D asked in Science & MathematicsPhysics · 4 weeks 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.

Relevance
• NCS
Lv 7
4 weeks ago

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!

• Whome
Lv 7
4 weeks 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