Calculation of Tensions on Strings that suspends an object

In summary, the conversation discusses finding the tensions (T1 and T2) in two strings holding an object of mass m in a steady state. Two methods are proposed, one involving solving equations using horizontal and vertical forces, and the other using the trigonometric relationship T=mgcosθ. The issue of finding the projection of the upward force along T1 is discussed, and it is concluded that this method is more difficult and requires taking into account the component of T2 in that direction. The conversation ends with a word of advice to not round off values too early.
  • #1
Samit Chak
10
0
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
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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  • #3
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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  • #4
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.
 
  • #5
First way meaning you put ##T_2 \cos 45^\circ = T_1 \cos 60^\circ## ? How do you find 29 N from that ?
 
  • #6
Samit Chak said:
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.
 
  • #7
BvU said:
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
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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  • #9
Graphical method and sine law.
 
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  • #10
azizlwl said:
Graphical method and sine law.

ooh! I missed it. Thanks a lot.
 
  • #11
Samit Chak said:
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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  • #12
BvU said:
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 :)
 

Related to Calculation of Tensions on Strings that suspends an object

1. How do you calculate the tension on a string that suspends an object?

To calculate the tension on a string that suspends an object, you need to consider two factors: the weight of the object and the angle of the string. The formula for tension is T = mg / cosθ, where T is the tension, m is the mass of the object, g is the acceleration due to gravity, and θ is the angle of the string with the vertical.

2. Why is it important to calculate the tension on a string that suspends an object?

Calculating the tension on a string that suspends an object is important because it helps determine whether the string is strong enough to hold the weight of the object. If the tension is too high, the string could break, causing the object to fall. If the tension is too low, the object may not stay suspended.

3. How does the angle of the string affect the tension?

The angle of the string affects the tension because it changes the direction of the force acting on the object. As the angle of the string increases, the tension on the string decreases, and vice versa. This is due to the cosine function in the tension formula.

4. What units are used to measure tension?

Tension is typically measured in units of force, such as newtons (N) or pounds (lbs). In the formula T = mg / cosθ, the mass (m) is typically measured in kilograms (kg) and the acceleration due to gravity (g) is measured in meters per second squared (m/s^2).

5. Can the tension on a string be greater than the weight of an object?

Yes, the tension on a string can be greater than the weight of an object. This can happen when the angle of the string is less than 90 degrees. As the angle approaches 90 degrees, the tension on the string becomes infinitely large. However, in most real-life scenarios, the tension on a string is not greater than the weight of an object because the angle of the string is typically less than 90 degrees.

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