Direction of a spring's force in more than one dimension

In summary, the torque around the red circle position is M*g*L*sin(theta)-k*x*y=I*w. (considering that the suspension bar is of negligible mass as the problem indicates)
  • #1
Est120
54
3
Homework Statement
Find the frequency of
vibration of the system for small amplitude values (small angle aproximation)
Relevant Equations
sum of the torques around the axis of rotation
1586490830513.png

adding all the torques around the red circle position (taking clockside direction as positive ):
-M*g*L*sin(theta)-k*x*y=I *w (considering that the suspension bar is of negligible mass as the problem indicates )here "x" is the normal "x" of hooke's law (I don't know exactly what it is for a problem in two dimensions) "y" is the lever arm between kx and the red circle

for me:
x=s=h*theta and y=h (because I think that the spring force is represented by the red vector at "L" and therefore this is its lever arm)
which ends at:
M*g*L*sin(theta)+k*h^(2)*theta= -I* d^(2)/dt^(2) [theta]

then using the small angle aproximation and factoring:
d^(2)/dt^(2) [theta]=-(theta)*[(M*g*L+k*h^(2))/I] so is in the form d^(2)/dt^(2) [theta]=-W^(2)*theta
and W is = sqrt[(M*g*L+k*h^(2))/I]= sqrt[(M*g*L+k*h^(2))/M*L^(2)]

which is the correct answer but i got two questions
1.-
what is "x"? in one-dimensional hooke's law "x" it is simply the horizontal distance.
here is the arc length? or the distance (straight line) between the point of application of the force and the point of balance?

2.-

what is the lever arm "y" for this it would help me to know what is the direction of the spring force as the red vector or as the black

i don't speak a lot of english sorry
 
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  • #2
with that W value i just used the formula W=2*pi* f and f is the correct answer but that's not my doubt
 
  • #3
I think that the problem wants you to assume a very small angle ##\theta##. In that case, your questions are not significant.
In a more general case, you would want to determine the direction of the spring, the amount it would be stretched or compressed in that direction, and apply the spring force exactly in that direction.
 
  • #4
FactChecker said:
I think that the problem wants you to assume a very small angle ##\theta##. In that case, your questions are not significant.
In a more general case, you would want to determine the direction of the spring, the amount it would be stretched or compressed in that direction, and apply the spring force exactly in that direction.
but where does the spring force points?
 
  • #5
Est120 said:
but where does the spring force points?
In this problem, I don't think that they want you to do any geometry with anything more than a tiny angle, so you can assume that the spring and the spring force are horizontal.

In the more general case, you would have to figure out where the ends of the spring are and the force would be along the straight line between the ends.
 
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  • #6
Est120 said:
... i got two questions
1.-
what is "x"? in one-dimensional hooke's law "x" it is simply the horizontal distance.
here is the arc length? or the distance (straight line) between the point of application of the force and the point of balance?
Imagine the bar in its vertical position and no tension on the spring.
The direction of the spring is perpendicular to the bar.
The end of the bar describes a red semi-circle of radius h around the red point.
The left end of the spring (with no tension) describes a blue semi-circle around its right end or anchoring point.

In this case, the distance "x" that the spring stretches is very close to the horizontal distance between these two points:
1) The original point of spring-bar connection when the bar was in its vertical position and there was no tension on the spring (bar and spring are perpendicular to each other).
2) The point of spring-bar connection when the bar is angled at full amplitude and tension on the spring is maximum (bar and spring form an angle a little less than 90 degrees). Blue semi-circle crosses red one.

However, the effective force in your equation that produces the torque that helps the mass and bar to swing back must be perpendicular to the bar (red vector in the schematic).
That force deviates the angle ##\Theta## from horizontal.
For that reason, you input ##kxcos(\Theta)## in your equation.
Since for that small angle, the value of ##cos(\Theta)## is so close to 1, the difference is very small, as previously explained.

Est120 said:
2.-
what is the lever arm "y" for this it would help me to know what is the direction of the spring force as the red vector or as the black
The lever arm for the red vector of magnitude ##kxcos(\Theta)## is always h for any position of the bar.

:cool:
 
Last edited:
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  • #7
There is so much wrong here that you should start again. In particular the equations are dimensionally inconsistent (and largely illegible).

First draw a free body diagram !
 

Related to Direction of a spring's force in more than one dimension

1. What is the direction of a spring's force in more than one dimension?

The direction of a spring's force in more than one dimension is dependent on the orientation of the spring and the direction of the applied force. In general, the force of a spring will be directed along the line connecting the two points of attachment, also known as the spring's axis.

2. How does the direction of a spring's force change in different dimensions?

In one-dimensional systems, the direction of a spring's force is simply along the axis of the spring. However, in two or three-dimensional systems, the direction of the force can vary depending on the orientation of the spring and the direction of the applied force. This can result in forces acting in multiple directions simultaneously.

3. Can the direction of a spring's force be changed?

Yes, the direction of a spring's force can be changed by altering the orientation of the spring or by changing the direction of the applied force. This can be achieved by rotating or moving the spring or by changing the angle at which the force is applied.

4. How does the direction of a spring's force affect its strength?

The direction of a spring's force does not directly affect its strength. The strength of a spring is determined by its spring constant, which is a measure of its stiffness. The direction of the force only affects the direction in which the spring can exert its strength.

5. Can the direction of a spring's force be calculated?

Yes, the direction of a spring's force can be calculated using vector mathematics. The direction of the force is dependent on the magnitude and direction of the applied force, as well as the orientation of the spring. By using vector addition and trigonometric functions, the direction of the spring's force can be determined.

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