Machine Dynamics, Solved Problems, Kinematics, Write velocity problem as a matrix equation
9#Mechanical_Engineering #Theory_of_Machines #Machine_Dynamics
#Kinematics
This video is part of a series of videos presenting solutions of problems related to the machine dynamics topic.
In this video, we are going to write velocity equations in matrix form for a mechanism involving 5 bars and 1 slider.
In this problem, we are asked to answer the following 5 questions related to the mechanism shown here
First, it is asked to write the loop-closure equations and then deduce the system of equations for the position problem.
Then, it is asked to deduce the system of equations for the velocity problem.
Finally, it is requested to write the velocity system of equations in the form of a matrix equation.
This mechanism is built from six links, the ground, four bars and one slider.
The six links are connected together using six rotating kinematic pairs and one prismatic kinematic pair.
In all seven lower pairs, are used.
In this mechanism no higher pair is used.
Thus, mobility is equal to 1.
As mobility is equal to 1, the number of degrees of freedom of the mechanism is one.
It means that only one independent parameter is required to define completely and uniquely the position of the mechanism.
It mean also that mechanism has one degree of freedom controlled by an external source of energy.
The mechanism should receive one input.
Consequently, one position parameter should be given in order to be able to solve the position problem.
For example, one bar angle should be considered as given.
Let theta 2, the angle of the bar A B, be the given angle.
Therefore, the position problem involves four unknowns,
Theta 3, the angle of the bar B D,
Theta 4, the angle of the bar D E,
Theta 5, the angle of the bar E F,
And, r 6, the linear position of the slider C.
In order to express the four unknowns in terms of the given angle theta 2, two loop-vector equations are required.
Here we will use the two loop-vector equations shown here.
A B + B C = A C
And
F E + E D = F C + C D
The first loop vector equation involves the bar A B or link 2, the bar B D or link 3, and the slider C or link 6.
The vector A B, is the vector position of link 2, it will be denotes as vector r 2.
The vector B C corresponds to a vector position of link 3, it will be denotes as r 3, A,.
We use here the subscript 3, A, for vector B C as the second loop vector equation involves another vector position for link 3.
The vector A C corresponds to a vector position of link 6, it will be denotes as r 6, a,.
Here again, we use the subscript 6, A, for the slider C as the second loop vector equation involves another vector position for the slider.
For the vector r 2, its angle theta 2 depends on time and is defined as shown here.
The angle theta 2 is here assumed given.
The modulus of the vector r 2 is constant and equal to the distance A B.
For the vector r 3 a, its angle theta 3 a, depends on time and is defined as shown here.
The angle theta 3 a, is here an unknown of the position problem.
The modulus of the vector r 3 a, is constant and equal to the distance B C.
For the vector r 6 a, its angle theta 6 a, is constant and equal to 0 degrees.
The modulus of the vector r 6 a, is time-dependent and is an unknown also of the position problem.
Let’s move now to the second question.
We are going to deduce the system of equations for the velocity problem starting from the system of equations of the position problem.
To this purpose, the position equations will be differentiated, with respect of time in order to establish the velocity equations
In order to differentiate these equations with respect of time we need to recall that:
The time derivative of r cosine theta of time, is equal to, minus r times d theta over d t times sine theta of time.
And, the time derivative of r sine theta of time, is equal to, r times d theta over d t times cosine theta of time.
The time differentiation of the first equation gives:
Minus r 2 d theta 2 over d t sine theta 2, minus r 3 A d theta 3 over dt sine theta 3, is equal to, d r 6 over d t.
Finally, we move the terms of v 6 and omega 3 from the right side to the left side.
Also, we move the terms of omega 3 from the left side to right side.
We have established this system of equations in the second question.
First organize the equations such as in each term the omega will be after the sine or cosine.
Thus the system writes:
In the first column of the matrix, we collect the factors of the variable omega 3.
The factor of omega 3 in the first equation corresponds, to the element of the first row and first column.
The factor of omega 3 in the second equation corresponds, to the element of the second row and first column.
The factor of omega 3 in the third equation corresponds, to the element of the third row and first column.
The factor of omega 3 in the fourth equation corresponds, to the element of the fourth row and first column.
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