Question

A meter stick balances horizontally on a knife-edge at the 51 cm mark. With two nickels stacked over the 6.0 cm mark, the stick is found to balance at the 45.0 cm mark. A nickel has a mass of 5.0 g. What is the mass of the meter stick

Answer 1

Answer:

65g

Explanation:

Two main conditions for equilibrium are:

I. The resultant force must be equal to zero. That is, sum of the forces acting in one direction about a point must be equal to the sum of the forces acting in the opposite direction about the same point.

II. The resultant moment must be equal to zero. That is, sum of the moments in one direction about a point must be equal to the sum of the moments in another direction about the same point.

For the above question,

the 51cm mark is the point where the resultant weight of the meter stick lies,

the pivot or point is the 45cm mark where the stick balanced when 2 nickels ( total mass (5.0g x 2) 10g were placed at the 6cm mark.

Using the conversion factor:

1000g(1kg) = 10N, we can convert mass to weight, calculate the weight of the meter stick then reconvert to mass.

That is,

mass of 2 nickels = 10g = 10/1000 = 0.01N.

Moment = Force x distance from line of force to pivot of rotation

Applying the principle of equilibrium,

Moment of left side = Moment of right side

0.01 x (45-6) = W x (51-45)

Where W = weight of the meter stick

W x 6 = 0.01 x 39

W x 6 = 0.39

W = 0.39/6

W= 0.065N

Therefore, mass of meter stick = 0.065 x 1000 = 65g.