EN3: Introduction to Engineering and Statics

 

 

 

                      

 

   Division of Engineering

    Brown University

 

 

3. Resultant of systems of forces

 

Machines and structures are usually subjected to lots of forces.  When we analyze force systems, we usually are interested in the sum of the forces.   The sum of a set of forces is known as the resultant of the force system.

 

We need to know how to add forces together.

 

3.1 Resultant of a set of discrete forces

 

This is straightforward. We know that forces are vectors, so they must add like vectors.  We can use the usual vector machinery to sum them.  The procedure is

1.      Choose convenient basis vectors  (or  )

2.      Using geometry or trigonometry, calculate the force component along each of the three reference directions  or .  You need to do this for each force.  Let’s suppose that there are N forces, with components ,  … .

3.      We can use the formula for a vector sum in terms of its components to show that the resultant force

 

In words, the components of the resultant force vector are equal to the sum of the components of all the individual forces.

 

Examples

 

A few simple examples illustrate the idea

 

1. The figure below shows a heavy box suspended from two cables.  The box is subjected to a vertical gravitational force W, and two forces of unknown magnitude ,  acting parallel to cables along OA, and OB, respectively.  Find an expression for the vector sum of the forces (in terms of ,  ), expressing your answer as components in the basis shown.

 

 

We start by expressing the forces in the basis, as outlined in the vector tutorial

 

 

 

so the sum is

 

 

 

 

2. The sailboat shown in the picture is subjected to the following forces

1.      A force of 500N acting perpendicular to the sail

2.      A weight force of 1000N acting vertically downwards

3.      A buoyancy force of unknown magnitude (denoted by B) acting vertically upwards

4.      A drag force of unknown magnitude (denoted by D) opposing forward motion of the boat

5.      A drag force of unknown magnitude (denoted by L) opposing sideways motion of the boat

Find an expression for the resultant force.

We follow the rules.  The {i,j,k} vectors are shown already.  Forces 2-5 are easy

The force acting on the sail is a bit more tricky – we need to find a unit vector perpendicular to the sail.  A view of the boat from above is helpful

 

Trigonometry shows that the components of a unit vector normal to the sail are

(remember that a unit vector has length 1, and resolve into i and k components)

 

Therefore

 

Finally we can sum the forces to get

 

 

 

 

3.2 Resultants of distributed loads

 

Many engineering problems involve distributed forces.  Examples include pressure forces acting on aerodynamic forces or structures, hydrostatic pressure acting on submarine structures or on ships, and gravity loading. 

 

We need to know how to calculate the resultant force exerted by continuously distributed forces (e.g. pressure distributions).  We’ll illustrate the general idea using 1D examples in this course, and defer more realistic 3D problems to future courses.

 

 

 

 

As a simple preliminary example, let’s the resultant force exerted by a uniform loading acting on a beam. The beam has length L and is subjected to a uniformly distributed load p per unit length.  We’ll work with the coordinate system shown in the picture.

 

Note that the resultant force due to the pressure acting on a small piece of the beam  width dx is

The total force can then be computed by summing (integrating) over the entire length of the beam

The same approach can be used to calculate resultant forces caused by more complex distributions of pressure. 

 

 

 

Let’s do a more complicated, realistic problem.  Part of a ship’s hull is modeled as a circular cylinder with radius R   and length L. The ship’s draft is R.  The hull at a depth h below the surface is subjected to a fluid pressure  per unit area acting normal to the surface of the hull, where  is the density of water and g is the acceleration of gravity.  Find an expression for the resultant force exerted by the pressure acting on the sides of the hull.

 

It’s best to work with the coordinate system shown in the picture.  We’ll do the integral over the surface using the angle  as the variable of integration.  

 

First calculate the force on a small segment of the hull’s surface between  and The length of the segment is and so its area is .  The pressure acting on this segment is

.

The magnitude of the force acting on the segment of the hull is therefore

 

 

The direction of the force is towards the center of the circle. Simple geometry shows that  is a unit vector with the correct direction.  Therefore the vector force acting on the segment is

 

The total force follows by summing (integrating) over the immersed portion of the ship’s hull.  Thus

We can quickly do the integral using MAPLE to find that

Of course, we could have anticipated this result using Archimedes’ principle (the buoyancy force on an object is equal to the weight of water it displaces).