What is the Tension in Each String for a Hanging Ball?

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The discussion centers on calculating the tension in two strings supporting a hanging ball with a mass of 0.850 kg, where String 1 forms a 30-degree angle and String 2 a 45-degree angle with the ceiling. The user initially calculated the tensions using incorrect angles and equations, leading to an incorrect tension value for String 2. Other participants pointed out the need to correctly apply trigonometric identities and equilibrium equations, emphasizing that the weight of the ball should be factored correctly in the equations. They suggested correcting the angle used for String 1 and re-evaluating the equations for equilibrium. The conversation highlights the importance of accurately setting up the problem to find the correct tension values.
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Homework Statement


A ball of mass 0.850 kg is dangling as shown. The angle created with String 1 and ceiling is 30 degrees. The angle created with the second and the ceiling is 45 degrees. What is the tension in each string?


Homework Equations


F=ma


The Attempt at a Solution


T1=tension in string 1
T2=tension in string 2
W=weight of ball

I determined the the x and y acceleration to be zero. T1y+T2y=0. and T1x+T2x+W=0. i then used trigonometric identities.

T1y=T1sin(120)
T2y=T2sin(45)

T1x=T1cos(120)
T2x=T2cos(45)
W=8.33N

i then did substitution with the formulas and i found the tension in string 1 to be 6.10 i know this to be correct. and substituting in for the second tension i get 4.31 but i am being told that this is wrong so what exactly am i doing wrong?
 
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dolerka said:
T1=tension in string 1
T2=tension in string 2
W=weight of ball

I determined the the x and y acceleration to be zero. T1y+T2y=0. and T1x+T2x+W=0. i then used trigonometric identities.

T1y=T1sin(120)
T2y=T2sin(45)

T1x=T1cos(120)
T2x=T2cos(45)
W=8.33N

i then did substitution with the formulas and i found the tension in string 1 to be 6.10 i know this to be correct. and substituting in for the second tension i get 4.31 but i am being told that this is wrong so what exactly am i doing wrong?

Suppose, for an experiment, the ball was substituted with one of twice the weight. What would be the tension in the strings?
 
welcome to pf!

hi dolerka! welcome to pf! :smile:

(have a degree: ° and try using the X2 icon just above the Reply box :wink:)
dolerka said:
I determined the the x and y acceleration to be zero. T1y+T2y=0. and T1x+T2x+W=0. i then used trigonometric identities.

T1y=T1sin(120)
T2y=T2sin(45)

T1x=T1cos(120)
T2x=T2cos(45)
W=8.33N

if y is up, then your W is in the wrong equation :redface:

and whyever are you using 120° ? :confused:
 


tiny-tim said:
hi dolerka! welcome to pf! :smile:

(have a degree: ° and try using the X2 icon just above the Reply box :wink:)if y is up, then your W is in the wrong equation :redface:

and whyever are you using 120° ? :confused:


bleh yes first time post. and the equation read T1y+T2y=0 should be T1x+T2x=0. and T1x+T2x+W=0 should be T1y+T2y+W=0. also i used 120o only because my book gives an example of a similar problem and also does this but that example on solved for one side.
 
Last edited:
the system is in equilibrium.
draw components of both the tensions...we get
T2sin30+T1sin45=ma...1
T2cos30=T1cos 45 ...2
solve de 2 eq. to get T1 and t2...please correct me...if i m wrong...i m new here!
 
welcome to pf!

hi Kartikc! welcome to pf! :wink:

(try using the X2 icon just above the Reply box :wink:)
Kartikc said:
the system is in equilibrium.
draw components of both the tensions...we get
T2sin30+T1sin45=ma...1
T2cos30=T1cos 45 ...2
solve de 2 eq. to get T1 and t2...

yes, that looks fine :smile:

(with "mg" of course)
 

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