- #1
Fattyman
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Six problems due Monday, and I have no idea what I'm doing on either of these.
Problem 1:
A group of people has a total mass of 1500kg and are standing on one end of a 20,000kg boat. They walk 6.5m to the other end of the boat. How much does the boat move? The water is frictionless.
rcom = (1/M)*[itex]\int[/itex](r)dm
I really have no idea where to begin, let alone how to try to solve it.
Problem 2:
A regular pentagon has sides of length a. Find the center of mass if you remove the triangle formed by the geometric center and the two vertices on the bottom of the pentagon.
rcom = (1/M)[itex]\int[/itex](r)dm
The book says to split the rest of the pentagon into four equal triangles, and I found the center of mass of the removed triangle. Since the five triangles that make up the pentagon are the same, their centers of mass should be the same distance from the geometric center of the pentagon, which I'm using as the origin.
Also, the book says the answer to this one is ".115a above the vertex of the removed triangle."
Problem 1:
Homework Statement
A group of people has a total mass of 1500kg and are standing on one end of a 20,000kg boat. They walk 6.5m to the other end of the boat. How much does the boat move? The water is frictionless.
Homework Equations
rcom = (1/M)*[itex]\int[/itex](r)dm
The Attempt at a Solution
I really have no idea where to begin, let alone how to try to solve it.
Problem 2:
Homework Statement
A regular pentagon has sides of length a. Find the center of mass if you remove the triangle formed by the geometric center and the two vertices on the bottom of the pentagon.
Homework Equations
rcom = (1/M)[itex]\int[/itex](r)dm
The Attempt at a Solution
The book says to split the rest of the pentagon into four equal triangles, and I found the center of mass of the removed triangle. Since the five triangles that make up the pentagon are the same, their centers of mass should be the same distance from the geometric center of the pentagon, which I'm using as the origin.
Also, the book says the answer to this one is ".115a above the vertex of the removed triangle."