Question

9. Lying on this board, Annabelle wants to determine how far her center of mass is from her feet.a) What is Annabelle’s mass?b) How far from her feet is Annabelles center of mass?

Answer 1

Given,

The mass of the board, m=5 kg

The reading on the weighing board near the feet of Annabelle, F₁=380 N

The reading of the weighing board near the head of Annabelle, F₂=280 N

The total distance between the two weighing boards, d=2.50 m

Let us assume that the centre of mass of the board lies exactly at its geometrical centre. Thus the distance of the centre of mass of the board from either of the weighing boards is l=2.50/2=1.25 m

a)

As the system is in equilibrium, the total downward gravitational force is equal to the upward reaction force applied by the boards. Thus,

`\begin{gathered} M* g+mg=F_1+F_2 \\ M=(F_1+F_2-mg)/(g) \end{gathered}`

Where M is the mass of Annabelle and g is the acceleration due to gravity.

On substituting the known values,

`\begin{gathered} M=(380+280-5*9.8)/(9.8) \\ =62.35\text{ kg} \end{gathered}`

Thus Annabelle's mass is 62.35 kg

b)

The entire system is also in rotational equilibrium and net torque is zero.

Let us assume that the weighing board near Annabelle's feet is the pivot point of the system.

Therefore,

`Mgd_c+mg* l=F_2* d`

Where d_c is the distance of the centre of mass of Annabelle from her feet.

On rearranging the above equation,

`d_c=(F_2* d-mg* l)/(Mg)`

On substituting the known values,

`\begin{gathered} d_c=(280*2.50-5*9.8*1.25)/(62.35*9.8) \\ =1.05\text{ m} \end{gathered}`

Thus the centre of mass of Annabelle is 1.05 m from her feet.