The uniform 100-kg beam is freely hinged about its upper end a and is initially at rest in the vertical position with theta = 0. determine the initial angular acceleration of the beam and the magnitude of the force fa supported by the pin at a due to the application of the force p = 300 n on the attached cable.

ac = 2.86 m / s²

Explanation:

Image can detail the system to determine the force in the FA to understand the system into the applicated force

m = 100 kg ,  L = 3 m

∑ F = 0 ⇒ Ay - 100 kg + P * cos (45) = 0

Ay = 768.86 N

∑ Mₐ = α * I ₐ

I ₐ = m * L² / 3  ⇒  I ₐ = 100 kg * 4² m / 3

Replacing

P * sin (45) * 3 = α * 100 kg * 4² m / 3  

α = 1.193 rad / s²

ac = α *2    ⇒  ac = 1.193 rad / s²  * 2

ac = 2.86 m / s²

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .