In Figure 5-43, a chain consisting of five links, each of mass 0.097 kg, is lift

Question

In Figure 5-43, a chain consisting of five links, each of mass 0.097 kg, is lifted vertically with a constant acceleration of magnitude a = 2.50 m/s2. Figure 5-43 Find the magnitude of the force link 1 and link 2. Find the magnitude of the force acting between 2 and link 3. Find the magnitude of the force acting between link 3 and link 4. Find the magnitude of the force acting between link 4 and link 5. Find the magnitude of the force F exerted on the top link by the person lifting the chain. Find the magnitude of the net force accelerating each link.

Explanation / Answer

This is a nice problem. Link 2 only has to pull on the last link, so this force will be the smallest. Link 5 needs to pull all the other links, so this force will be bigger, and the person will have to exert the biggest force. One thing we need to decide, does this problem take place on earth, or is the person out in space somewhere. I will solve assuming the person is on the earth. This will add an extra 9.8 m/s^2 to the acceleration, since it takes a force just to hold the links up. F = ma a) For link 2 the mass being accelerated is link one, which is 0.097 kg. The force is overcoming g in addition to accelerating the link up, so a = (2.5+9.8)m/s^2 F = 0.097kg12.3m/s^2 = 1.2 N. Link two is pulling up on link one and link one is pulling down on link two. The problem only asks for magnitude, so we don't need to write in the direction. b) For link 3 there are now two links being accelerated by link 3 instead of just one. Two links have twice the mass as one, so the force is twice as much F = 2.4 N. (this is true since mass is to the first power in F=ma. If it were squared in the formula, then we couldn't just double the force when the mass doubled) Link three is pulling up on link two with a force of 2.4 N. Link 2 is pulling down on link 3 with a force of 2.4 N c) For link 4, there are now three links being accelerated by link four. Three links have three times the mass as one, so the force is three times as much as in part a) or 3.6 N. d) For link 5, There are now four links being pulling up by link five. Four links have, you guessed it, four times the mass as one, so the force is four times as much as in part a) or 4.8 N. e) The person has to accelerate all five links. The mass of five links is 5*0.097kg, so the force is F = 5*0.097kg(2.5 +9.8)m/s^2 = 6.0 N (we could have done it out like this for each, but you can see how the five-times-as-much comes into the equation here) f) Now, each of the links (except for link one) is being pulled on and also pulling. Let's look at link 4. Link four is being pulled up with 4.8N and it is pulling on link three with 3.6N. The net force is 1.2 N. We could do this for all of the links and we would get the same answer. The net force is just the force needed to accelerate that particular link. The net force to accelerate a link is the same force that it takes to accelerate link 1. So the net force for each link is 1.2 N.

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