Tension in the strings (with diagram)

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SUMMARY

The discussion focuses on calculating the tension in two strings (T1 and T2) in a static equilibrium scenario using the equations of motion. The derived tensions are T1 = 150 N and T2 = 259.8 N, based on the equilibrium conditions for both vertical and horizontal forces. The calculations utilize trigonometric functions for angles of 30° and 60°, ensuring accurate results. A correction was noted regarding the sign in the horizontal force equation, emphasizing the importance of direction in vector analysis.

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AlphaRock
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Homework Statement


Find the tension in each string given the diagram.

Homework Equations



Fx = ma.
Fy = ma.

The Attempt at a Solution



Fy = 0. Because it's at equilibrium.
T1(sin30) + T2(sin60) - mg = 0.
T1 = [mg - T2(sin60)]/(sin30)

Fx = 0.
T1(cos30) + T2(cos60) = 0.
T1 = [-T2(cos60)]/(cos30)

T1 = T1
[mg - T2(sin60)]/(sin30) = [-T2(cos60)]/(cos30)
150/(sin30) = -T2[(cos60)/(cos30)] + T2[(sin60)/(sin30)]
150/(sin30) = T2[[(sin60)/(sin30)] - [(cos60)/(cos30)]]
T2 = 259.8 N

Substitute T2 into
T1 = [-T2(cos60)]/(cos30)
T1 = 150 N.

Therefore,
T1 = 150 N.
T2 = 259.8 N
Is this the right answer?
 

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AlphaRock said:
T1(cos30) + T2(cos60) = 0.
You need a minus sign here. One component is positive and the other is negative.
 

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