Question

Suppose the student in (Figure 1) is 68kg, and the board being stood on has a 12kg mass. What is the reading on the left scale? What is the reading on the right scale?

Answer 1

The equilibrium conditions allow to find the results for the balance forces are:

• F₁ = 225.4 N

• F₂ = 558.6 N

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$\frac{W_1 \ 1 + W_2 \ 1.5}{2}$

Let's calculate F₂

         F₂ = $\frac{W_1 \ 1 + W_2 \ 1.5 }{2}$  

         F₂ = (m g + M g 1.5)/ 2

         F₂ = $\frac{(12 + 68 \ 1.5 ) \ 9.8}{2}$  

         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:

• F₁ = 225.4 N

• F2 = 558.6 N

Learn more here:  brainly.com/question/12830892