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Australopithecus

  • 2 years ago

A bag of cement of weight 375 N hangs from three wires as suggested in Figure P5.24. Two of the wires make angles q1 = 55.0° and q2 = 31.0° with the horizontal. If the system is in equilibrium, find the tensions T1, T2, and T3 in the wires. can someone show me how to solve this problem?

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  1. Australopithecus
    • 2 years ago
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  2. Australopithecus
    • 2 years ago
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    It says the answer is: T1 = 322 N, T2 = 216 N, T3 = 375 N but why

  3. edr1c
    • 2 years ago
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    the mass is in equilibrium, so the net force on the mass is zero. so your T3 will be equal to mg, given in question = 375N

  4. edr1c
    • 2 years ago
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    the knot or the point where the 3 wires meet is also in equilibrium, so the summation of force in the horizontal direction will be equal to zero, and also applies for the same for the force in the vertical direction, \[\sum_{}^{}F _{X}=0\]\[\sum_{}^{}F _{Y}=0\]

  5. edr1c
    • 2 years ago
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    so after deriving the horizontal and vertical components you will be able to find the tension on the other 2 wires

  6. Australopithecus
    • 2 years ago
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    how do I do that?

  7. edr1c
    • 2 years ago
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    |dw:1342549591897:dw| 1 of the horizontal components

  8. edr1c
    • 2 years ago
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    bear in mind that the horizontal component for the tension made by the wire with angle 55 deg is in the opposite direction and will be a negative value. after obtaining the horizontal component equation with 2 unknowns T1 and T2, solve simultaneously with the vertical component.

  9. Australopithecus
    • 2 years ago
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    Not to be an ingrate but can you please just show me how to solve this?

  10. edr1c
    • 2 years ago
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    |dw:1342550687656:dw| substitute T2 into Fy

  11. Australopithecus
    • 2 years ago
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    oh I see you use system of equations

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