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Example:  Sliding and Hanging Weights, and a Ramp

Figure 1 -- Weights and ramp:
weights m1 and m2, ramp, and pully
This is a very simple example of the use of the Lagrangian formulation of Newtonian mechanics to solve a problem.  We have two weights, m1 and m2, attached together with a string of length l  (figure 1).  One is on a ramp; the other is hanging from the string.  The string goes over a pully. Both weights are initially stationary.  We'll start by neglecting friction, then do it again with friction later in the page.

With x units of string between the pulley and weight m1, it drops file:///nfs/dodo/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_Q3zwDt.png units below the level of the pulley.  So, the (gravitational) potential energy of weight m1 is

(1)    file:///nfs/dodo/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_agHD2H.png

When weight m1 slides x units down the ramp, weight m2 must rise x units.  So, the (gravitational) potential energy of weight m2 is

(2)   file:///nfs/dodo/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_IWMaiX.png

The kinetic energy of the system is, of course,

(3)    file:///nfs/dodo/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_8wHdfJ.png

So, following equation (lagrange.4), the Lagrangian for the system is:

(4)    file:///nfs/dodo/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_06UPVy.png

The partial derivatives we need are:

(5)    file:///nfs/dodo/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_7QkI1s.png

As soon as we start to write the equation of motion, we realize this substitution helps a lot:

(5b)    file:///nfs/dodo/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_5PNHB7.png

And with substitions (5b), the equation of motion, following (lagrange.3), is:

(6)    file:///nfs/dodo/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_hvbmup.png

We can see immediately from (6) that there are three cases.  In case (7a), weight  m1 slides down the ramp, and weight m2 rises.

(7a)    file:///nfs/dodo/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_4GsM8Q.png

In case (7b), the weights balance, and nothing happens.

(7b)    file:///nfs/dodo/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_oB2e4b.png

And in case (7c), weight  m1 slides up the ramp, and weight m2 falls.

(7c)    file:///nfs/dodo/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_ZYBorQ.png

We can integrate (6) twice to obtain a closed-form solution.  Assuming the weights are stationary to start with, and m1 starts at x0:

(8)    file:///nfs/dodo/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_XvYG7Q.png

Once Again, with Friction

We'll assume there's friction at work when m1 slides along the ramp, and we'll use the friction model given in (lagrange.9).  In this case there's just one dimension, so it's very simple:

(9)    file:///nfs/dodo/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_KQwAaF.png

We will assume f1 is positive for case (7a), and negative for case (7c).  (In case (7b) nothing moves and friction doesn't matter!)   The "cos θ" term is due to the fact that the force of gravity is straight down, which is at angle θ to a line perpendicular to the ramp's surface.  Note that the dimensions of f2 are time/distance, while f1 is dimensionless.

Plugging equations (5) and friction force (9) into the equation of motion (lagrange.8), we obtain

(10)    file:///nfs/dodo/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_TGF6fu.png

or, slightly rearranged, and again substituting (5b):

(11)    file:///nfs/dodo/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_bh5IyT.png

We can see a couple of things immediately by looking at (11).  First, if the "constant friction" is too large, the weights won't move at all; the initial acceleration will be zero.  In order for them to move, we must have:

(12)    file:///nfs/dodo/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_NFVK1x.png

As the velocity increases, the viscous friction term becomes larger, and eventually the acceleration again goes to zero; the terminal velocity, when acceleration is zero, must be:

(13)    file:///nfs/dodo/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_h2BIMs.png

Finally, we can solve for the velocity and position.  Let's rearrange (11) a bit:

(14)    file:///nfs/dodo/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_9bpNli.png

That looks like an exponential.  The coefficient of file:///nfs/dodo/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_Xh2GXn.png is positive; the right hand side may be either positive or negative depending on which way the system slides (case (7a) or case (7c)).  Let's substitute a and b for the constants, to reduce clutter, with a>0:

(15)    file:///nfs/dodo/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_PuwYdZ.png

A solution to the homogeneous equation (with b=0) is:

(16)    file:///nfs/dodo/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_VG8WBF.png

and with a little fiddling we obtain the particular solution (for b≠0):

(17)    file:///nfs/dodo/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_vKUskY.png

Adjusting this to match the initial conditions at t=0 (which were x=x0, file:///nfs/dodo/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_Xh2GXn.png=0), we obtain:

(18)    file:///nfs/dodo/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_c3MAPw.png

Taking a firm grip on our pencil and looking back at (14), we substitute the definitions for a and b back into (18), differentiate, and divide out some common factors to obtain the equations for the position and velocity:

(19a)    file:///nfs/dodo/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_GkPMjn.png

(19b)    file:///nfs/dodo/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_R79h9k.png

At time t=0, (19b) is certainly zero, and as t->file:///nfs/dodo/home/slawrence/website/physics_insights/physics/formulas/eqe_temp_image_O06U0k.png, we see that (19b) approaches the terminal velocity we found in (13), which encourages us to think that these rather ugly equations might actually be correct.  As is often the case, introducing a general friction function made the problem a lot messier.  Keep in mind that, in this case, the sign we assumed for f1 depends on the ratio of the masses and the angle of the ramp.

Page created on 12/1/06.  Minor corrections, 12/2/06