The rotational and translational equilibrium conditions allow finding the results for the questions about the forces in the system are;
a) The cable tension is; T = 415.8 N
b) The x component of the wall reaction is: R = 371.8 N
Given parameters
To find
a) The tension of the rope.
b) The x component of the wall reaction.
Newton's second law for linear and rotational movements relate forces and torques are the accelerations of bodies, in the cases where the accelerations are zero, they are called equilibrium conditions:
τ = F x r = F r sin θ
Where F is the force, τ the torque, r the position and θ the angle.
In a free-body diagram you have the forces without the details of the bodies. In the attachment we see a diagram of the forces and the reference system for this case.
Let's use trigonometry to break down stress.
Sin θ = [tex]\frac{T_y}{T}[/tex]
cos θ = [tex]\frac{T_x}{T}[/tex]
[tex]T_y[/tex] = T sin θ
Tₓ = T cos θ
Using distance values.
tan θ = [tex]\frac{y}{x}[/tex]
θ = tan⁺¹ [tex]\frac{y}{x}[/tex]
θ = tan-1 [tex]\frac{0.9}{1.8}[/tex]
θ = 26.6º
A) We use the rotational equilibrium condition, they indicate that the turning point is the point of union of the post with the wall, we will take the counterclockwise turns as positive.
W₁ 1.8 + W₂ [tex]\frac{1.8}{2}[/tex] - [tex]T_y[/tex] 1.8 = 0
[tex]T_y=(m_1+\frac{m_2}{2} )\ g[/tex]
[tex]T_y[/tex] = [tex](14+ \frac{10}{2} ) \ {9.8[/tex]
[tex]T_y[/tex] = 186.2 N
[tex]T_y[/tex] = T sin 26.6
T = [tex]\frac{T_y}{sin \ 26.6}[/tex]
T = [tex]\frac{186.2 }{sin \ 26.6}[/tex]
T = 415.8 N
b) Let's apply the translational equilibrium condition for each axis.
x- axis
R - Tₓ = 0
R = T cos tea
R = 415.8 cos 26.6
R = 371.8 N
In conclusion using the rotational and translational equilibrium conditions we can find the results for the questions about the forces in the system are;
a) The cable tension is; T = 415.8 N
b) The x component of the wall reaction is: R = 371.8 N
Learn more about equilibrium conditions here: brainly.com/question/7031958
