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

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SUMMARY

The discussion focuses on the mechanics of a uniform ladder AB, weighing W and measuring 2.5m in length, positioned against a smooth vertical wall OA at a height of 2m and a distance of 1.5m from the wall. The tension T in the supporting rope OC is derived from the equilibrium of moments, leading to the formula T=(3W)/(8cos(theta)-6sin(theta)). Participants clarify the contributions of the normal force and the tension's vertical component, emphasizing the importance of accurate moment calculations around points A and B.

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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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Show what you have tried, and where you are stuck, so we can help you.
 
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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