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/
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