Ladder against wall. (If you help me, you are a legend).

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The discussion revolves around calculating the tension in a rope supporting a ladder resting against a wall. The ladder's dimensions and the forces acting on it, including weight and tension, are analyzed using moments about points A and B. The confusion arises regarding the presence of two terms involving theta in the tension equation, specifically the -6sin(theta) component. It is clarified that the normal force at point B must balance both the weight of the ladder and the vertical component of the tension, leading to the correct formulation of the tension equation. Understanding these forces and their interactions is crucial for solving the problem accurately.
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Homework Statement



A uniform ladder AB, of weight W and length 2.5m rests against a smooth vertical wall OA with its foot on smooth horizontal ground OB. The ladder is in a vertical plane perpendicular to the wall. It is kept in position with OA=2m and OB=1.5m by a light rope OC joining O to a point C on the ladder such that angle COB=theta. Show that the tension T in the rope is given by

T=(3W)/(8cos(theta)-6sin(theta))

Homework Equations



This is just a moments question.

The Attempt at a Solution



I've tried taking moments about B, but don't seem to get why there are two terms in theta in that equation.
 
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I'm really not sure how to proceed. If I take moments about B, there seem to be two forces I need to include: the tension in the rope, and the weight force. Am I correct?
 
Use the moment about A.
 
When I try moment about A,
Clockwise moment is due to the tension, with value moment=2Tcostheta. as well as weight.
Anticlockwise moment is due to the normal force at B (which is twice the weight force), hence moment here is 0.75W.
Then solving I get T=3W/(8costheta). So where does the -6sintheta come from?
 
Does the problem state anything about the location of point C? (I doubt that it does.)
 
Nope, C can vary.
 
xduckksx said:
When I try moment about A,
Clockwise moment is due to the tension, with value moment=2Tcostheta. as well as weight.
Anticlockwise moment is due to the normal force at B (which is twice the weight force), hence moment here is 0.75W.
Then solving I get T=3W/(8costheta). So where does the -6sintheta come from?
Why do you say that the normal force is twice the weight force?
 
I'm not sure. Should it be the same?

And can you explain where that -6sin theta comes from?
 
  • #10
A component of T is vertical. The normal force at B must cancel both the force of gravity and the vertical component of T.
 
  • #11
Isn't this accounted for in calculating the clockwise moment of the overall tension?

Oh, I see. The normal force at B also takes into account this vertical component. FML.
 
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