### Fall 2007

Fall 2007 1 Mech

Fall 2007 2 Mech

Fall 2007 3 EM

Fall 2007 4 EM

Fall 2007 5 QM

Fall 2007 6 QM

Fall 2007 7 SM

Fall 2007 8 SM

Fall 2007 9 Misc

Fall 2007 10 Misc

Fall 2007 11 Mech

Fall 2007 12 Mech

Fall 2007 13 EM

Fall 2007 14 EM

Fall 2007 15 QM

Fall 2007 16 QM

Fall 2007 17 SM

Fall 2007 18 SM

Fall 2007 19 Misc

Fall 2007 20 Misc

A rope of mass *M* and length *L* is suspended in the earth’s gravitational field, *g*, with the bottom end of the rope touching a surface. The rope is then released from rest and falls limply on the surface (i.e., without the elements bouncing upwards). Find the force *F(t)* on the surface as a function of time, 0 < t < $\infty$, and make a sketch of it. At what time does *F(t)* reach its maximum? What is the value of this maximum force?

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Answer

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All the elements of the rope are in free fall. It takes time *T* = *2L/g* for the last element to reach the surface. Hence, at *t > T* we have

Consider now 0 < *t* < *T*. Here *F* is a sum of two terms, *F = F _{1} + F_{2}*. The first one is the the weight $F_1 = \mu l g$ of the part that has already fallen. Here $l(t) = gt^2/2$ is the length of the fallen piece and $µ = M/L$ is the mass per unit length. The second term is the transfer of momentum

*F*= dP/dt from the element of length

_{2}*dl = vdt*that comes to rest during the time (t, t+dt).

The velocity of this segment is *v = gt*, and so $F_2 = \mu v(vdt)/dt = \mu g^2 t^2$.

Accordingly,

We see that *F* reaches a maximum *F = 3Mg* at *t = T*, then experiences a sudden drop to the three times smaller value, after which it remains constant.

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