How Do You Calculate the Tension in Cords with Varying Angles and Loads?

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SUMMARY

The discussion focuses on calculating the tension in three cords supporting loads in an elevator moving at constant velocity. The angles provided are θ1 = 40°, θ2 = 50°, and θ3 = 62°, with masses m1 = 3 kg and m2 = 6 kg. The key equation used is F = mg(sin θ), but participants emphasize the importance of considering the y-components of the tensions to ensure equilibrium. The correct approach involves setting T1(sin 62) equal to T3 to solve for T1, and using T2 = T1(cos 28) to find T2, confirming that sin 28 is the correct trigonometric function for the angle in question.

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Homework Statement



Figure P4.24 shows loads hanging from the ceiling of an elevator that is moving at constant velocity. Find the tension in each of the three strands of cord supporting each load, given that θ1 = 40°, θ2 = 50°, θ3 = 62°, m1 = 3 kg, and m2 = 6 kg.

p4-24alt.gif


Homework Equations



F = mg(sin theta)

The Attempt at a Solution



it asks for the answers for all of the T's but i found it for T1 - 3 for (a), and T3 for (b). I tried doing T1 = 6(9.8)(sin62) for part a, image B but it told me i have the wrong answer. and I'm completely lost on T2 for part b.

any suggestions?
 

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anteaters said:

Homework Statement



Figure P4.24 shows loads hanging from the ceiling of an elevator that is moving at constant velocity. Find the tension in each of the three strands of cord supporting each load, given that θ1 = 40°, θ2 = 50°, θ3 = 62°, m1 = 3 kg, and m2 = 6 kg.

p4-24alt.gif


Homework Equations



F = mg(sin theta)

The Attempt at a Solution



it asks for the answers for all of the T's but i found it for T1 - 3 for (a), and T3 for (b). I tried doing T1 = 6(9.8)(sin62)

This is close, but notice that it would make T1 be smaller than T3, which can't be true here.

Instead, you want the y component of T1 to cancel out the y component of T3 (since T2 has no y component). What are the y components of those two strings?
 
okay i was looking through my book and realized what to do. i was just plugging and chugging with numbers instead of looking at the problem and actually solving it. so for T1 i needed to realize that m(a) = 0 because the acceleration of the system = 0. so the only other force in the y direction was T3. so i found T1(sin 62) - T3 = 0. then i solved for T1 by doing 58.8/sin 62 = 66.59...
now for T2, would i set it up as T2 - T1(cos 28) = 0? i got 28 because in the x direction the angle would be 28 degrees below the axis in a negative direction. or would it be sin 28? any help would be appreciated
 
Last edited:
nevermind, i figured it out. it was sin 28. thanks for your time alphysicist.
 

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