation of a force vector along its line

It is important to note that you can translate force along its line while solving mechanical problems, since resulting equations are not influenced by this. We consider the car above as a rigid body. However, this is not the case if we need to deal with car's internal forces caused by applied external force F. In the first case the rear bumper is impacted by force, in second case the forward bumper is under load. The study of internal stress factors caused by application of external forces is subject of the discipline called "Strength of materials".

Introducing a Force into the model

The solving of problems of mechanics starts with decision of what is our target system to calculate. When we explore a specific system, we isolate it from external world, and substitute the impact of external objects with forces.

fig. 1.3.3 Idealization of mechanical system

On the top of fig. 1.3.3 we see one carriage inside the train, which is impacted by neighbor carriages and by the crane on top. The carriage is also impacted by railway, which forces it to move in specific direction and do not allows the carriage to fall through the ground. And finally carriage is impacted by its weight force, which is acting towards the ground.

Now we decide that balance of the carriage is the subject of our investigation and the carriage itself is our target system.

On the bottom of fig. 1.3.3 we are isolating our target carriage from the bodies outside, and substitute the impact of external bodies with according forces. The impact of neighbor carriages is substituted with forces F1 and F2, the impact of the crane with force F3, the impact of railway with forces N1 and N2.

This is essential part of solving mechanical problems - to focus on a specific system (a single body or a set of bodies) and mentally separate it from the outer world.

The forces mentioned above are concentrated loads. They are applied to the body in a single point. In the real world interacting bodies always create some area of contact, which has a finite non-zero value. If contact area is small compared to the extents of the bodies, impact can be represented with concentrated load in order to simplify further calculations. As you see, concentrated load is just another model, which approximates real physical behavior.

On fig. 1.3.4 you see the contact area and contact load distribution between the rail and the wheel. Despite both are made from steel there is still some finite contact area. However, contact area is small compared to wheel dimensions and resulting impact can be represented with a single force N.

fig. 1.3.4 Representing contact interaction with a concentrated load

Another type of load widely used in calculations is distributed load. It is used when the contact area is not small enough, and replacing it with concentrated load will lead to high calculation errors. For instance, snow provides a distributed load to the rooftop.

fig. 1.3.5 Distributed load example

Here q is an evenly distributed load, i.e. intensity of such load remains constant in every single surface point. This is a particular case. In general, intensity of a distributed load can vary with position. In our case we can't just represent the way snow impacts the rooftop with a concentrated load. If we do that, more load will come to the central supports, which is not the case.

Distributed load is being measured in N/m, when force is distributed along some direction, or in N/m2, when force is distributed on the surface. First case typically occurs when dealing with planar problems (all forces are in plane and thus we have 2D problem). Second case can be present when solving three-dimensional problem.

Sometimes we have to compute the resulting force of the distributed load. If load is evenly distributed along some linear section, resulting force is distributed load multiplied by section's length:

F=qL[N]F = q\cdot L\:[N]

The resulting force is applied in the center of the section in this case. In the problem mentioned above (rooftop and snow) we can substitute distributed load with a set of concentrated ones, applied to each single span (section between neighbor supports). This yields the following equivalent scheme:

fig. 1.3.6 Substitute distributed load with a concentrated one


Q1=qL1;Q2=qL2;Q3=qL3;\left | Q1 \right |=q\cdot L1;\quad\left | Q2 \right |=q\cdot L2;\quad\left | Q3 \right |=q\cdot L3;

In the case of surface-distributed load the resulting force is distributed load multiplied by surface area:

F=qA[N]F=q\cdot A\:[N]

For instance, one square meter of rooftop surface is covered with 3 kg of snow. In order to transform it to force (compute weight) mass should be multiplied by acceleration of gravity g = 9.81 m/s2.

g=mg=39.81=29.43 Nm2g=m\cdot g=3\cdot 9.81=29.43\:\ \frac{N}{m^{2}}

Now if total rooftop area is 45 m2, the resulting force of snow pressure is:

Q=qA=29.4345=1324.35NQ=q\cdot A=29.43\cdot 45=1324.35\: N

Now you are familiar with the concept of "Force" as a measure of mechanical interaction between bodies. Force applied to a body tries to translate the body in the direction of force application.

Another important concept is a Moment of Force (or Torque), which is a kind of load which tries to twist body.

Moment of Force (Torque)

The Moment of Force F relatively to origin O is defined as a product of Force magnitude and distance between Force impact line and origin O:

fig. 1.3.7 Moment of Force around the origin
M=dF[Nm]M=d\cdot F\: [N\cdot m]

The distance d is a lever arm. The presented formula is quite simple. It says: Moment of Force is Force times lever arm.

Moment of force is measured in Newton-meters (N*m).

More formally, Moment of Force is a vector, which is a cross-product of vectors OB and F:

mo(F)=OB×F\vec{m_{o}}(\vec{F})=\vec{OB}\times \vec{F}

Moment of force is directed perpendicular to OBC plane, which goes through vectors OB and F. If we translate the force F along its application line (as we might do), the vector's OB direction will also change, but the cross-product of OB and F will remain constant.

In similar manner a Moment of Force around an axis can be defined. In order to find it, Force should first be projected on the plane, which is perpendicular to the axis. Then the projected force is multiplied by the distance between the impact line and axis. It's better to understand from the following figure:

fig. 1.3.8 Moment of Force around the axis

Simply put, Moment of force is a measure of twisting impact applied to the body. The direction of Torque vector coincides with the axis, which the body will be twisted around. Every plane perpendicular to the axis will be a twisting plane (rotation occurs in this plane)

In order to increase Torque, either Force of lever arm should be increased.

fig. 1.3.9 Force and lever arm

Let's consider the Force F on the figure above remains constant. Now increasing lever arm L leads to increasing the Torque (M2 is greater than M1). You feel how it becomes easier to twist the nut. To tighten the nut properly specific amount of torque should be applied. If torque is to low - the nut will release. If torque is to high - the nut's thread will be damaged. The correct torque required for specific nut is typically given in car manuals. A special Torque Wrench is being used to ensure the correct torque.

Torque is a free vector in opposite to Force. It can be parallel translated to arbitrary point of the body (while direction remains the same), the resulting twisting impact will not change.


Force and Torque are two fundamental load types in mechanics. All interactions can be reduced to these two types. Force tries to apply a translational motion to a body, while torque tries to apply a twisting motion.