This topic shows how Force-Vectors are calculated and displayed for a Slider.

A Slider = Part (sliding-part) + Slide-Joint + Motion-Dimension FB.

Note: Motions for this Tutorial Step: Download the ZIP file, extract the MTD file for this tutorial. Use MotionDesigner > Open and Append. Motions for Tutorial13-1B. to load the motions into MotionDesigner.

The Slider


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

The CAD-Line is at the top.

A Line -

40mm long is parallel to the CAD-Line
is 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 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:

Line 1 (defined by Point 1 and 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 Point 3 and Point 4) of the Slide-Joint, in the Sliding-Part: Its length is 40mm(20+20). The Line is offset by 50mm from the Part's CAD-Line.
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.


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


2.Click Force toolbar > Display Forces 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 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.

The 'Force Vectors' start each of the four Points that define the two Lines for the Slide-Joint.
The four Points that make up a Slide-Joint can ONLY exert a force that is perpendicular(⊥) to the Lines.
If the sliding-part is a Slider(sliding-part+Motion-Dimension FB), and it is a Power Source, then it can exert a 'Motive Force' that acts along the axis of the Slide-Joint.

CASE 1: Stationary or Constant Velocity Slider


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.


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.


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.


Case 2: Forces on Accelerating Slider.


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 ()

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)

∑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 Moment about Point '1' 0.020(m)*17.405(N) - 0.060(m)*7.595(N) - 0.100(m)*R4(N) = 0 ; R4 = -1.076N. (Upwards)
Summing Vertical Forces: ∑FV: R1 + R2 + R3 + R4 = 0; R1 = 10.9N (Downwards).

These force vectors ACT ON the Base-Part.

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