How Do You Calculate Tension in Two Ropes Holding a Suspended Object?

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SUMMARY

The discussion focuses on calculating the tension in two ropes holding a suspended object with a mass of 5 kg. The ropes, measuring 3 m and 5 m in length, make angles of 52 degrees and 40 degrees with the horizontal. Participants emphasize the importance of separating horizontal and vertical components of tension using equations derived from Newton's second law, specifically ATsin(52) + BTsin(40) = mg for vertical forces and ATcos(52) + BTcos(40) = 0 for horizontal forces. The length of the ropes and the height difference of the ceiling are deemed irrelevant to the calculation of tension.

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Homework Statement


Ropes 3m and 5m in length are fastened to an object that is suspended over a cieling. the object has a mass of 5kg. the ropes, fastened at different heights, make angles of 52 degree and 40 degree with the horizontal. Find the tension in each wire, and the magnitude of each tention.

Homework Equations


The Attempt at a Solution


I kno that the tension on one rope, A and the tension on the other rope, B equal to the mass times gravity.
so AT+BT = mg
and i want to find the a equation for AT and BT

Just wondering, since i kno the angle and length of the rope, i know the vertical component of the rope. But then i don't kno where to go after that
 
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What you need to do is draw a force diagram and write down all the horizontal and vertical components of tension as a result of the mass.

Next, you need to use Newton's second Law and find the net force in both horizontal and vertical directions.

Just remember that net force (vector sum of components) = ma, and if the object is not moving, then the net force is zero.
 
Hi k1point618! :smile:
k1point618 said:
so AT+BT = mg

No, that's wrong … that only works if both ropes are vertical.

As jaseh86 says, you must treat the horizontal and vertical components separately.

The vertical ones will add to mg, and the horizontal ones to … ? :smile:
 
Horizontal adds to Zero!

Well, then i have
ATsin52 + BTsin40 = mg, and also


ATcos52 + BTcos40 = 0 ?
 
k1point618 said:
Horizontal adds to Zero!

Well, then i have
ATsin52 + BTsin40 = mg, and also


ATcos52 + BTcos40 = 0 ?

Yes! :smile:
 
So i was wondering, why does the problem give the length of the ropes? and how does the high difference of the cieling affects the problem?
 
k1point618 said:
So i was wondering, why does the problem give the length of the ropes? and how does the high difference of the cieling affects the problem?
k1point618 said:
Ropes 3m and 5m in length are fastened to an object that is suspended over a cieling. the object has a mass of 5kg. the ropes, fastened at different heights, make angles of 52 degree and 40 degree with the horizontal. Find the tension in each wire, and the magnitude of each tention.

hmm … it's a weird question, because you suspend things under a ceiling, not over it. :confused:

Are you sure you have given us the full correct question?
 
Sorry, Over a ceiling, but the ropes are fastened at different heights.
 
k1point618 said:
So i was wondering, why does the problem give the length of the ropes? and how does the high difference of the cieling affects the problem?
k1point618 said:
Sorry, Over a ceiling, but the ropes are fastened at different heights.

Then I've no idea why they give you that information. :redface:
 
  • #10
so the different hieghts of the ceiling does not matter to the problem?
 
  • #11
k1point618 said:
so the different hieghts of the ceiling does not matter to the problem?

No!
 
  • #12
K, THANK YOU!:smile:
 

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