Moment of force : $M = F \times d \implies F = \dfrac{M}{d}$ where, F is the force applied and d is the distance from origin. As all measuring scales are identical they have same moment of force when they are of same length.

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From second relation we can come to a conclusion that force applied is inversely proportional to distance from the origin that means lesser the distance from origin more is the force applied. For our convenience let us extend all the scales to 75 cm. The following are the positions of fingertip measured from the table : $\begin{array}{c}~A : 100~cm \\ B : 75~cm \\ C : 50~cm \\ D : 75~cm \\ E : 25~cm \\ \end{array}$ As you can see the least distance is measured in case of E so the greatest force must be applied in case of E and least in case of A.

The point is figure out what direction and size of force to stick applied by the table? let me see three kind of force applied by table. For A,D, force direction is uppward, which applied by table to left side end of stick. The B show size of the force is zero, for C,E show the force is downward, which applied by table to left side end of stick, conclude by Newton's first law, since the sticks at rest, here must be a force pushing upwards from the fingertip to cancel out mg of sticks and force applied by table. the E is answer, which force applied by fingertips is 8/3 times of the sticks's weight

Consider balancing the moment of the five cases we have:

$Fl_f = mgl_c \quad ...(1)$

where

• $F$ --- the force applied by the fingertip
• $l_f$ --- the distance between the fingertip and the hinge at the edge of the table
• $m$ --- the mass of the stick
• $g$ --- acceleration due to gravity
• $l_c$ --- the distance between the center of mass of the stick and the hinge at the edge of the table

Let the length of the stick be $i$ and the mass per length of the stick be $\sigma$ . Then $l_c = \dfrac l2$ and $m = \sigma l$ and equation (1) becomes:

$Fl_f = \sigma l g \cdot \frac l2 \implies F = \sigma g \cdot \frac {l^2}{2l_f}$

Since case E has the largest $l =100$ and the smallest $l_f = 25$ , $F_E = 200\sigma g$ is the greatest .

Without any formula it is obvious that $F_A<F_B<F_C<F_E$ . Also obvious is $F_D<F_E$ which leads to E as solution.

Let m be the mass of each identical stick. If a stick is balanced by a force f then as per the law of moments f* distance of the force from the hinge = weight of the stick * distance of center of mass to the hinge. The five cases presented above in the diagram translate to the equations below

CASE A

fx50 = 2m x 25

f = m

CASE B

f*25 = 2m * 25

f= 2m

CASE C

f*25 = 3m * 75/2

f = 4.5m

CASE D

f*75 = 4m * 50

f = 2.67m

CASE E

fx25 =4m * 50

f = 8m

We can see that the highest force has to be applied in case E and the lowest in case A.

The force must make a torque equal to the done by the stick itself. So we look for the stick which makes the largest torque and choose the finger which is closest to the stick's support, which is $E$ .

To apply the greatest force the stick needs to be long and apply the force closer to the table.

Figure E has a long stick with a force applied near the fulcrum/axle. More the force is applied near to the fulcrum, more force it needs to hold the stick.

As E is the longest of all the sticks......

balance the torques acting by forces about the hinge . for first torque about point on table is zero so

f 0.50 = mg 0.25

f=mg/2...

for second similarly

f 0.25=mg 0.25

f=mg

for third

f 0.25=(mg 0.75)/2

f=(mg*3)/2

.....and so on.