Thus the center of mass of the system lies at x = 1.7.

Problem :

Calculate the center of mass of the following system: A mass of 10 kg lies at
the point (1,0), a mass of 2 kg lies at the point (2,1) and a mass of 5 kg lies
at the point (0,1), as shown in the figure below.

To find the center of mass in a two dimensional system, we must complete two
steps. First we must find the center of mass in the x-direction, and then in the
y-direction. We know that the total mass of the system is 17 kg. Thus:

x_{cm}

=

(m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3})

=

= = .824

Also, then

y_{cm}

=

(m_{1}y_{1} + m_{2}y_{2} + m_{3}y_{3})

=

= = .412

Thus the center of mass of the system lies at the point (.824, .412).

Problem :

Consider the system from problem 2, but now with forces acting upon the system.
On the 10 kg mass, there is a force of 10 N in the positive x direction. On the
2 kg mass, there is a force of 5 N inclined 45^{o} above horizontal.
Finally, on the 5 kg mass, there is a force of 2 N in the negative y direction.
Find the resultant acceleration of the system.

Since we already know the position of the center of mass and the total mass of
the system, we can use the equation F_{ext} = Ma_{cm} to find the
acceleration of the system. To do so, we must find the net force by breaking
each force acting on the system into x and y components:

F_{x} = 10 + 5 cos 45 = 13.5 NF_{y} = 5 sin 45 - 2 = 1.5 N

Thus the magnitude of the net force is given by:

F = = 13.6 N

And the force is inclined above the horizontal by an angle of:

θ = tan^{-1} = 6.3^{o}

The resultant force has a magnitude of 13.6 N and an inclination of 6.3 degrees,
as shown
below:

Now that we have the resultant force on the system, we can find the acceleration
of the system. To conceptualize this, we imagine that all the mass of the system
is placed at the spot of the center of mass, and the net force acts on that
spot. Thus:

F_{ext} = Ma_{cm}

Implying that

a_{cm} = = = .8 m/s^{2}

The center of mass of the system accelerates at a rate of .8 m/s^{2} in the same
direction as the net force (6.3^{o} above horizontal). Of course, since
outside forces are acting on the individual particles, they will not move in the
same direction as the center of mass. The motion of the individual particles
can be calculated simply using Newton's Laws.

Problem :

Two masses, m_{1} and m_{2}, m_{1} being larger, are connected by a spring.
They are placed on a frictionless surface and separated so as to stretch the
spring. They are then released from rest. In what direction does the system
travel?

We can regard the two masses and the spring as an isolated system. The only
force felt by the masses is the spring force, which lies inside the system. Thus
no external force acts on the system, and the center of mass of the system is
never accelerated. Thus, because the velocity of the center of mass is
initially zero (as neither block is moving before they are released) this
velocity must remain at zero. Though each block is accelerated by the spring in
some way, the velocity of the center of mass of the system never changes, and
the position of center of mass of the system never moves. The blocks will
continue oscillating on the spring, but will not cause any translational motion
of the system.

Problem :

A 50 kg man stands at the edge of a raft of mass 10 kg that is 10 meters long.
The edge of the raft is against the shore of the lake. The man walks toward the
shore, the entire length of the raft. How far from the shore does the raft move?

You may ask what this problem has to do with center of mass. Let's examine
closely exactly what is going on. Since we are talking about systems of
particles in this section, let's visualize this situation as a system. The man
and the raft are two separate objects, and mutually interact when the man walks
across the boat. Initially the boat is at rest, so the center of mass is a
stationary point. When the man walks across the boat, no external force acts on
the system, as the boat is allowed to glide across the water. Thus while the man
walks across the raft, the center of mass must stay in the same place. In
order to do so, the raft must move out from the shore a certain distance. We can
calculate this distance, which we shall denote by d, using center of mass
calcualtions.

We start be calculating the center of mass when the man is at point A. Remember
that we can choose our origin, so we shall choose x = 0 to be at the shoreline.
For this problem we can assume that the raft has a uniform density, and thus can
be treated as if all its mass were at its midpoint, of x = 5. Thus the center of
mass is:

x_{cm} = m_{1}x_{1}+m_{2}x_{2} = = 9.2 m

The center of mass of the system is, and must always be, 9.2 m away from the
shore. Next we calculate the center of mass when the man is at point B,
introducing our variable, d. The man is a distance d from the shoreline, while
the raft is a distance d + 5 from the shoreline. Thus:

x_{cm} = =

This quantity must equal our original center of mass, or 9.2 m. Thus:

= 9.2

60d + 50 = 552

d = 8.4 m

Thus as the man moves from point A to point B, the raft gets displaced 8.4
meters from the shore.