6.11.4. A Statics Problem Solved With Projections

Another application of projections comes from the geometry. Projections relate two vectors as sides of a right triangle. Thus projections can be used to solve problems that could also be solved using trigonometry. With projections, it is not necessary to compute angles between the vectors or to use the trigonometry functions.

Consider an example of a wagon sitting on an incline as shown in Fig. 6.27. The handle of the wagon is held at an angle. We want to calculate the force on the handle required to keep the wagon from rolling down the incline. We won’t worry about inefficiency of the wheels. The force from the handle need only counter the force from gravity.

../../_images/wagon_fig.png

Fig. 6.27 A wagon on an incline

NOTE: All vectors and forces are calculated as upward and to the right. The force from gravity, f_g, will be negative. Then, the important statics equation is that the sum of forces equals zero. In the calculations, projection matrices are denoted as \bf{P}. For the sake of clarity in the code, vector variables begin with a capital letter, while scalar variables are lower case.

../../_images/wagon_forces.png

The force towards the left (down the incline) is the projection of the gravity force onto the incline, \bf{A}.

\bf{F_l} = \bf{P_a}\,\bf{F_g}

The sum of the forces to the left and right is zero.

\bf{F_r} + \bf{F_l} = 0 \longmapsto \bf{F_r} = -\bf{F_l}

The force to the right is a projection of the handle force vector onto the incline. The handle force vector is the product of the scalar force vector and the unit vector in the direction of the handle, \bf{F_h} = f_h\,\bf{H}. We need to calculate the inverse of the projection, so we separate the scalar from the unit vector to include the vector direction in the calculation.

\bf{F_r} = \bf{P_a}\,\bf{F_h} = \bf{P_a}\,\bf{H}\,f_h = - \bf{P_a}\,\bf{F_g}

f_h = (\bf{P_a}\,\bf{H})^{-1}\,(- \bf{P_a\,F_g})

%% File: wagon.m
%  Statics problem of a wagon on an incline solved with vector
%  projections

P = @(a) a*a'/(a'*a);   % Projection matrix equation
V = [0 1]';             % Vertical vector
A = [8 3]';             % vector of the incline slope
H = [1 1]';             % vector of handle direction
H = H/norm(H);          % unit vector of handle direction
                        % Fh = fh*H
fg = 23 * -9.81;        % Gravity force down (F = m*a) in newtons
                        % 23 kg is about 50 pounds
Fg = fg * V;            % Gravity force vector
Pa = P(A);              % Projection matrix to incline A
Fl = Pa * Fg;           % force down hill to left
Fr = -Fl;               % force to right, Fr + Fl = 0
fh = (Pa * H) \ Fr;     % Fr = Pa*Fh = Pa*H*fh
fprintf('Force on handle = %.3f newtons or %.3f kg\n', ...
    fh, fh/9.81);

Challenge Question

By changing the vectors of the incline and the handle, one can verify the correctness of the solution. What are the boundary conditions? Try to determine the relationship of the slope of the incline to the mechanical advantage for raising the wagon vertically using a rolling wagon or cart on an incline?

Now complete the Projection of Static Force Vectors homework assignment.