Associated Topics || Dr. Math Home || Search Dr. Math

Lifting an Object of Changing Mass

```
Date: 7/28/96 at 20:28:34
From: Ramon Ponte
Subject: Lifting an Object of Changing Mass

Suppose you grab the end of a chain that weights 3lb/ft and lift it
straight up off the floor at a constant speed of 2ft/s.

(a) Determine the upward force F you must exert as a function of S
(height)

(b) How much work do you do in lifting the top of the chain to
s = 4ft?

Strategy: Treat the part of the chain you have lifted as an object
that is gaining mass

0      -     hand
0      |
0      |3ft
00000    |
0000000   -    chain

-Ramon Ponte
```

```
Date: 8/1/96 at 11:23:40
From: Doctor Anthony
Subject: Re: Lifting an Object of Changing Mass

Newton's second law in its general form is

F = d(mv)/dt  This is normally written m*dv/dt because mass is
usually constant. When mass is varying we need to look at what is
happening in more detail.

Consider a body moving so that at time t, its mass is m, its velocity
is v and the resultant force acting on the body is F.

Suppose that, at this instant, an extra mass dm moving at velocity u
is added to the body so that, at time t+dt, the body has mass m+dm and
velocity v+dv

Just before the addition of dm the momentum of the body is  mv  and
the momentum of dm is dm.u

Immediately afterwards, the momentum of the augmented body is
(m+dm)(v+dv)   So the increase in momentum in time dt is

(m+dm)(v+dv) - mv - dm.u  = mv + m.dv + v.dm + dm.dv - mv - dm.u
= m.dv + dm(v-u) + dm.dv

The impulse of the force acting on the body at the same time interval
is F.dt,  and so we have:

F.dt = m.dv + (v-u)dm + dm.dv

as dt -> 0 this becomes      F = m.dv/dt + (v-u)dm/dt

If the mass increment, dm, has no velocity of its own before becoming
part of the body, i.e. u = 0, the equation becomes

F = m.dv/dt + v.dm/dt

Now consider what happens as we lift the end of the chain at a
constant velocity of 2 ft/sec.  In above Formula F is the RESULTANT
force, so it is the combination of P the pull exerted by your hand
upwards, and mg, the force of gravity acting downwards.  At time t, a
length s feet is off the floor so the mass of this is 3s lbs, and the
gravitational pull is 3s.g lbals

So F = P - 3s.g = 3s.dv/dt + 2.dm/dt     But dv/dt = 0 so this becomes

P - 3s.g = 2.dm/dt  and dm/dt = 6 (2 ft/sec = 6 lbs/sec mass increase)

P - 3s.g = 2*6 = 12

P = 3s.g + 12   lbals

If we want the force in lbs we divide by g = 32 to get

P = 3s + 12/32  lbs

P = 3s + 3/8  lbs

(b) Work done in lifting the end to  4 ft.

W.D = INT(0 to 4)[P.ds]

= INT(0 to 4)[(3s + 3/8)ds]

=  (3/2)s^2 + (3/8)s   between 0 and 4

=  (3/2)*16 + (3/8)*4

= 24 + 3/2   lbs.ft

= 25.5  lbs.ft

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus
High School Physics/Chemistry

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search