Respuesta :
Answer:
F=√[(1.5(14.58L+11.96))² + (3.2(2.97L - 157.03) + 62.72)²]
Explanation:
Given data,
The mass of the bar AB, m = 6.4 kg
The angular velocity of the bar, θ˙ = 2.7 rad/s
The angle of the bar at A, θ = 24°
Let the length of the bar be, L = l
The angular moment at point A is,
∑ Mₐ = Iα
Where, Mₐ - the moment about A
α - angular acceleration
I - moment of inertia of the rod AB
[tex]-mg(\frac{lcos\theta}{2})=\frac{1}{3}(ml^{2})\alpha[/tex]
[tex]\alpha=\frac{-3gcos\theta}{2l}[/tex]
Let G be the center of gravity of the bar AB
The position vector at A with respect to the origin at G is,
[tex]\vec{r_{G}}=[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}][/tex]
The acceleration at the center of the bar
[tex]\vec{a_{G}}=\vec{a_{a}}+\vec{\alpha}X\vec{r_{G}}-\omega^{2}\vec{r_{G}}[/tex]
Since the point A is fixed, acceleration is 0
The acceleration with respect to the coordinate axes is,
[tex](\vec{a_{G}})_{x}\hat{i}+(\vec{a_{G}})_{y}\hat{j}=0+(\frac{-3gcos\theta}{2l})\hat{k}\times[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}]-\omega^{2}[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}][/tex]
[tex](\vec{a_{G}})_{x}\hat{i}+(\vec{a_{G}})_{y}\hat{j}=[-\frac{cos\theta(2l\omega^{2}+3gsin\theta)}{4}\hat{i}+(\frac{2l\omega^{2}sin\theta-3gcos^{2}\theta}{4})\hat{j}][/tex]
Comparing the coefficients of i
[tex]=-\frac{cos\theta(2l\omega^{2}+3gsin\theta)}{4}[/tex]
Comparing coefficients of j
[tex](\vec{a_{G}})_{y}=\frac{2l\omega^{2}sin\theta-3gcos^{2}\theta}{4}[/tex]
Net force on x direction
[tex]F_{x}=(\vec{a_{G}})_{x}[/tex]
substituting the values
[tex]F_{x}[/tex]=1.5(14.58L+11.96)
Similarly net force on y direction
[tex]F_{y}=(\vec{a_{G}})_{y}+mg[/tex]
= 3.2(2.97L - 157.03) + 62.72
Where L is the length of the bar AB
Therefore the net force,
[tex]F=\sqrt{F_{x}^{2}+F_{y}^{2}}[/tex]
F=√[(1.5(14.58L+11.96))² + (3.2(2.97L - 157.03) + 62.72)²]
Substituting the value of L gives the force at pin A
