<< 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 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. STEP 1: Edit the Base-Part to add a Line
TOP-TIP: You can use Tools menu > Show other Kinematic and Sketch-Elements to see the Rocker when you edit other Parts in the Part-Editor. |
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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 Anchor-Points for the Spring Fb. |
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STEP 3: Enable the Spring FB
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The image to the left shows the Spring Function-Block The Spring FB does not exert a Force, yet. Thus, MechDesigner does not show Force Vecotrs at the Spring's anchor-points. Constant Force We can balance the Gravitational Force if we enter a Constant Force in the Spring Force dialog-box. A positive Constant-Force pulls the Spring's anchor-points. (o>>>+<<<o) A negative Contact-Force pushes the Spring's anchor-points (o<<<–>>>o) STEP 4: Edit the Spring FB to enter a Constant-Force= -9.80665N
-9.80665N exerts a Force upwards against the Rocker. |
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![]() -9.8066N |
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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 (Counter-clockwise +ve) ∑M1=0 ; M2 + 0.2(m)*9.81(N) - 0.1(m)*9.81(N) = 0 . Thus M2= -0.981Nm.
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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. |
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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 anchor-points 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 |
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Note: The Spring FB has: Two output-connectors : •Top output-connector = Distance, velocity and acceleration of one anchor-point relative to the other. •Bottom output-connector =Total Force, X-Force and Y-Force (relative to the coordinate axes of the Mechanism-Plane). One input Connector : Optionally, a Force Function at its input-connector from a Math FB. The Output Data-Type of the Math FB must be set to Force. |