What is the minimum force required to keep two cables taut at a given point?

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To keep two cables taut at a point, the force P must be greater than zero to ensure the horizontal component of tension is maintained. If P is too small, the weight will pull the connection point A toward the wall, while if P is too large, it will pull A upward. The discussion emphasizes resolving forces into x and y components to determine the conditions under which the tension in the cables remains positive. Understanding these principles is crucial for analyzing the system effectively. Engaging in such discussions can enhance knowledge and problem-solving skills in physics.
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The cables are tied at A and loaed as shown
Determine the range of values of P for which both cables remain taut !
http://www.freeimagehosting.net/uploads/27fcfeba9a.jpg

I have an idea on the AC rope. Since the tension force T must not be zero on the horizontal line (or the line along AC), therefore, the min force of P > 0. Because when P>0, the horizontal component is not zero. Is it true? How about AB
 
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Welcome to PF!

Hi fociboy ! Welcome to PF! :smile:

If P is too small, the 960N weight will pull A toward the wall.

If P is too large, it will pull A upward.

Hint: the question is really asking you find the formulas for the tension in each cable.

Then you choose the values for which the tension in either cable is zero. :smile:
 
Dear tiny-tim,
I've known how to solve it. We resolve all the force into x and y component. And determine the conditions that the tension force on the rope is more than zero
 
Hi fociboy! :smile:
fociboy said:
Dear tiny-tim,
I've known how to solve it. We resolve all the force into x and y component. And determine the conditions that the tension force on the rope is more than zero

(you mean I know how to solve it, or I've found how to solve it :wink:)

That's right … resolve into components. :smile:

Is everything ok now?
 
I see. But you know, I've started studying in university and there're still a lot of things to know. I hope I can join and become an active member of the forum. Thanks :)
 
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