Question

What could be the possible answer to the question ?thankyou ~​

Answer 1

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Answer 2

The value of the force, F₀, at equilibrium is equal to the horizontal

component of the tension in string 2.

Response:

• The value of F₀ so that string 1 remains vertical is approximately 0.377·M·g

How can the equilibrium of forces be used to find the value of F₀?

Given:

The weight of the rod = The sum of the vertical forces in the strings

Therefore;

M·g = T₂·cos(37°) + T₁

The weight of the rod is at the middle.

Taking moment about point (2) gives;

M·g × L = T₁ × 2·L

Therefore;

$T_1 = \mathbf{\dfrac{M \cdot g}{2}}$

Which gives;

$M \cdot g = \mathbf{T_2 \cdot cos(37 ^{\circ})+ \dfrac{M \cdot g}{2}}$

$T_2 = \dfrac{M \cdot g - \dfrac{M \cdot g}{2}}{cos(37 ^{\circ})} = \mathbf{\dfrac{M \cdot g}{2 \cdot cos(37 ^{\circ})}}}$

F₀ = T₂·sin(37°)

Which gives;

$F_0 = \dfrac{M \cdot g \cdot sin(37 ^{\circ})}{2 \cdot cos(37 ^{\circ})}} = \dfrac{M \cdot g \cdot tan(37 ^{\circ})}{2} \approx \mathbf{0.377 \cdot M \cdot g}$

• F₀ ≈ 0.377·M·g

Learn more about equilibrium of forces here:

brainly.com/question/6995192