A student suspends a chain consisting of three links, each of mass m 0.250 kg, from a light rope. The rope is attached to the top link of the chain, which does not swing. She pulls upward on the rope, so that the rope applies an upward force of 9.00 N to the chain. a. Draw a free-body diagram for the entire chain, considered as a body, and one for each of the three links. (b) Use the diagrams of part (a) and Newton's laws to find (i) the acceleration of the chain, (ii) the force exerted by the
top link on the middle link, and (iii) the force exerted by the middle link on the bottom link. Treat the rope as massless.

1. There are four objects of interest in this problem: the chain as a whole and the three individual links. For each of these four objects, make a list of the external forces that act on it. Besides the force of gravity, your list should include only forces exerted by other objects that touch the object in question.

2. Some of the forces in your lists form actionâreaction pairs (one pair is the force of the top link on the middle link and the force of the middle link on the top link). Identify all such pairs.
3. Use your lists to help you draw a free-body diagram for each of the four objects. Choose the coordinate axes.
4. use your lists to decide how many unknowns there are in this problem. Which of these are target variables?

EXECUTE

5. Write a Newton's second law equation for each of the four objects, and write a Newton's third law equation for each actionâreaction pair. You should have at least as many equations as there are unknowns (see step 4). Do you?
6. Solve the equations for the target variables.

EVALUATE
7. You can check your results by substituting them back into the equations from step 6. This is especially important to do if you ended up with more equations in step 5 than you used in step 6.
8. Rank the force of the rope on the chain, the force of the top link on the middle link, and the force of the middle link on the bottom link in order from smallest to largest magnitude. Does this ranking make sense? Explain.
9. Repeat the problem for the case in which the upward force that the rope exerts on the chain is only 7.35 N. Is the ranking in step 8 the same? Does this make sense?

B) a= 2.2 m/s²,  T₁ '= 5.45 N, T₂ ’= 3 N

8)     F> T₁ ’> T₂’

Explanation:

A) In the adjoint we can see the free-body diagrams of each element and of the set

B)

i) The acceleration of the chain is

           F - W = M a

           M = m₁ + m₂ + m₃

           a =

           a = - g                            (1)

           a = 9 / (3 0.250) - 9.8

           a = 2.2 m / s²

the positive sign indicates that the system is rising

ii) the outside of the top link over the middle

            T₁ '- W₂ -T₃ = m a

the acceleration of all the links is the same because they are united

Tension T3 is the lower link force that must be equal to its weight

            T₃ = W₃

we substitute

            T₁ ’= m a + W₂ + W₃

            T₁ ’= m a + m g + m g

            T₁ ’= m (a + 2g)                   (2)

            T₁ ’= 0.250 (2.2 + 2 9.8)

            T₁ '= 5.45 N

iii) the force on the lower link

             T₂ ' -W₃ = m a

             T₂ '= m a + m g

             T₂ '= m (a + g)               (3)

             T₂ '= 0.25 (2.2 + 9.8)

              T₂ ’= 3 N

1) The external forces are

set

* the force of the cueda (F) upwards

* the force of the drawer (W) down

Top link

* the strength of the rope (F)

* the pso of the link (W1)

* the tension of the second link (T2)

Middle link

* The tension of the first link (T1 ')

* Weight (W2)

* the tension of the last link (T3)

Lower link

* The tension of the intermediate link (T2 ')

* Weight (W3)

2) the action and reaction forces are

* T₂ and T₁ '

* T₃ and T₂ '

this are forces of equal magnitude and direction, applied in two bodies

3) in the attachment you have the free body diagram of the body, the vertical upward direction is considered positive

4) The inconnitas are the acceleration of the body and the internal tensions between each link bone 2 tensions

5, 6 and 7)

      F - W₁ -T₂ = m a

      T₁'- W₂ -T₃ = m a

      T₂ '- W₃ = m a

      T₂ = T₁ '

      T₃ = T₂ '

One way to check that the sign is correct is that the action and reaction force between two bodies can cancel out when adding the equations

        F -W₁ - W₂ - W₃ = (m + m + m) a

for which we can evaluate and find the acceleration of the systems

8) order the strengths from greatest to least

     F = 9 N,

     T₁ '= 5.45 N

     T₂ ’= 3 N

           F> T₁ ’> T₂’

9) repeat the problem for an exeran force of F = 7.35 N

the acceleration we substitute in equation 1

                        a = F / m -g

                        a = 7.35 / 0.75 - 9.8

                       a = 0

the system is in equilibrium

the voltages using 2 and 3 are

                   T₁ ’= m (a + 2g)

                   T₁ ’= 0.25 (0+ 2 9.8)

                   T₁ ’= 4.9 n

                    T₂ '= m (a + g)

                    T₂ '= 0.25 (0 +9.8)

                    T₂ ’= 2.45 N

the order of force is

                      F> T₁ ’> T₂’

glass has been rated on the mohs scale as 5.5