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A weightless bar is pinned at the top of a triangle with N weight. Both rests at the horizontal plane.

There is a horizontal force F acting on the weightless bar, and a given coefficient of friction μ for both the triangle and the bar with the horizontal plane.

Should a normal and friction force be applied at the base of the weightless bar? Or should the reaction only be at the pinned joint?

Or what other reactions should there be, other than the normal and friction force on the triangle.

enter image description here

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  • $\begingroup$ Please define 'weightless bar'. In the real world, there is no 'weightless bar'. For more information, see: en.wikipedia.org/wiki/Wriction and in Dutch nl.wikipedia.org/wiki/Zelfremmigheid I gave some explanation. $\endgroup$ Jan 15 at 6:47
  • $\begingroup$ Is the triangle attached to the horizontal plane, or is there friction? Also what is the height of the Force wrt to the horizontal plane? $\endgroup$
    – NMech
    Jan 15 at 20:24
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Lets call

  • length of the inlined bar=L

  • Y of force = y

  • base of triangl= b

  • height of triangle=h

  • angle of bar= a

There will be a normal force $F*y/h*cos(a)$

and a horizontal force $F-F*y/h*cos(a)* \mu - N\mu $.

But the system will accelerate if

If $F_{ef}>N*\mu_{triangle}$

The triangle will topple if $F_{ef}*y>N*b/2 $

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UPDATE (I think I misunderstood the constraints in my previous answer)

Assuming

  • the greyed out triangle beneath B is rigid and its resting on the horizontal plane with a coefficient of friction $\mu$
  • the beam is pinned on B and simply resting on A.
  • The force F is large enough, so that the triangle is about to move/rotate.

Then it depends on the coefficient of friction. If the coefficent of friction is:

  • $\mu>\frac{W}{2H}$: then the triangle will tip around D
  • $\mu<\frac{W}{2H}$: then the triangle will translate horizontally

where:

  • H is the height of the triangle ($L_{BE}$)
  • W is the breadth of the triangle at the base ($2\cdot L_{DE}$)

High coefficient of friction $\mu>\frac{W}{2H}$: Tipping

In the first case ($\mu>\frac{W}{2H}$) the forces on the beam are presented in the following image:

enter image description here

  • The red are reaction forces (perpendicular to contact surfaces).
  • The green arrows correspond to friction forces and are perpendicular to the corresponding reaction forces.

The reaction on the triangle is at point D, because the force F will create a turning moment which will tend to overturn the triangle. The result is that the triangle will pivot around point D (assuming the triangle is rigid).

If the triangle is deformable, then things are more complicated, and the friction should be distributed along the bottom of the triangle in a non uniform way .

Low coef. of friction $\mu<\frac{W}{2H}$ : Sliding

In the second case the friction force will not be enough to tip the triangle.

Forces on edges of the Beam

You can calculate:

  • from equilibrium of moments about B that:

$$N_A = \frac{(L_{AB}-L_{AF})\sin\phi}{L_{AB}\cdot (\cos\phi- \mu \sin\phi)}F$$

where:

  • $L_{AB}$ is the length between points A and B
  • $L_{AF}$ is the length between points A and F

- from equilibrium of forces on X:

$$B_x = F - \mu N_A= \left(1-\mu\frac{L_{AB}-L_{AF})\sin\phi}{L_{AB}\cdot (\cos\phi- \mu \sin\phi)}\right)\cdot F$$

- from equilibrium of forces on Y:

$$B_y = - N_A= -\frac{(L_{AB}-L_{AF})\sin\phi}{L_{AB}\cdot (\cos\phi- \mu \sin\phi)}\cdot F$$

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