This model opens with a ladder resting on the floor, leaning against the wall. The floor and wall are represented by the forces they exert on the ladder. The only force we know everything about is the weight of the ladder. Its magnitude and direction are fixed by gravity. We know the direction of the other forces because reactions must be perpendicular to the surface from which they arise. Friction must lie parallel to the surface and point away from the movement tendency of the object. We find friction between the ladder and the floor and not the ladder and the wall because the wall is described in the problem as smooth and the floor as rough. We may set any convenient initial magnitude for the unknown (variable) forces. I usually choose a magnitude that shows up OK on the free-body diagram. When you click the Action button, the correct magnitudes for equilibrium will be displayed.

Let's run the model at this point by clicking on Action. Notice that the in the solution, the floor reaction force exactly balances the weight and the floor friction force exactly balances the wall reaction. Always do a quick reality check on model output. Mathematically correct solutions do not always make physical sense. Suppose for example we change the ladder angle from 70 degrees to 10 degrees. Enter 10 in the ladder angle control to accomplish this. Then click Action again. As you can see the model cheerfully shows a solution that is mathematically correct, but look at the relative magnitudes of the floor friction force and the floor reaction force. The ratio, which is the coefficient of friction, is about 2.8. The floor must be not only rough but also sticky to keep the ladder from falling.

Next reset the vectors to their initial state using the Clear button and responding Yes to the question about restoring the pre-run condition. Then click the Vector button to add a wall friction vector. on the vector setup dialog, enter y=1, magnitude = variable, attachment point = green end and vector sort = friction. In the vector label field enter Wall Friction = i*. The i* at the end of the label tells the model to automatically append and update as required the vector's components to the Wall Friction name. Now run the model again, still with the 10 degree ladder angle.

Notice that rather than the struggle by floor friction and wall reaction we saw in the previous setup to keep the ladder from falling, the model uses enough wall friction to do most of the supporting of the green end. In general this algorithm will come up with the solution that minimizes the total change in the vectors' initial magnitudes. Again you might look at the ratio of friction force to reaction force to get coefficients of friction. I find the wall coefficient to be 0.45 and the floor coefficient to be about 1.25. The addition of a bit of wall friction relieves the floor of a lot of the friction load.

We will do one more trick with the ladder model. Set the angle back to 70 degrees and clear the model to the pre-run conditions. Then click on the Wall Friction label and when the vector-editing dialog comes up, click the Delete button. Then click the Vector button to add a new force to the system. This time enter y = -8, magnitude = fixed, attachment point = movable and sort = weight. For the label enter Man Weight = i*. This force represents the weight of a man that will be climbing the ladder. Start the man at the bottom of the ladder by entering 0 in the percent position field. What will be the maximum height to which the man may climb without causing the ladder to slip, if the floor friction coefficient is 0.2?

To find the equilibrium condition, the model first shifts any movable forces to try to balance the system. If the maximum available movement of any movable vectors leaves the system out of equilibrium, then we go to work adjusting the remaining variable forces. The total down force in this system is 10 units so in the solution the total up force must be 10 units. The only up force available is the floor reaction force so at equilibrium we know that the floor reaction force will be 10 units. Now the 0.2 coefficient of friction tells us that the maximum floor friction force will be 2 units. Make the floor friction vector fixed magnitude at x = 2 units. Then just click Action to run the model.

You should see that placing the man at the 56.18 position creates equilibrium with the maximum available friction force. If the man climbs beyond this point, the torque resulting from his weight will overcome the limited floor friction force and the foot of the ladder will kick out with serious consequences. If you are inexperienced in the use of ladders, you might not have known that the higher you climb, the more friction you need between the ladder and the floor.

You might want to play around with fixed and variable wall friction forces or vary the weight of the man or the coefficient of friction between the ladder and the floor... or possibly not. In any event we are done with the ladder model for now.

For details on the operation of the controls, see the Model Controls help page.