192k views
4 votes
For the last part of the lab, you should have found the mass of the meter stick. So if a mass of 85 g was placed at the 2 cm MARK and the pivot point was moved to the 38.6 cm MARK, what would have been the mass of the meter stick

User Guidot
by
4.4k points

1 Answer

5 votes

Answer:

272.89g

Step-by-step explanation:

Find the diagram to the question in the attachment below;.

Using the principle of moment to solve the question which states that the sum of clockwise moment is equal to the sum of anticlockwise moment.

Moment = Force * Perpendicular distance

Taking the moment of force about the pivot.

Anticlockwise moment:

The 85g mass will move in the anticlockwise

Moment of 85g mass = 85×36.6

= 3111gcm

Clockwise moment.

The mass of the metre stick M situated at the centre (50cm from each end) will move in the clockwise direction towards the pivot.

CW moment = 11.4×M = 11.4M

Equating CW moment to the ACW moment we will have;

11.4M = 3111

M = 3111/11.4

M = 272.89g

The mass of the metre stick is 272.89g

For the last part of the lab, you should have found the mass of the meter stick. So-example-1
User Icecream
by
4.3k points