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marissa [1.9K]
3 years ago
9

Under what conditions would a rope remain in equilibrium during a tug of war

Physics
2 answers:
Andrei [34K]3 years ago
7 0
<h3>Answer;</h3>

B)  When the net force acting on the rope is zero.  

<h3><u>Explanation;</u></h3>
  • <u><em>Equilibrium refers to a state of balance where there is no net force. The forces acting at point in equilibrium are opposite and equal.</em></u>
  • <u><em>Therefore, for a body at equilibrium the vector sum of all the forces acting on that body must be zero and also the vector sum of torques on the body must be zero</em></u>.
  • <em><u>In a tug of war for example, the equilibrium will be achieved if the two teams involved on either side apply equal forces in opposite direction, such that there will be no net force on the rope, that is the net force on the rope is zero.</u></em>
miskamm [114]3 years ago
3 0
When the teams on each end of the rope exert exactly the
same force ... in opposite directions ... the net force on the
rope is zero, and it doesn't accelerate in either direction.
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PART ONE
stepladder [879]

Answer:

1. A1, B2, C3

2. 47.1°

Explanation:

Sum of forces in the x direction:

∑Fₓ = ma

f − Fᵥᵥ = 0

f = Fᵥᵥ

Sum of forces in the y direction:

∑Fᵧ = ma

N − W = 0

N = W

Sum of moments about the base of the ladder:

∑τ = Iα

Fᵥᵥ h − W (b/2) = 0

Fᵥᵥ h = ½ W b

Fᵥᵥ (l sin θ) = ½ W (l cos θ)

l Fᵥᵥ sin θ = ½ l W cos θ

The correct set of equations is A1, B2, C3.

At the smallest angle θ, f = Nμ.  Substituting into the first equation, we get:

Nμ = Fᵥᵥ

Substituting the second equation into this equation, we get:

Wμ = Fᵥᵥ

Substituting this into the third equation, we get:

l (Wμ) sin θ = ½ l W cos θ

μ sin θ = ½ cos θ

tan θ = 1 / (2μ)

θ = atan(1 / (2μ))

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A car going initially with a velocity 13.5 m/s accelerates at a rate of 1.9 m/s for 6.2 s. It then accelerates at a rate of-1.2
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Answer:

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    Maximum speed = 25.28 m/s

b) We have maximum speed = 25.28 m/s, then it decelerates 1.2 m/s² until it stops.

         v = u + at  

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3 years ago
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