The equilibrium conditions allow to find the results for the balance forces are:
When the acceleration is zero we have the equilibrium conditions for both linear and rotational motion.
∑ F = 0
∑ τ = 0
Where F are the forces and τ the torques.
The torque is the product of the force and the perpendicular distance to the point of support,
The free-body diagrams are diagrams of the forces without the details of the bodies, see attached for the free-body diagram of the system.
We write the translational equilibrium condition.
F₁ - W₁ - W₂ + F₂ = 0
We write the equation for the rotational motion, set our point of origin at scale 1, and the counterclockwise turns are positive.
F₂ 2 - W₁ 1 - W₂ 1.5 = 0[tex]\frac{W_1 \ 1 + W_2 \ 1.5}{2}[/tex]
Let's calculate F₂
F₂ = [tex]\frac{W_1 \ 1 + W_2 \ 1.5 }{2}[/tex]
F₂ = (m g + M g 1.5)/ 2
F₂ = [tex]\frac{(12 + 68 \ 1.5 ) \ 9.8}{2}[/tex]
F₂ = 558.6 N
We substitute in the translational equilibrium equation.
F₁ = W₁ + W₂ - F₂
F₁ = (m + M) g - F₂
F₁ = (12 +68) 9.8 - 558.6
F₁ = 225.4 N
In conclusion using the equilibrium conditions we can find the forces of the balance are:
Learn more here: brainly.com/question/12830892
