<< Click to Display Table of Contents >> Navigation: Getting Started Tutorials  MechDesigner > Tutorial 13: Forces: Introduction > Step 13.2: Rocker and a Spring 
In this topic, we balance the forces that act on the Rocker with a Spring.
•Downward Force and Clockwise Moment on the Rocker that result from Gravitational Force
is 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. STEP 1: Edit the BasePart to add a Line
TOPTIP: You can use Tools menu > Show other Kinematic and SketchElements to see the Rocker when you edit other Parts in the PartEditor. 

STEP 2: Click Force Toolbar > Add Spring FB To add the Spring FB you must click a Point in two different Parts
The Points become the AnchorPoints for the Spring Fb. 

STEP 3: Enable the Spring FB


The image to the left shows the Spring FunctionBlock and the Symbol that represents the Spring FB. The Spring FB does not exert a Force, yet. Thus, MechDesigner does not show Force Vecotrs at the Spring's anchorpoints. Constant Force We can balance the Gravitational Force if we enter a Constant Force in the Spring Force dialogbox. A positive ConstantForce pulls the Spring's anchorpoints. (o>>>+<<<o) A negative ContactForce pushes the Spring's anchorpoints (o<<<–>>>o) STEP 4: Edit the Spring FB to enter a ConstantForce= 9.80665N
9.80665N exerts a Force upwards against the Rocker. 

9.8066N 

Addition of Vertical Forces acting on the Rocker (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 Rotational Center of the Rocker (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 anchorpoints of the Spring FB as 100mm. Therefore, the Spring FB is deflected by the model by 10mm Therefore, the Spring FB exerts a Force = 10mm * 0.98066N/mm = 9.81N upwards 

Note: The Spring FB has: Two outputconnectors : •Top outputconnector = Distance, velocity and acceleration of one anchorpoint relative to the other. •Bottom outputconnector =Total Force, XForce and YForce (relative to the coordinate axes of the MechanismPlane). One input Connector : Optionally, a Force Function at its inputconnector from a Math FB. The Output DataType of the Math FB must be set to Force. 