A one-meter rod is suspended at its middle so that it balances. Suppose one-gram on the rod at the following distances from one end.

20cm 27cm 38cm 44cm 58cm 70cm 83cm 94cm

Where must a one gram weight be hung so that the rod will balances at the 50 centimeter mark
Where must two gram weight be hung so that the rod will balance at the 50 centimeter mark

general principle of a fulcrum:

If a mass of p units is placed x units from the fulcrum and it balances a mass of q units at y units on the oppose side from the fulcrum , then
px = qy

so make a sketch placing the 1 gram weights at the given distances
The the fourth weight on the right side be k g.
then 1(30) + 1(23) + 1(22) + 1(6) = 1(8) + 1(20) + 1(33) + 1(k)
81 = 61 + k
k = 20

so the weight should be placed 20 cm from the centre or at the 70 cm mark.

To find where a one-gram weight must be hung so that the rod balances at the 50 centimeter mark, we need to consider the moments or torques created by the weights on the rod.

The moment of a weight is given by its weight value multiplied by its distance from the fulcrum point (where the rod is suspended).

First, let's calculate the total moments on the rod. We know that the rod balances when the total sum of moments on one side of the fulcrum is equal to the total sum of moments on the other side.

Given the one-gram weights at the following distances from one end of the rod:
20cm 27cm 38cm 44cm 58cm 70cm 83cm 94cm

Now, let's calculate the total sum of moments on one side of the fulcrum. We can use the formula:

Sum of moments = (weight1 * distance1) + (weight2 * distance2) + ... + (weightn * distancen)

On one side of the fulcrum (to the left), we have:
(20g * 20cm) + (27g * 27cm) + (38g * 38cm) + (44g * 44cm) + (58g * 58cm) + (70g * 70cm) + (83g * 83cm) + (94g * 94cm)

Next, we need to find the total sum of moments on the other side of the fulcrum. Since the rod is in balance, the sum of moments on both sides should be equal.

Now, we can set up an equation where the sum of moments on one side is equal to the sum of moments on the other side:

Sum of moments on one side = Sum of moments on the other side

Plugging in the values we calculated, we have:

(20g * 20cm) + (27g * 27cm) + (38g * 38cm) + (44g * 44cm) + (58g * 58cm) + (70g * 70cm) + (83g * 83cm) + (94g * 94cm) = (xg * 50cm)

Where 'x' represents the unknown number of grams to be hung at a certain distance to balance the rod at the 50 centimeter mark.

By solving this equation, we can find the value of 'x' and the corresponding distance at which the one-gram weight must be hung.

To find where a two-gram weight must be hung so that the rod balances at the 50 centimeter mark, the process is the same. However, in this case, we set up the equation:

(20g * 20cm) + (27g * 27cm) + (38g * 38cm) + (44g * 44cm) + (58g * 58cm) + (70g * 70cm) + (83g * 83cm) + (94g * 94cm) = (2xg * 50cm)

The only difference is that we multiply the weight 'x' by 2, as we are dealing with a two-gram weight.