Step 13.1B: A Slider

A View of the Sliding-Part in the Part-Editor.

The CAD-Line is at the top.

A Line -

The kinematic-chain, to the left, is a Slider. [To help you see the dimensions and Point numbers in the image, I am not showing the Motion-Dimension].

As is the definition of a Slide-Joint, the Slide-Joint is constructed with 2 Lines; a Line in two different Parts. Each Line is defined by its start-Point and its end-Point - 4 Points in total.

In the image, I have added numbers [1,2,3,4] to identify the 4 Points:

•	Point Mass in the Sliding-Part : [1kg]. Centre-of-Mass: 50mm along the CAD-Line.

See CAD-Line dialog-box | Mass Properties tab > User Mass Properties to add Mass. Edit its Centre-of-Gravity along the X-axis to locate the 1kg Mass, as in the image, to the left.

Notes: This is a schematic of a typical Linear Slider Rail and Linear Slide Block that you might use in your design.

The Points 1, 2, 3, and 4 are approximately equivalent to the four Points in the image of the model above.

In your model and design you should make sure that the Points [1,2,3,4] of the Lines that you select for the Slide-Joint, are in equivalent positions to those indicated in the schematic image, to the left.

If a minimum of one Part has Mass, MechDesigner displays Force Vectors. The Force Vectors show the direction and magnitude of each Force.

Note: you may need to click the Vector Scales buttons to increase or decrease the length of the Force and Torque Vector Arrows . These buttons are below the graphic area, in the middle of the Feedback Area.

If necessary, change the background colour of the graphic area to a Dark-Grey so that you can see the magnitude of each vector next to the arrowhead of each Force Vector.

When the Part is moving at Constant-Velocity, then inertia-forces are not present. The Part is not rotating, and thus there is no Coriolis or Centrifugal Force.

Sliding-Part.

The Gravitational Force must be 'balanced' [put into equilibrium] by reaction forces that act on the sliding-part.

In the image, you can see that there are two, upwards acting force-vectors, each equal to 4.90N,that ACT ON the sliding-part, by the Base-Part. At Points 3 and 4]

These two Force Vectors are equal because of the symmetry of the mass with the short 40mm line in the sliding-part, and that the Slide-Joint is horizontal.

Base-Part.

Refer to the dimensions and point numbers in the image at the top of this topic.

These Force Vectors ACT ON the Base-Part, by the Sliding-Part.

Note: Motion name for this video is CV.MTD

The Lines in the Slide-Joint should be thought of as a Linear Slide Rail and a Linear Slide Block, as you would purchase from THK

The Force Vectors on the Base-Part [Machine-Frame] as a sliding 'Block' moves along it with Constant-Velocity.

In reality, the forces would be distributed along the Lines in the two Parts.

However, in the forces vectors are at the Points at each end of each Line in the Slide-Joint.

Note: Download the motions [see top of this topic], set the Edit menu > Machine Settings > Cycling Parameters > Cycles/Min to 300. Add a Motion FB for the Slider and link ConstAcc to the Motion FB

The acceleration of the sliding-part to the Right is 10m/s/s [Taking Acceleration to the Right [→] as Positive]

Consider the Sliding-Part.

Summing Horizontal Forces

∑FH : inertia 10N[←] + Motive-Force = 0,. Thus, Motive-Force = -Inertia 10N, or 10N [→].

These force vectors must be 'balanced' by the reaction force acting on the sliding-part.

Taking Moments about Point 3: [Counter-clockwise +ve] [see image at top for point position]

∑M3= 0 ; [0.050[m]*10[N]] - [0.020[m]*1[kg]*9.81[m/s/s]] - [0.04[m]*R4[N] = 0 ; R4 = 7.595N

Summing Vertical Forces [ ↑+ve]

Consider the Base-Part.

These force vectors ACT ON the Base-Part.