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.

STEP 1: Edit the Base-Part to add a Line  

1.Add the Line so that one of its Points is 100mm below the end of the Rocker Part, when the Rocker is horizontal. Close the Part-Editor

TOP-TIP: You can use Tools menu > Show/Hide other Kinematic and Sketch-Elements to see the Rocker (in Grey) when you edit other Parts in the Part-Editor.

STEP 2: Click Force Toolbar > Add Spring FB

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

STEP 3: Click two Points – as indicated to the left.

Click the point that is the end-Point of the CAD-Line in the Rocker and not the center-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 4,5,6 may not be necessary, as the Spring-Force FB should be enabled automatically.

Enable the Spring FB

We must use the Spring FB dialog-box.

1.Double-click the Spring FB in the graphic-area to open the Spring FB dialog-box

2.Click the Spring Parameters separator to see its parameters.

3.Click the Enable Spring FB check-box

The tick should show in the check-box

Do not edit the Spring or Velocity Factors

4.Click to close the Spring FB dialog-box

The image to the left shows the Spring Function-Block 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 anchor-points.

STEP 8: Double-click the Spring FB in the graphic-area to open the Spring FB dialog-box, again

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.

STEP 9: Enter a Constant Force = -9.80665N in the Spring FB dialog-box to exert 9.80665N upwards against the Rocker Part.

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.

When we make the Free-Length = 110mm and the Spring-Rate = 0.980665N/mm, the Spring FB will exert 9.80665N upwards.

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:

•Distance, linear velocity and linear acceleration between the two Points

From the bottom output-connector

•Total Force, X-Force and Y-Force. The X and Y-Force components are relative to the coordinate axes of the Mechanism-Plane.

One input Connector: A Force Function can be added to the input from a Maths FB.