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} (0.5)/(1.5)F_R=W-F_R \\ F_R+(0.5)/(1.5)F_R=W \\ (1.5+0.5)/(1.5)F_R=W \\ F_R=(1.5W)/(1.5+0.5) \\ F_R=(1.5W)/(2) \\ F_R=(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.