D asked in Science & MathematicsPhysics · 1 month ago

Physics Suspended Mass?

The image above shows a situation where a mass is suspended by two cables. The mass of the item is 44 kg. The horizontal distance between the anchor points is 14 meters. Angle theta is 35 degrees and angle phi is 31 degrees as measured from the horizontal. The dimension h is 5.412 meters.

What is the magnitude of the force labeled T1 in Newtons?

What is the magnitude of the force labeled T2 in Newtons?

Find the value of the height H in meters.

Find the value of the distance x in meters.

Attachment image

2 Answers

Relevance
  • NCS
    Lv 7
    1 month ago
    Favourite answer

    horizontal equilibrium:

    T₁*cos31º = T₂*cos35º

    T₂ = T₁*1.0464

    vertical equilibrium, substituting for T₂

    T₁*sin31º + (T₁*1.0464)*sin35º = 44kg * 9.8m/s²

    T₁ = 387 N ≈ 390 N ◄

    T₂ = T₁*1.0464 = 405 N ≈ 400 N ◄

    x = h / tanφ = 5.412m / tan31º = 9.007 m ≈ 9 m ◄

    H = (D-x)*tanΘ = (14 - 9.007)m * tan35º = 3.5 m ◄

    Hope this helps!

    Source(s): Please revisit this working: https://answers.yahoo.com/question/index?qid=20201...
  • Whome
    Lv 7
    1 month ago

    Statics, so forces in any direction must sum to zero.

    Horizontal analysis

    Τ₂cosθ - T₁cosφ = 0

    Τ₂ = T₁cosφ/cosθ

    Vertical analysis

    Τ₂sinθ + T₁sinφ - mg = 0

    (T₁cosφ/cosθ)sinθ + T₁sinφ = mg

    T₁(cosφtanθ + sinφ) = mg

    T₁ = mg / (cosφtanθ + sinφ)

    T₁ = 44(9.8) / (cos31tan35 + sin31)

    T₁ = 386.645... ≈ 390 N                ANSWER

    Τ₂ = T₁cosφ/cosθ

    Τ₂ = 386.645cos31/cos35

    Τ₂ = 404.589... ≈ 400 N                ANSWER

    tanφ = h/x

    x = 5.412/tan31 = 9.007 m          ANSWER

    tanθ = H / (14 - 9.007)

    H = 4.993tan35 = 3.496 m          ANSWER

Still have questions? Get answers by asking now.