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Three teams engage in a three-way tug-of-war. Each team is pulling on a rope that is attached to a tire.
The first year the three teams compete each person is of equal strength and therefore pulls with the
same force. The losing team from this competition vows to win the competition the following year. If
the position and strength of the first and second place teams don’t change for the following year’s
competition, how much stronger should the losing team be to win if they don’t change their position
either?

Three Teams Engage In A Threeway Tugofwar Each Team Is Pulling On A Rope That Is Attached To A Tire The First Year The Three Teams Compete Each Person Is Of Equ class=

Sagot :

Ajeigbeibraheemidn

The schematic view of the free-body diagram for the given image in the question is attached in the image below.

From the image;

  • The tension acting on the rope by teams 1, 2, and 3  are F₁, F₂, and F₃ respectively.

Resolution of Forces:

Component of forces along the x-axis = Fcosθ

Component of forces along the y-axis = Fsinθ

Thus, the resolution of forces into x-direction and y-direction is shown in the second image below.

From the second image, the resolution of force along the horizontal direction(x-component) of the force of team A is:

  • Fcos (30) = 0.8660F

Vertical direction(y-component) is:

  • Fsin (30) = 0.5F

However, T₃ (tension in force F₃) counters the x-component of T₁ and it is higher than the component of T₁.

∴

T₃ wins here.

Similarly, the T₂ force counters the y-component of T₁, and the T₂ force is greater than T₁.

∴

T₂ wins here.

This implies that in the same force case, T₁ lost.

Now, we are left with 0.5F of T₂, and 0.134 of T₃ which are perpendicularly working to each other.

Hence, in the first year, the resultant force along T₂ is the winner.

However, For T₁ to win next year provided the team doesn't change their position;

T₁ has to do the following:

  • F' cos 30 ≥ F
  • F' sin 30 ≥ F

Taking the minimum condition;

F'cos(30) = F

[tex]F' = \dfrac{F}{cos (30)}[/tex]

F' = 1.15 F

F'sin(30) = F

[tex]F' = \dfrac{F}{sin (30)}[/tex]

F' = 2F

Therefore, we can conclude that for team 1 to win in the next year if they did not change their position, they need to apply a force greater than or equal to 2F or twice the force of the other team.

Learn more about resultant force here:

https://brainly.com/question/17232218?referrer=searchResults

View image Ajeigbeibraheem
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