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

The CAD-Line is at the top. The X-axis is on top of the CAD-Line!

A Line -

•40mm long is parallel to the CAD-Line

•50mm from the CAD-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, a 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, the numbers (1,2,3,4) identify the 4 Points:

•Line 1 (defined by start-Point 1 and end-Point 2) of the Slide-Joint, in the Base-Part: Its length is 100mm. This Line is horizontal. It supports the sliding-Part.

•Line 2 (defined by start-Point 3 and end-Point 4) of the Slide-Joint, in the Sliding-Part: Its length is 40mm(20+20). The Line is 50mm below the Part and CAD-Line.

•Point Mass in the Sliding-Part : (1kg). center-of-Mass: 50mm along the CAD-Line.

See CAD-Line dialog-box | Mass Properties tab > User Mass Properties to add Mass. Edit its center-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 with a Slide Block, that you might use in a machine design.

1, 2, 3, and 4 - in the image - are approximately equivalent to the start-Points and end-Points of the two Lines in the Slide-Joint of the model above.

In your model and design you should make sure that the Points (1,2,3,4) of the Lines in Slide-Joint, have the length equal to the length of the Slide-Rail and Sliding-Block.

1.Click Force toolbar > Calculate Forces so that the icon becomes the same as the icon to the left.

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

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

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

If necessary, change the background color 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.

•There is a Force Vector from the start-Point and end-Point of each Line that defines the Slide-Joint - four(4) Points , and four(4) Force-Vectors

•The four Points that make up a Slide-Joint can ONLY exert a force that is perpendicular(⊥) to o each Line.

•If the sliding-part is a Slider, and it is a Power Source, then there is a Motive Force-Vector in the direction of the Slider.

When the Part moves 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.

•Gravitational-Force: The Slider has a Mass of 1kg. Thus, there is a gravitational force vertically downwards = 9.81N. This force vector is not shown.

•Inertia-Force: It is stationary, or moving at a constant-velocity. Thus the Inertia Force = 0N

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.

•The total force acting upwards by the BasePart on the sliding-Part is 9.81N.

•Taking Moment about 'Point 1': 0.020(m)*4.9(N) + 0.060(m)*4.9(N) - 0.100(m)*R4(N) = 0 ; R4 = 3.92N.

•Summing Vertical Forces: ∑FV: R1 + R2 + R3 + R4 = 0;  R1 = 5.88N

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.

•Gravitational Force, vertical, down: : 1kg*9.81m/s/s = 9.81N Thus there is a gravitational force vertically downwards = 9.81N. This force vector is not shown.

•Inertia Force, horizontal, (-) = 1(kg)*-10(m/s/s) = 10N ()

Add 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

Add Vertical Forces ( ↑+ve)

•∑FV: R3 + R4 - mg(N);  R3 +(-7.595(N)) - 9.81(N) ; R3 = 17.405(N)

Consider the Base-Part.

•The total force acting upwards by the Base-Part on the sliding-Part is 9.81N.

•Taking Moments at Point '1 0.020(m)*17.405(N) - 0.060(m)*7.595(N) - 0.100(m)*R4(N) = 0 ; R4 = -1.076N. (Upwards)

•Add Vertical Forces: ∑FV: R1 + R2 + R3 + R4 = 0; R1 = 10.9N (Downwards).

These force vectors ACT ON the Base-Part.