In this topic, we balance the forces that act on the Rocker with a Spring.
• | The Downward Force and Clockwise Moment on the Rocker that result from Gravitational Force |
...will be balanced by...
• | An Upward Force and Counter-Clockwise Moment on the Rocker from a Spring |
The Rocker is stationary.
A Spring FB exerts a force between two Points in two different Parts. In this case, a Point in the Rocker and a Point in the Base-Part.
Add the Line so that one of its Points Step 1 is complete. TOP-TIP: Use Tools menu > Show/Hide other Mechanisms to make the Rocker visible (in Grey) when you edit Parts in the Part-Editor.
|
|||||||
To add the Spring Fb, you must click a Point in two different Parts |
|||||||
Click the point that is the end-Point of the CAD-Line in the Rocker and not the centre-Point of the Arc(+). Click the end-Point of the Line in the Base-Part that is 100mm below the Rocker The order you select the Points is not important. The Points become the Anchor-Points for the Spring Fb. |
|||||||
Steps 5,6,7 may not be necessary, as the Spring-Force FB should be enabled automatically. If necessary, we must enable the Spring FB We must use the Spring-Force FB dialog-box.
If necessary, click the Spring Parameters separator to see its parameters.
The tick should show in the check-box. Do not edit the Spring or Velocity Factors.
|
|||||||
The image to the left shows the Spring Function-Block The Spring FB does not exert a Force, yet. Thus, MechDesigner does not show Forces at the Spring's anchor-points.
|
|||||||
Constant Force |
We can balance the Gravitational Force if we edit the Constant Force parameter in the Spring Force dialog-box. A positive 'Constant-Force' pulls the Spring's anchor-points together. A negative 'Contact-Force' pushes the Spring's anchor-points apart.
|
||||||
Addition of Vertical Forces acting on the Rocker (Point 2) : (upwards +ve). ∑FV=0 : R2(N) -1(kg)*9.807(m/s/s) + 9.807(N) = 0; R2 = 0N Addition of Horizontal Forces acting on the Rocker (Point 2) : (upwards +ve). ∑FH=0 : R2H(N) = 0; R2H = 0N Take Moments about 1, acting on the Rocker (Point 2): (Counter-clockwise +ve) ∑M1=0 ; M2 + 0.2(m)*9.81(N) - 0.1(m)*9.81(N) = 0 . Thus M2= -0.981Nm.
|
|||||||
Spring-Rate and Free-Length You can use the Spring-Rate and Free-Length Parameters to obtain the same result. We will enter a 'Free-Length' and 'Spring-Rate' so that the Spring exerts a force equal and opposite to the Gravitational Force. Also, edit the Constant Force = 0N again. |
|||||||
Why? The model defines the actual distance between the Spring FB's two anchor-points = 100mm. Thus, the Spring FB is deflected by 10mm Therefore, the Spring FB exerts a Force = 10mm * 0.98066N/mm = 9.81N upwards on the Rocker. |
|||||||
Note: The Spring Force FB has: Two output-connectors. From the top output-connector:
From the bottom output-connector
One input Connector: A Force Function can be added to the input from a Maths FB. |