String tension question (1 Viewer)

[Karldahemster]

I need help answering this question.

View attachment Tension in strings puzzle.pdf

 

[InteGrand]

I need help answering this question.

View attachment 33998

Let the tensions be T1 and T2, and obtain two simultaneous equations by decomposing forces in the horizontal and vertical direction (using right-angle trigonometry), and using that the acceleration in each of these directions should be 0 (assuming we are in equilibrium).

 

[Karldahemster]

Let the tensions be T1 and T2, and obtain two simultaneous equations by decomposing forces in the horizontal and vertical direction (using right-angle trigonometry), and using that the acceleration in each of these directions should be 0 (assuming we are in equilibrium).

I don't quite follow. What would be the equation?

 

[pikachu975]

Horizontal:
T1 cos30 - T2 cos10 = 0
T1 = T2 cos10/cos30

Vertical:
mg = 10x9.8 = T1 sin30 + T2 sin10
98 = T2 cos10 tan30 + T2 sin10
T2 = 98/(cos10 tan30 + sin10)
T2 = 132.035 N (tension in string B)

T1 = 132.035 cos10/cos30
T1 = 150.145 N (tension in string A)

 

[Karldahemster]

Horizontal:
T1 cos30 - T2 cos10 = 0
T1 = T2 cos10/cos30

Vertical:
mg = 10x9.8 = T1 sin30 + T2 sin10
98 = T2 cos10 tan30 + T2 sin10
T2 = 98/(cos10 tan30 + sin10)
T2 = 132.035 N (tension in string B)

T1 = 132.035 cos10/cos30
T1 = 150.145 N (tension in string A)

Where did you get the 20 degree angle from? The 10 I understand.

 

[thejimster_73]

There was no 20-degree angle. Or did he edit his post?

Pikachu is correct btw, Karldahemster.

 

[Karldahemster]

yeah he did

 

[pikachu975]

There was no 20-degree angle. Or did he edit his post?

Pikachu is correct btw, Karldahemster.

yeah he did

I had 20 originally but realised it was 30 so I edited it

 

[Karldahemster]

I had 20 originally but realised it was 30 so I edited it

Thank you for your help.

 

[Andy005]

Hi pikachu975, how did you get T2cos10tan30? I'm a bit confused understanding on how to get tan30.

Thanks

 

[jazz519]

Hi pikachu975, how did you get T2cos10tan30? I'm a bit confused understanding on how to get tan30.

Thanks

Horizontal:
T1 cos30 - T2 cos10 = 0
T1 = T2 cos10/cos30

Vertical:
mg = 10x9.8 = T1 sin30 + T2 sin10
98 = T2 cos10 tan30 + T2 sin10

He subbed the expression T1=T2cos10/cos30 into line 1 of the vertical which gives mg = T2cos10/cos30 * sin30 + T2sin10

remember tan x = sin x /cos x, so sin30/cos30 = tan30

 

[Andy005]

Thanks