Question

(a) In Example 4, how much of the chain would you intuitively expect the constant 5 -pound force to be able to lift?

(b) What is the initial velocity of the chain?

(c) Why is the time interval corresponding to $x\left(t\right)⩾0$given in Figure 5.3 .7 not the interval of definition of the solution (21)? Determine the interval . How much chain is actually lifted? Explain any difference between this answer and your prediction in part (a).

(d) Why would you expect to be a periodic solution?

Short answer

$$\begin{gathered} a)5\text{ft} \\ b)v\left(0\right)=4\sqrt{10}\text{ft}/\text{s} \\ c)0⩽t⩽\frac{3\sqrt{10}}{8}and{x}_{\max }=7.5\text{ft} \end{gathered}$$

Step-by-step solution

To find the pound force

(a) In Example 4 , you would intuitively expect the constant 5-pound force to be able to lift pounds of chain, that is, 5 feet (one feet of chain has one pound of weight).

To find the initial velocity of chain

The height of the chain at the time $t,x\left(t\right)$is given by

$x\left(t\right)=\frac{15}{2}-\frac{15}{2}{\left(1-\frac{4\sqrt{10}}{15}t\right)}^2$

The velocity is then

$v\left(t\right)=x^{\text{'}}\left(t\right)=-15\left(1-\frac{4\sqrt{10}}{15}t\right)·\frac{-4\sqrt{10}}{15}=4\sqrt{10}-\frac{160}{15}t$

Therefore, the initial velocity is

$v\left(0\right)=4\sqrt{10}\text{ft}/\text{s}$

To find the Time interval

c)

The time interval given in the Figure 5.3.7 is not the interval of definition of the solution. The time interval in the Figure 5.3 .7 is the graph of the solution where is non-negative. It also includes the interval where is decreasing.

But, we need to take the physical interpretation into consideration. The task said that one end of the chain is pulled vertically upward by constant force of 5 pounds. Because of that, we do not look at the part where is decreasing. Therefore, the interval of the solution is interval between zero and the time for which reaches its maximum point.

Let us determine the interval . For that, let us $t\ne 0$find such that From (1), you can see that this is equivalent to:

$$\begin{gathered} 1={\left(1-\frac{4\sqrt{10}}{15}t\right)}^2=1-\frac{8\sqrt{10}}{15}t+\frac{160}{225}{t}^2 \\ 0=\frac{4\sqrt{10}}{15}t\left(\frac{4\sqrt{10}}{15}t-2\right) \end{gathered}$$

The solution of the last equation, beside $t=0$, is $t=\frac{15}{2\sqrt{10}}=\frac{15\sqrt{10}}{20}=\frac{3\sqrt{10}}{4}$Since $x\left(t\right)$is actually translated quadratic function whose graph goes through the origin, its maximum point is exactly at half of $\frac{3\sqrt{10}}{4}$,which is $\frac{3\sqrt{10}}{8}$. Therefore, the interval $I$ is $0⩽t⩽\frac{3\sqrt{10}}{8}$To find maximum point, simply substitute $\frac{3\sqrt{10}}{8}$into (l):

$x\left(\frac{3\sqrt{10}}{8}\right)=\frac{15}{2}-\frac{15}{2}{\left(1-\frac{4\sqrt{10}}{15}·\frac{3\sqrt{10}}{8}\right)}^2=\frac{15}{2}$

Therefore, the chain is lifted

The difference in this answer and in prediction in part (a) exists because we assume, without thinking, that the force and the height have linear connection. That is not true. You can see it from the formula$F=ma$where is the force, is the mass and is the acceleration of the mass. Also, is given by$a=\frac{d^{2}x}{d{t}^2}$. That is not linear.

To solve the periodic solution

d)

We can expect$x\left(t\right)$ to be a periodic solution. Here is why.

The pound force will not be able to lift the chain more than feet. Then, the chain will go below the maximum height because of the momentum gained by the force. (We could also explain the last part of the part (c) using the momentum). Then, because the force is constant, the chain will again go upward to its maximum height, and so on.

Step 5:Final proof

$$\begin{gathered} a)5\text{ft} \\ b)v\left(0\right)=4\sqrt{10}\text{ft}/\text{s} \\ c)0⩽t⩽\frac{3\sqrt{10}}{8}and{x}_{\max }=7.5\text{ft} \end{gathered}$$