Question

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Answer 1

Answer:

58.8 N

Explanation:

Let 'F₁' be the force by front arm and 'F₂' be the force by back arm.

Given:

Mass of the rod (m) = 3.00 kg

Length of the pole (L) = 4.50 m

Acceleration due to gravity (g) = 9.8 m/s²

Distance of 'F₁' from one end of pole (d₁) = 0.750 m

'F₂' acts on the end. So, distance between 'F₁' and 'F₂' = 0.750 m

Now, weight of the pole acts at the center of pole.

Now, distance of center of pole from 'F₁' is given as:

d₂ = (L ÷ 2) - d₁

$d_2=\frac{4.50}{2}-0.75=1.5\ m$

Now, as the pole is held horizontally straight, the moment about the point of application of force 'F₁' is zero for equilibrium of the pole.

So, Anticlockwise moment  = clockwise moment

$F_2\times d_1=mg\times d_2\\\\F_2=\frac{mg\times d_2}{d_1}$

Plug in the given values and solve for 'F₂'. This gives,

$F_2=\frac{3.00\ kg\times 9.8\ m/s^2\times 1.5\ m}{0.75\ m}\\\\F_2=\frac{44.1}{0.75}=58.8\ N$

Therefore, the force exerted by the back arm on the pole is 58.8 N vertically down.