A set of weights were hung onto a meter stick where one of them has a mass of 100 g located right on the edge of the meter stick, another one with a mass of 80 g on the other edge, and another one with a mass of 150 g placed 15 cm from the 100 g weight. Find the center of mass of the meter stick with its hanging

Principles of Physics: A Calculus-Based Text
5th Edition
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Author:Raymond A. Serway, John W. Jewett
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Chapter8: Momentum And Collisions
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SOLVE FOR THE FOLLOWING AND DON'T USE SHORCUTS USE THE SAME PROCESS ON THE SAMPLE PROBLEM BEING PROVIDED. 

A set of weights were hung onto a meter stick where
one of them has a mass of 100 g located right on the
edge of the meter stick, another one with a mass of
80 g on the other edge, and another one with a mass
of 150 g placed 15 cm from the 100 g weight. Find
the center of mass of the meter stick with its hanging
weights.
Transcribed Image Text:A set of weights were hung onto a meter stick where one of them has a mass of 100 g located right on the edge of the meter stick, another one with a mass of 80 g on the other edge, and another one with a mass of 150 g placed 15 cm from the 100 g weight. Find the center of mass of the meter stick with its hanging weights.
Solution
0.59m
(m, d1) + (m2 d2)
r =
0.7kg
1.2 kg
Gravitational
center is around
here
m1 + m2
(0.7 kg x 0 m) + (1.2 kg × 0.59 m)
r =
0.7 kg + 1.2 kg
r = 0.37 m
Sample Problem
• Find the center of mass on a rod
with 3 mass points.
0.85m
0.39m
0.2kg
0.5kg
0.4kg
Solution
0.85m
0.39m
0.2kg
(m, d1) + (m2 d2) + (m3 dz)
r 3=
0.5kg
0.4kg
mị + m2 + m3
Gravitational center is around here
(0.5 kg x 0 m) + (0.2 kg x 0.39 m) + (0.4 kg x 0.85 m)
r =
0.5 kg + 0.2 kg + 0.4 kg
r = 0.38 m
Transcribed Image Text:Solution 0.59m (m, d1) + (m2 d2) r = 0.7kg 1.2 kg Gravitational center is around here m1 + m2 (0.7 kg x 0 m) + (1.2 kg × 0.59 m) r = 0.7 kg + 1.2 kg r = 0.37 m Sample Problem • Find the center of mass on a rod with 3 mass points. 0.85m 0.39m 0.2kg 0.5kg 0.4kg Solution 0.85m 0.39m 0.2kg (m, d1) + (m2 d2) + (m3 dz) r 3= 0.5kg 0.4kg mị + m2 + m3 Gravitational center is around here (0.5 kg x 0 m) + (0.2 kg x 0.39 m) + (0.4 kg x 0.85 m) r = 0.5 kg + 0.2 kg + 0.4 kg r = 0.38 m
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