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 1).
What is the reading on the left scale?
What is the reading on the right scale?
Can someone help me with this question?

Answer 1

Step-Step Explanation:-

As the system is at equilibrium, this means that the reading on the scales have to add to the weight of the student, which means that:-

$F^{l} + F^{r} = W$

or

$F^{l} =W - F^{r}$As we know that momentum 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 forces, we have:-

$\frac{0.5}{1.5}F^{r} =W - F^{r}$

$= > F^{r} + \frac{0.5}{1.5} = W$

$= > \frac{1.5 + 0.5}{1.5}F^{r} = W$

$= > F^{r} = \frac{1.5 W}{1.5 + 0.5}$

$= > F^{r} = \frac{1.5 W}{2}$

$= > \frac{1.5}{2} \times 62 \times 9.8$

$= > F^{r} = 455.7$

Now we know the reading on the right scale, we plug it on the equation for the left scale.

$F^{l} = W - F^{r}$

$= > F^{l} = 62 \times 9.8 - 455.7$

$= > F^{l} = 151.9$

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

Answer 2

The force of reaction at the left will be  $F_c=152.057\ N$ and at the right side will be $F_D=456.165\ N$

What will be the force on both sides?

It is given that

Mass of the body m=62kg

Now the Force of the body

$F_{body} =mg=62\times 9.81=608.22\ N$

Now to find the effect of this force on the scale we will suppose the left point is C and the right point will be D.

Now to find the at Point D (right side ) Take a moment about point A

So the equation will become

$F_D\times2=F_{body}\times 1.5$

$F_D=\dfrac{F_{body}\times 1.5}{2}$

$F_D=\dfrac{ 608.22\times 1.5}{2}$

$F_D=456.165\ N$

Now to find the force on the left side by the equilibrium of the forces

The algebraic sum of all the forces will be zero

$F_D+F_C-F_{body}=0$

$F_C=F_{body}-F_D$

$F_C=608.22-456.16=152.05N$

Thus the force of reaction at the left will be  $F_c=152.057\ N$ and at the right side will be $F_D=456.165\ N$

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