**Two objects of equal mass m are connected by a massless rope that
does not stretch. One object lies on a flat table, while the other hangs
over the edge, as shown in the diagram. The coefficient of kinetic friction
between block and table is 0.5.**

Draw a free body diagram for each object showing all the forces acting on each.

Block on table
hanging block

Choose a coordinate system for each diagram.

Fill out the table showing the components for the forces
and of the acceleration in each diagram.

Block on table |
Hanging block |
||||

Force |
x_{1}-component |
y_{1}-component |
Force |
x_{2}-component |
y_{2}-component |

normal | 0 | n_{1} |
tension | 0 | T_{2} |

friction | -f | 0 | weight | 0 | -mg |

weight | 0 | -mg | |||

tension | T_{1} |
0 | |||

Acceleration |
a_{1} |
0 | Acceleration |
0 | -a_{2} |

Now apply N2 to each coordinate direction in each diagram.

n_{1} - mg = 0;
T_{1} - f = ma1; T_{2}-mg
= -ma_{2}

How many equations do you have? 3

How many unknowns?
n_{1}, f, a1, a_{2}, T_{1}, T_{2}:
6 total.

Use the connection between the two objects to write 2 more equations relating the accelerations of the two particles (1 equation) and the tension forces acting on each block (1 equation).

T_{1} = T_{2}
(Notice these are force *magnitudes*.)

a_{1} = a_{2}
(Similarly these are acceleration *magnitudes*.)

Finally use the relation between normal force and kinetic friction force to write another equation. Solve to find the acceleration of the hanging block.

f = m_{k}n_{1}

Now we have 6 equations in 6 unknowns: **Bingo! **Let's
write T_{1} = T_{2} = T and a_{1} = a_{2}
= a. Then:

Then: n_{1} = mg;
f = m_{k}mg;
T = m(g-a) = m(a + m_{k}g)

So:
g(1 - m_{k}) = 2a.

Thus
a = g/4.