A0.15 kg meter stick is balanced with the pivot point at the 18 cm mark. a weight of 3.2n is hung from the shorter end. where should the weight be positioned in order to balance the meter stick?

At 15cm to the left of the pivot

Explanation:

The weight 0.15kg(1.5N) and 3.2N are balanced on a meter stick. This weights are parallel to each other on the stick. Questions on parallel forces acting on a body are solved using the *principle of moment."

Moment is the turning effect of force about a point.

Principle of moment states that the sum of clockwise moment is equal to the sum of anticlockwise moments.

According to the question, the meter (100cm) stick is balanced with a pivot @ 18cm mark {this is the shorter end}

If the 3.2N weight is hung at xcm from the shorter end, this weight will turn about the pivot in the anticlockwise direction.

Since moment = Force × perpendicular distance from the force

Moment of the 3.2N weight = 3.2N × x = 3.2x

The weight of the stick (1.5N) will be positioned at the center of the stick i.e @ the 50cm mark which is at 32cm to the right of the pivot.

The weight of the stick will turn in the clock wise direction

The moment of the weight in the clockwise direction = 1.5×(50-18) = 1.5×32 = 48Ncm

Using the principle of moment, we will equate both moments to have;

3.2x = 48

x = 48/3.2

x = 15cm

This means the 3.2N weight will be positioned at the 15cm mark to the left if the pivot to balance the meter stick.

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form