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 CounterClockwise 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 BasePart.
Add the Line so that one of its Points is 100mm below the end of the Rocker Part, when the Rocker is horizontal. Step 1 is complete. TOPTIP: Use Tools menu > Show/Hide other Mechanisms to make the Rocker visible (in Grey) when you edit Parts in the PartEditor.


To add the Spring Fb, you must click a Point in two different Parts 

Click the point that is the endPoint of the CADLine in the Rocker and not the centrePoint of the Arc(+). Click the endPoint of the Line in the BasePart that is 100mm below the Rocker The order you select the Points is not important. The Points become the AnchorPoints for the Spring Fb. 

Steps 5,6,7 may not be necessary, as the SpringForce FB should be enabled automatically. If necessary, we must enable the Spring FB We must use the SpringForce FB dialogbox.
If necessary, click the Spring Parameters separator to see its parameters.
The tick should show in the checkbox. Do not edit the Spring or Velocity Factors.


The image to the left shows the Spring FunctionBlock and the Symbol that represents the Spring. The Spring FB does not exert a Force, yet. Thus, MechDesigner does not show Forces at the Spring's anchorpoints.


Constant Force 
We can balance the Gravitational Force if we edit the Constant Force parameter in the Spring Force dialogbox. A positive 'ConstantForce' pulls the Spring's anchorpoints together. A negative 'ContactForce' pushes the Spring's anchorpoints 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): (Counterclockwise +ve) ∑M1=0 ; M2 + 0.2(m)*9.81(N)  0.1(m)*9.81(N) = 0 . Thus M2= 0.981Nm.


SpringRate and FreeLength You can use the SpringRate and FreeLength Parameters to obtain the same result. We will enter a 'FreeLength' and 'SpringRate' so that the Spring exerts a force equal and opposite to the Gravitational Force. Also, edit the Constant Force = 0N again. 

When we make the FreeLength = 110mm and the SpringRate = 0.980665N/mm, the Spring FB will exert 9.80665N upwards. Why? The model defines the actual distance between the Spring FB's two anchorpoints = 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 outputconnectors. From the top outputconnector:
From the bottom outputconnector
One input Connector: A Force Function can be added to the input from a Maths FB. 