Calculation of Tensions on Strings that suspends an object

  • Thread starter Samit Chak
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Suppose an object of mass m is hung in a steady state by two strings. One string makes an angle 60 deg. at the point of contact with the object with the horizontal line. The other string makes 45 deg.
To find the tensions in these strings (T1 and T2) we take all Horizontal Forces or their horizontal components and equate it to 0. Same thing repeated for Vertical. Then solve the equations.
But if I consider that there is a net force in upward direction with magnitude mg which is countering gravitational force. Then can't we find the T1 and T2 as T1 = mg cos 30 and T2=mg cos 45 ? I am getting different results with these two methods. Can you pls. let me know what could be the issue here?
 

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  • #2
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Hello Samit,
Yes we can :smile: ! If you post what you have thus far, someone will be able to point you to a discrepancy, I am sure !
 
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Thanks for your reply.
A diagram is attached. Now, let us evaluate T1. Given m = 4 Kg and g = 10 m/s^2. If we solve it the first way, we get T1 = 29 N.
But if I take the second method then T1 = mg cos 30 = 34.6 N.
 

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I think I am doing something wrong in the second method....not fully getting hold of it. mg is the vector sum of T1 and T2. Rest I am stuck.
All I am trying to do is to see if I can solve without breaking individual forces in horizontal and vertical components and then follow method 1.
 
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First way meaning you put ##T_2 \cos 45^\circ = T_1 \cos 60^\circ## ? How do you find 29 N from that ?
 
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mg is the vector sum of T1 and T2.
Correct. Now, don't over-hurry: vectors are equal if the components are equal. So write down the equations for the components just like you described in the first post.
 
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First way meaning you put ##T_2 \cos 45^\circ = T_1 \cos 60^\circ## ? How do you find 29 N from that ?
Horizontal Forces are : T1 cos 60 - T2 cos 45 = 0, i.e. 0.5 (T1) - 0.7 (T2) =0
Vertical forces : T1 sin 60 + T2 sin 45 - mg = 0, i.e. 0.86 (T1) + 0.7 (T2) - 40 = 0 where m = 4 and g = 10
Adding the two equations we get 1.36 T1 = 40, so T1 = 29.4

Now, if I think T1 = mg cos 30 (30 degree is the angle T1 makes with Vertical). Calculating: T1 = 4*10*0.86 = 34
I am finding the upwards force's projection along T1. Something is conceptually wrong I think.
 
  • #8
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I think I got the issue. mg cos 30 does not give the value of T1. It gives the projection as shown in the picture.

Apart from method 1 (solving for 2 equations - Horizontal and Vertical Forces) - are there any other method? Using Vectors in any other way?
 

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Graphical method and sine law.
 
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Horizontal Forces are : T1 cos 60 - T2 cos 45 = 0, i.e. 0.5 (T1) - 0.7 (T2) =0
Vertical forces : T1 sin 60 + T2 sin 45 - mg = 0, i.e. 0.86 (T1) + 0.7 (T2) - 40 = 0 where m = 4 and g = 10
Adding the two equations we get 1.36 T1 = 40, so T1 = 29.4 N .
Well done. Now I understand your 29.4. And it comes from using both equations.
Now, if I think T1 = mg cos 30 (30 degree is the angle T1 makes with Vertical). Calculating: T1 = 4*10*0.86 = 34
I am finding the upwards force's projection along T1. Something is conceptually wrong I think .
Yes! What is wrong here is that you are now analyzing in the direction along T1. You can do that (if you want to make things difficult), but then you do have to take into account the component of T2 in that direction as well !

Last word of advice: don't round off ##\ \sqrt {1\over 2}\ ## too early. Same with g = 9.81 m/s2. Unless instructed to do so, of course.

All clear now ? :rolleyes:

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Well done. Now I understand your 29.4. And it comes from using both equations.
Yes! What is wrong here is that you are now analyzing in the direction along T1. You can do that (if you want to make things difficult), but then you do have to take into account the component of T2 in that direction as well !

Last word of advice: don't round off ##\ \sqrt {1\over 2}\ ## too early. Same with g = 9.81 m/s2. Unless instructed to do so, of course.

All clear now ? :rolleyes:

--
All clear! Thanks :)
 

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