Why generalized force equals to zero?

AI Thread Summary
Generalized force Q1 is determined to be zero due to the nature of the forces acting on the system, specifically frictional and normal reaction forces. The equations of motion in generalized coordinates simplify the analysis by focusing on independent parameters that correspond to the system's degrees of freedom. The discussion highlights that if a generalized potential is constant, the generalized forces will also be zero. The transformation equations for coordinates are essential for understanding how generalized coordinates relate to physical constraints. Overall, the choice of generalized coordinates is crucial for effectively solving the equations of motion in mechanics.
harmyder
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Homework Statement


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Lecturer says that:
Generalized Coordinate: θ = q1
Generalized Force: Q1 = 0


I don't understand why Q1 equals to zero.

Homework Equations


Q_i = \sum_j \underline F_j \frac{\delta \underline r_j}{\delta q_i}

The Attempt at a Solution


First i need to find partial derivatives:
\frac{\delta \hat x}{\delta q_1} = \frac{\delta \theta r}{\delta \theta}= r (i have lost vector here, but i don't know where to put it in \theta r)
\frac{\delta \hat y}{\delta q_1} = \frac{\delta r}{\delta \theta} = 0 (same here)

Now forces at x and y directions:
F_x \frac{\delta \hat x}{\delta q_1} = -F + m\bf g \sin\psi
F_y \frac{\delta \hat y}{\delta q_1} = N - m\bf g \cos\psi

Now add them:
Q_1 = r (-\bf F + m\bf g \sin\psi) is it zero?
 
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harmyder said:
Now forces at x and y directions:
Fxδ^xδq1=−F+mgsinψF_x \frac{\delta \hat x}{\delta q_1} = -F + m\bf g \sin\psi
Fyδ^yδq1=N−mgcosψ

what are F and N ?
if they are forces acting on the body and if its frictional and normal reaction forces then check its values ?
 
drvrm said:
what is F and N
Friction and reaction forces. You can see it form the image.
 
then the magnitude of N should be equal to the component of weight in the direction of normal to the inclined plane and something similar for F .
what is the equation of motion in generalized coordinate , generalized velocities and time-...what is the advantage of such a description?
see
Although there may be many choices for generalized coordinates for a physical system, parameters which are convenient are usually selected for the specification of the configuration of the system and which make the solution of its equations of motion easier. If these parameters are independent of one another, the number of independent generalized coordinates is defined by the number of degrees of freedom of the system.
 
My question is why Q1 equals to zero, I'm aware of how to choose generalized coordinates. Thanks.
 
Actually generalized forces are transformed from the transformation relations operating between the two sets of coordinates - If a generalized potential can be defined. in that situation a constant potential will mean generalized forces to be zero - in your case the constraining equation is of the type say dx - r. d(theta) =0
In using Lagrangian method the forces of constraint do not appear as the virtual displacements are such that they are consistent with the forces of constraint.
What is the transformation equation of coordinates in your example?
 
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drvrm said:
What is the transformation equation of coordinates in your example?

harmyder said:

The Attempt at a Solution


First i need to find partial derivatives:
\frac{\delta \hat x}{\delta q_1} = \frac{\delta \theta r}{\delta \theta}= r (i have lost vector here, but i don't know where to put it in \theta r)
\frac{\delta \hat y}{\delta q_1} = \frac{\delta r}{\delta \theta} = 0 (same here)

So, x = r*theta and y = r.
 
as r is a constant only theta is your generalized coordinate with variation
so delta/delta(theta) = d/dx. dx/dtheta + d/dy. dy/d(theta)

Q(theta) = F(x).r =0
 
drvrm said:
...
Sorry, could you please use itex tag, it is hard to understand your equations. Thanks.
 
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