Equivalent Force Calculation for Fixed Quadrant Block with Sliding Rope Tensions

  • Thread starter Thread starter kingsgambit
  • Start date Start date
  • Tags Tags
    Forces Rope Slope
Click For Summary

Homework Help Overview

The discussion revolves around calculating the equivalent force acting on a fixed quadrant block with sliding rope tensions, specifically T1 and T2, which differ due to belt friction. The original poster seeks to understand the forces applied to the block and how to determine an equivalent force for further calculations.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the concept of an equivalent force as the vector sum of T1 and T2, questioning how this force interacts with the static equilibrium of the system. The original poster expresses uncertainty about the equivalent force's location and its role in calculating the reaction force on the welded block.

Discussion Status

Some participants have provided insights regarding the vector nature of the equivalent force, while others have shared updates and additional images for verification. The discussion appears to be ongoing, with various interpretations being explored without a clear consensus.

Contextual Notes

The problem involves static equilibrium and the effects of belt friction on rope tensions, which are critical to understanding the forces at play. There may be constraints related to the specific angles and forces that have not been fully defined in the discussion.

kingsgambit
Messages
3
Reaction score
0

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.
 
Last edited by a moderator:
Physics news on Phys.org
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.
 
bumpity bump
 

Similar threads

Replies
2
Views
1K
Replies
46
Views
6K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
Replies
24
Views
3K
Replies
7
Views
2K
Replies
6
Views
12K
  • · Replies 13 ·
Replies
13
Views
10K
Replies
6
Views
15K
  • · Replies 3 ·
Replies
3
Views
6K