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Re: [Phys-l] A different normal force. Was: Re: The Normal Force

Wow!

The book gives * the Acceleration of M as m ax / M and of m, as ax = g sin cos/ (1 + m sin^2 /M)

Substituting, multiplying by M (to obtain horiz. force) and dividing by sin (to obtain normal component from the horiz.) gives your result!

Which leaves the partially desimplified problem of the concave wedge. It's Marian's problem # 34 in Ch. 6 (LaGrangian and Hamiltonian dynamics). The answer in the back of the book is lambda = -3mg sin(theta) + 2mgsin (theta sub zero)

I presume theta zero is the initial position. This answer is remarkably similar to the tension of a pendulum's rod
mg[3cos(theta) - 2 cos(amplitude)] Cliché warning: In fact if one measures theta from the horiz; clock wise increasing instead of the usual WRT the vertical anteclock increasing, they are the same (even the sign) Note: In both cases the length (radius) "cancels out".

I presume one may solve the concave wedge, also, by Bob's method, "my **" tension method (E / work principle), and finding the L'ian, etc. I presume the latter is what Marian wanted us to do. Would some one give me a hint, i.e. what are the coordinants?

* I believe it now.
** in every algebra and +, text.

bc, shamed.



LaMontagne, Bob wrote:

Hi,

I'm attempting to solve this as a diversion while I'm judging a figure skating competition - so please be kind if there are errors.

I will use M and Ax as the mass and acceleration of the wedge, m and ax and ay as the mass and accelerations of the particle. N is the normal force. I will simply use sin, cos, and tan as the trig functions for the angle, theta, of the wedge.

1. N sin = m ax
2. N sin = M Ax
3. ay = (ax + Ax) tan

The vertical forces acting on the paticle give

N cos - mg = m ay

Substitute 1., 2., and 3. into the above to get

N = m g cos / ( 1 + m/M sin^2)
Seems to work OK for all the limiting values of m, M, and theta that I substituted.

Bob at PC




-----Original Message-----
From: phys-l-bounces@carnot.physics.buffalo.edu on behalf of Bernard Cleyet
Sent: Thu 5/24/2007 5:59 PM
To: Forum for Physics Educators
Subject: [Phys-l] A different normal force. Was: Re: The Normal Force

I've been working on a normal force problem that evidently is beyond my skill at 70. One method appears will work, but is too harry. the other, I suppose, I shouldn't give up so quickly, but I'm lazy, hoping one of you will solve it for me.

A wedge sits on a frictionless surface and a particle sits on the slopping surface of the wedge, w/ mass m (wedge is mass M). Find the reaction force as a function of the masses and the angle of the wedge. I tried the PE - KE (looks as it'll work, but just got too harry (70 repeating). So I switched to trying to fine the general cords. for the L. And got stuck. Anyone help?

bc


p.s. This has a practical app. if one desimplifies it. The particle is a cylinder that rolls down a curve. The "wedge" is attached to the pendulum of a clock. For my purpose, all I need is to use the reaction force as a torque and how does it (the torque) varies w/ an initial speed of the wedge (pendulum). The pendulum's amplitude is limited to about 30 milliradian, but the wedge is rather near the suspension.

Those of you, horologically inclined, will recognize this as the Synchronome escapement. The gravity escapement is "activated" every 30' very near BDC (The Airy criterion)

John Denker wrote:


On 05/24/2007 10:45 AM, Jeffrey Schnick wrote:




I'm just looking for a slightly more complete explanation, at
the same level--one that might also account for why stepping onto the
surface of a pond is so much different when the water is in its liquid
state than it is when the water is in its solid state.





cut
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_______________________________________________
Forum for Physics Educators
Phys-l@carnot.physics.buffalo.edu
https://carnot.physics.buffalo.edu/mailman/listinfo/phys-l