Let us now apply the procedure outlined above to slightly more difficult examples.

Example 1: There are three massless and frictionless pulleys P1, P2 and P3. P1 and P2 are fixed and P3 can move up and down, as shown in figure 5. A massless rope R1 passes over the pulleys as shown and two masses m₁ and m₂ attached at its ends. A third mass m₃ is hanging from P3 by a rope R2 of fixed length. Find the acceleration of the three masses.

In figure 5 we have also shown the distances of different pulleys and masses from the ground, with the vertically up direction taken to be positive. The heights h₁ and h₂ of pulleys P1 and P2, respectively, are fixed whereas height y_p of pulley P3 can change. We go about solving the problem according to the steps given above.

Step 1: We identify two constraints and the forces of constraints as: rope R1 has fixed length with the force of constraint being tension T₁ in the rope. The other constraint is that rope R2 has fixed length with the tension T₂ in the rope as the constraint force. Because of massless pulleys and ropes and frictionless surfaces T₁ is the same throughout rope R1.

Step 2 : Make free-body diagrams of the subsystems. We consider only those subsystems that can move. Thus we make free-body diagram of each mass and the pulley P3 as shown in figure 6.

Step 3 : By looking at the free-body diagrams, write equations of motion for each subsystem. In terms of the distances shown in figure 5, we get

and because the pulley is massless

Thus equations of motion give four equations. However there are six unknowns viz. . Their number exceeds the number of equations obtained so far by two.

Step 4 : The additional two equations are provided by the constraint equations. The constraint that rope R1 is of fixed length is expressed as (see figure 5 for the variables used)

Differentiating this equation twice with respect to time gives

The second constraint that rope R2 is of fixed is equivalent to

which upon differentiating gives

Thus the equations that describe the motion of the system fully are:

I will leave Step 5 – that is solving the equations - for you to do but give you partial answer. It is

I would now like you to try a similar problem but with slight difference. Let us attach the centre of the third pulley to a spring of spring constant k (see figure 7). Then find the equations of motion for the two masses and solve them.

Example 2 : As another example of constrained motion we take a small block of mass m sliding down on a cylindrical surface from its top (figure 8). The question we ask is at what angle from the horizontal would the mass slide off the surface of the cylinder.

Since this problem involves motion along a circular path I would use planar polar coordinates. I take the origin at the centre of the cylinder and let the x-axis be along the horizontal and y-axis along the vertical. Assume that the radius of the cylinder is R . The constraint in this problem is that r  = constant = R. The corresponding constraint force is the normal reaction N of the cylindrical surface on the block. The free-body diagram of the mass on the cylinder is shown in figure 9.

We now write the equations of motion in the planar polar coordinates. That gives in the direction

and in the direction

We again have three variables but only two equations. The third equation is provided by the equation of constraint i.e.

r = constant = R

which gives

With this the equations to be solved are

To solve these we use

Substituting this in the equation for above gives

This when substituted in the equation for leads to

The point when the mass slips off the cylinder is where N becomes zero. So the corresponding is given by

Example 3: Let us take one more example of constrained motion when two bodies are involved. I put a block of mass m on a wedge of mass M with wedge angle θ (see figure 10). The wedge is free to move on a frictionless plane. There is no friction also between m and M . We wish to find the resulting motion.

There are clearly two subsystems, the masses m and M . There are two constraints in the system. Constraint one is that the mass m moves along the edge of the wedge so its x and y components are not independent. The other constraint is that the wedge moves only in the x direction. The constraint forces are obviously the normal reaction N₁ on mass m by the wedge and the normal reaction N₂ on the wedge by the ground. The free-body diagrams for the two subsystems are as shown in figure 11.

Notice that in the free-body diagram of the wedge, there is no mg of block. It is all accounted for by N₁. To set up the equations of motion, let us choose our co-ordinates system a follows (see figure 12): Let the coordinate of the right-hand side lower corner of the wedge be given the co-ordinates (x₁ y₁ ) and let the co-ordinates of the block be (x₂ y₂ ).

The equations of motion in terms of these coordinates are:

For the six variables - - of the system, we need two more equations, which are provided by the constraints equations. These are

and

which gives

Thus the equations to be solved are

These equations can now be solved to get all the variables as a function of time. That task is left for you. I'll leave you with answers for N₁ :