Physics 220 Class Exercise October 4 1999

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-component |
y-component |
Force |
x-component |
y-component |

normal |
0 |
n1 |
tension |
0 |
T2 |

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

weight |
0 |
-mg |
|||

tension |
T1 |
0 |
|||

Acceleration |
a1 |
0 |
Acceleration |
0 |
-a2 |

Now apply N2 to each coordinate direction in each diagram.

n1 - mg = 0; T1 - f = ma; T2-mg = -ma2

How many equations do you have? 3

How many unknowns? n1, f, a1, a2, T1, T2: 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).

T1 = T2 (Notice these are force *magnitudes*.)

a1=a2 (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}n1

Now we have 6 equations in 6 unknowns: **Bingo! **Let's write T1 = T2 = T and a1 = a2 = a. Then:

Then: n1 = mg; f = m
_{k}mg; T = m(g-a) = m(a + m
_{k}g)

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

Thus a = g/4.