Step 13.1A: A Stationary Rocker

The kinematic-chain, to the left, is a kinematically-defined chain [ Green Part-Outlines].

It is a Rocker that is horizontal and 200mm long.

The Rocker has a mass. The mass is calculated by MechDesigner from the Extrusion Density, Extrusion Depth [see: Extrusion dialog-box] and the shape of the Profile. The shape of the Profile is specified by the sketch-loop [see Pink Profile, sketch-loop].

To add the Profile and specify its shape:

In addition to its Mass[kg], MechDesigner finds its Moment-of-Inertia[kg.m2] and the position of the Centre-of-Mass relative to the from the Origin of the CAD-Line.

In this case, because the oval shape is symmetrical and equal to the length of the Part, the centre-of-mass is 100mm from the Pin-Joint; the mid-point along the Part.

MechDesigner identifies the centre-of-mass of the Part with a small 'position' symbol.

Forces Toolbar > Calculate Forces . Click the icon so that it becomes the same as the icon to the left.

Visibility toolbar > Display Force Vectors Click the icon so that it becomes the same as the icon to the left.

Force refers to a 'generalised force', which includes: gravitational, inertia, centrifugal, Coriolis, and Spring forces, and also 'moments'.

If a minimum of one Added Part has a mass, MechDesigner shows the Force Vectors. The Force-Vectors give the direction and magnitude of each Force. [Note if the Force Vectors do not show, click the 'Rebuild' tool, and also you may need to Cycle ['C' Key] the model.

If necessary, increase or decrease the length of the Force and Torque Vectors with the Force and Torque Vector Scales buttons.

If necessary, change the background colour of the graphic-area to Grey , or a dark colour, so that it is easier to see the textual force magnitude, at the end of each Force-Vector.

In this force analysis, there are two Points at the Pin-Joint :they are Point 1 on the Base-Part, Point 2 on the Rocker.

Move your mouse-pointer to the Part-Outline of the Rocker Part. You will see that the Force-Vectors that ACT ON the Rocker become RED.

The summation of the Vertical Forces acting on the Rocker [Point 2] : [upwards is +ve].

∑FV=0 : R2[N] - 1[kg]*9.807[m/s/s] = 0; R2 = 9.807N [upwards]

Take Moments about Point 1, acting on the Rocker, Point 2:  [Counter-clockwise is +ve]

∑M1=0 ; M1 - 0.020[m]*1[kg]*9.81[m/s/s] = 0 ; M1 = 0.9807Nm

Gravitational Vector - not shown. There is, however, a vertical gravitational force of [1[kg]*9.807[kg/m/s/s]]=9.81N

 Vertical Force of 9.81N  acts upwards. The Base-Part's reaction force acts on the Rocker. The Rocker would fall freely if the Base-Part did not resist Rocker with the vertical Force.

Counter-Clockwise Moment (Torque) of 0.98Nm.. The Rocker would rotate freely clockwise if the Base-Part did not resist the Rocker with this Torque.

Now move your mouse-pointer to the vertical Force-Vector that acts on the Base-Part. All vectors that ACT ON the Base-Part turn Red.

There is:

Vertical Force 9.81N downwards, that acts on the Base-Part. This is the Gravitational Force that acts-on the Base-Part.

Clockwise Moment (Torque) of 0.98Nm that acts on the Base-Part.

Now, the same Rocker is vertical. [To do this, edit the Base-Value in the Motion-Dimension FB to +90º].

My mouse-pointer is above the Part-Outline of the Rocker, so that the Part-Outline and the Force-Vector have become RED.

You can see that the Vertical Force, 9.807N, is the same as before. This is not a surprise. The Part has the same mass and it is the only joint through which the force can act.

The Moment is now 0Nm. This is because the Centre-of-Mass of the Rocker is vertically above the Pin-Joint. The Rocker 'balances' directly above the Pin-Joint, and hence, a Torque is not necessary to hold the Rocker.

Edit the Motion-Dimension FB again to make the Base-Value 0º.