Answer:
a) a = 2.383 m / s², b) T₂ = 120,617 N
, c) T₃ = 72,957 N
Explanation:
This is an exercise of Newton's second law let's fix a horizontal frame of reference
in this case the mass of the sleds is 30, 20 10 kg from the last to the first, in the first the horizontal force is applied.
a) request the acceleration of the system
we can take the sledges together and write Newton's second law
T = (m₁ + m₂ + m₃) a
a = T / (m₁ + m₂ + m₃)
a = 143 / (10 +20 +30)
a = 2.383 m / s²
b) the tension of the cables we think through cable A between the sledges of 1 and 20 kg
on the sled of m₁ = 10 kg
T - T₂ = m₁ a
in this case T₂ is the cable tension
T₂ = T - m₁ a
T₂ = 143 - 10 2,383
T₂ = 120,617 N
c) The cable tension between the masses of 20 and 30 kg
T₂ - T₃ = m₂ a
T₃ = T₂ -m₂ a
T₃ = 120,617 - 20 2,383
T₃ = 72,957 N
The tension in the cable is 23.2 N
<h3>What is the tension in the string?</h3>
The tension in the cable can be resolved into horizontal and vertical forces Tcosθ and Tsinθ respectively.
Tcosθ, is acting perpendicularly, Tcosθ = 0
Taking moments about the pivot:
Tsinθ * 2.2 = 4 * 9.8 * 0.7
Solving for θ;
θ = tan⁻¹(1.4/2.2) = 32.5°
T = 27.44/(sin 32.5 * 2.2)
T = 23.2 N
In conclusion, the tension in the cable is determined by taking moments about the pivot.
Learn more about moments of forces at: brainly.com/question/23826701
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Formula from physics to get the answer.