1
Forces
1.1 Terminology and Notation
A force exerted on a body tends to change the state of the body, that is, if the body is rigid the force tends to move the body, but when the body is elasto-plastic the force tends to deform the body.
A force can be defined as a vector quantity that is defined by magnitude and direction. The direction of a force is specified by its orientation (also known as the line of action) and sense. The magnitude of a force is a positive scalar. A scalar is a number expressed in specific units of measure.
Vectors (forces) are usually denoted by boldface letters. If the starting point and the end point of a vector (force) are given, the vector (force) could be denoted by or more simply . It is also usual to denote the magnitude of the vector (force) by or by . Some other notations for vectorial quantities could be , , or .
Graphically a force is represented by a straight arrow as shown in Figure 1.1. The point is named the application point or the origin of the force and the line passing through and is named the action line of .
There are some possible operations regarding vectors.
Equality of forces
Two forces and are equal to each other when they have the same magnitude and direction, that is
(1.1) If the forces and are equal but are acting at different locations on the same body it will not cause identical motion.
Multiplication of a Force by a Scalar
The product between a force and a scalar written as , is a force having the same orientation as , the same sense as if if and opposite sense if , and the magnitude .
Figure 1.1 Vector representation.
Zero Force
A zero force, usually denoted by , has a zero magnitude and an undefined direction.
Unit Vector (Force)
A unit vector has its magnitude equal to unity, that is, . Any force can be written as a product of a unit vector having the same orientation and sense as the force and its magnitude or equivalent
(1.2) Addition of Forces
The sum of a two forces and is a new force named resultant. The sum of the forces and is the force represented graphically by the diagonal of the parallelogram shown in Figure 1.2 with its tail connecting the tail of the force and head connecting the head of the force .
The sum is named the difference of the two forces as shown in Figure 1.3.
Figure 1.2 Parallelogram law of vector addition.
Figure 1.3 Parallelogram law of vector subtraction.
1.2 Resolution of Forces
If the unit vectors , , have the same application point (origin) and are perpendicular to each other, as shown in Figure 1.4, they form a Cartesian reference frame.
Any force can be expressed with respect to the unit vectors , , by where , , and are the , , components of the force.
The magnitude of can be written as
Addition and subtraction of forces could be easily manipulated using the resolution of forces into components. Considering the forces and , one can calculate
and
1.3 Angle Between Two Forces
The angles between the forces and , and respectively and - in the range between and - are usually denoted by Greek letters such as and , as shown in Figure 1.5.
The direction of a force in a Cartesian frame is given by the direction cosines (Figure 1.6) of the angles between by the force and the associated unit vectors , , , written as
A unit force (of magnitude 1) having the same direction as can be written as
Figure 1.4 Resolution of a force.
Figure 1.5 The angles and between the forces and , and respectively and .
Figure 1.6 Direction cosines.
1.4 Force Vector
The position force (vector) shown in Figure 1.7 of a point relative to a point can be written as
(1.3) The position force (vector) shown in Figure 1.7 of the point relative to a point is calculated with
(1.4) Figure 1.7 Position forces (vectors).
1.5 Scalar (Dot) Product of Two Forces
Definition. The dot product of two forces and is
(1.5) where is the angle between the forces and .
1.6 Cross Product of Two Forces
The cross product of two forces and is another force defined by (Figure 1.8)
(1.6) where is a unit force normal to and having its direction given by the right-hand rule.
The magnitude of the cross product is given by
When and the cross product can be calculated using
(1.7) Figure 1.8 Cross product of two forces and .
1.7 Examples
Example 1.1
Figure 1.9 shows three forces , , and , and the angles of the forces with the horizontal , , and . The forces have the magnitudes kN, kN, and kN. Find the resultant of the planar forces and the angle of the resultant with the horizontal.
Solution
The input data are introduced in MATLAB with:
clear; clc; close all F1 = 1; % kN F2 = 3; % kN F3 = 2; % kN % angle of force F1_ with x-axis theta1 = pi/6; % angle of force F2_ with x-axis theta2 = pi/3; % angle of force F3_ with x-axis theta3 = pi; Figure 1.9 Graphical representation the forces , and .
The components of the forces on and axes are
(1.8) or in MATLAB:
% components of forces F1_, F2_, and F3_ F1x = F1*cos(theta1); F1y = F1*sin(theta1); F1_ = [F1x F1y]; F2x = F2*cos(theta2); F2y = F2*sin(theta2); F2_ = [F2x F2y]; F3x = F3*cos(theta3); F3y = F3*sin(theta3); F3_ = [F3x F3y]; The numerical values are:
F1_ = [ 0.866 0.500] (kN) F2_ = [ 1.500 2.598] (kN) F3_ = [-2.000 0.000] (kN) The resultant is calculated with
(1.9) and the angle of the horizontal with the horizontal axis is
(1.10) With MATLAB the resultant and the angle are calculated with:
R_ = F1_+F2_+F3_; phi = atand(R_(2)/R_(1)); and the results are
% R_ = F1_+F2_+F3_ = [ 0.366 3.098] (kN) % phi = atan(Ry, Rx) = 83.262 (deg) The MATLAB representation of the forces is shown in Figure 1.10 and it is obtained with:
sa = 4; hold on axis([-sa sa -sa sa]) axis square quiver(0,0,F1_(1),F1_(2),0,'Color','k','LineWidth',1.2) quiver(0,0,F2_(1),F2_(2),0,'Color','k','LineWidth',1.2) quiver(0,0,F3_(1),F3_(2),0,'Color','k','LineWidth',1.2) quiver(0,0,R_(1),R_(2),0,'Color','r','LineWidth',2) text(F1_(1),F1_(2),' F_1',... 'fontsize',14,'fontweight','b') text(F2_(1),F2_(2),' F_2',... 'fontsize',14,'fontweight','b') text(F3_(1),F3_(2),' F_3',... 'fontsize',14,'fontweight','b') text(R_(1),R_(2),' R',... 'fontsize',14,'fontweight','b') grid on xlabel('x'), ylabel('y'), Example 1.2
Figure 1.11 shows a system of spatial forces with the magnitudes N, N, N, and N. The parallelepiped has the dimensions m, m, and m. Find:
- (a) the resultant of the system of forces;
- (b) the angle between the forces and ;
- (c) the projection of the force on the force ; and
- (d) calculate .
Figure 1.10 MATLAB representation of the forces , and .
Solution
- 1)
- (a) The input data in MATLAB are:
clear; clc; close all a = 2; % m b = 3; % m c = 5; % m F1 = 15; % N F2 = 30; % N F3 = 10; % N F4 = 15; % N A Cartesian reference frame is selected as shown in Figure 1.11. The position force (vector) of the application point of the force is
(1.11) The position vector of the application point of the force is
(1.12) The position (vector) of the application point of the force is
(1.13) The position vector of the application point of the force is
(1.14) Figure 1.11 System of four spatial forces , , and...
