Equivalent Force Calculation for Fixed Quadrant Block with Sliding Rope Tensions

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The discussion focuses on calculating the forces acting on a fixed quadrant block with sliding rope tensions T1 and T2, which differ due to belt friction while the system remains in static equilibrium. An equivalent force is proposed to act at the center of the quadrant, represented as the vector sum of T1 and T2, which should align with their intersection point. The user seeks confirmation on their equations for vertical and horizontal forces, relating normal force (N), weight (W), and reaction forces (R1 and R2). The calculations aim to determine the reaction force on the welded block based on these tensions. Overall, the thread emphasizes the need for clarity in deriving the equivalent force and its implications for the system's stability.
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Homework Statement



http://server7.pictiger.com/img/480530/other/forces.png

http://server7.pictiger.com/img/480530/other/forces.png

I would like to know the forces applied on the quadrant block as shown in the pic. Is there an equivalent force I can take for the whole quad? The block is fixed and allows the rope to slide freely. T1 and T2 (rope tensions) are different due to belt friction, but system is in static equilibrium.


I think there is an equivalent force that acts at the centre of the quad. I need this force to calculate reaction force on welded block. I'm not sure about this though.
 
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The "equivalent" force is the vector T1 + T2, and it must pass through the intersection of T1 and T2.
 
update needs checking

http://img502.imageshack.us/my.php?image=forcesax4.png

Is the following correct? Please refer to image link.

Vertical : N*cos(A1) + W = R2
Horizontal: N*sin(a1) = R1



T1 and T2 are taken into account by N.
 
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