Question

A 62 kg student stands on a very light, rigid board that rests on a bathroom scale at each end as shown in Figure.( i ) What is the reading on left scale?(ii) What is the reading on right scale?

Answer 1

Since the system is at equlibrium this means that the reading on the scales have to add to the weight of the student, this means that:

$F_L+F_R=W$

or:

$F_L=W-F_R$

We also know that the moment of the system have to be zero, this means that the momentum for each scale is equal and then we have:

$1.5F_L=0.5F_R_{}$

Plugin this in the equation for the forces we have:

$\begin{gathered} \frac{0.5}{1.5}F_R=W-F_R \\ F_R+\frac{0.5}{1.5}F_R=W \\ \frac{1.5+0.5}{1.5}F_R=W \\ F_R=\frac{1.5W}{1.5+0.5} \\ F_R=\frac{1.5W}{2} \\ F_R=\frac{1.5}{2}(62)(9.8) \\ F_R=455.7 \end{gathered}$

Now that we know the reading on the right scale we plug it on the equation for the left scale:

$\begin{gathered} F_L=W-F_R \\ F_L=(62)(9.8)-455.7 \\ F_L=151.9 \end{gathered}$

Therefore the reading on the left scale is 151.9 N and the reading on the right scale is 455.7 N.