Q. 90A - Calculus

Question

Ian is a bit worried about taking a fall into a crevasse while carrying a heavy pack and towing a heavy sled. He does some tests on an old rope, dropping from a tree in his backyard as shown below. 

He cannot measure the force on the rope, but he takes a video, from which he can find his position (in feet below the tree limb) at time $t>1.2$ seconds as

$x\left(t\right)=20+7e^{=0.25t}\sin (4.7t-5.8)$

(a) Weight is a force, given by mass times acceleration. Ian weighs $160$ pounds, and the acceleration due to gravity that causes his weight is $32$ feet per square second. What is Ian’s mass? (The units are called “slugs.”)

(b) Recalling that acceleration is the second derivative of position $x\left(t\right)$, what is the force on Ian at any time $t>1.2$?

(c) Use a graphing calculator or other graphing utility to make a graph of Ian’s acceleration over time. When is the upward force on Ian the greatest? Note that since we are measuring distance below the tree limb, in this situation an upward force is negative. What is that force? 

Step-by-Step Solution

Answer

Part (a) The mass of The person I is $7.4$ in kilogram and $16.4$ in pounds.

Part (b) $F\left(t\right)=-7.4 [154.2e^{-0.25t}\sin (4.7t-5.8)+16.5e^{-0.25t}\cos (4.7t-5.80]$ in Newton.

Part (c) Graph

Part (a) Step 1. Given information

The position on the rope is given by $x\left(t\right)=20+7e^{-0.25t}\sin (4.7t-5.8)$ at time $t>1.2$ seconds. The force on rope is not measured. The person I’s weight $160$ pounds, and acceleration due to gravity $32$ feet per square second.

Part (a) Step 2. Calculation

Let us take the weight of The person I as and mass of The person I as .

$$\begin{gathered} M=72.57 kilogram=160 pounds, \\ g=32 feet per square meter=9.7536 meters per square meter \end{gathered}$$

Since weight is a force given by mass times acceleration.

$$\begin{gathered} F=M=mg \\ m=\frac{M}{g} \\ m=\frac{72.57}{9.7536} \\ m=7.44 \end{gathered}$$

Thus, the mass of The person I is $7.4$ in Kilogram and $16.4$ pounds.

Part (b) Step 1. Calculation

Consider the acceleration which is the second derivative of $x\left(t\right).$

$$\begin{gathered} x\left(t\right)=20+7e^{-0.25t}\sin (4.7t-5.8) \\ x\text{'}\left(t\right)=7e^{-0.25t}[4.7\cos (4.7t-5.8\left)\right]+7(-0.25)\sin (4.7t-5.8)e^{-0.25t} \\ x\text{'}\left(t\right)=-7{(4.7)}^2{e}^{-0.25t}\sin (4.7t-5.8)-7(4.7)(0.25)\cos (4.7t-5.8)e^{-0.25t}-7(0.25)(4.7)e^{-0.25}\cos (4.7t-5.8)+7{(0.25)}^2\sin (4.7t-5.8)e^{-0.25} \\ x\text{'}\left(t\right)=-154.63e^{-0.25t}\sin (4.7t-5.8)-8.225\cos (4.7t-5.8)e^{-0.25t}-8.225e^{-0.25t}\sin (4.7t-5.8)[-154.63+0.4375]+e^{-0.25t}\cos (4.7t-5.8)[-8.225-8.225] \\ x\text{'}\left(t\right)=-154.2e^{-0.25t}\sin (4.7t-5.8)-16.5e^{-0.25t}\cos (4.7-5.8) \end{gathered}$$

Then, the force on The person I at any time is given by mass times acceleration. i.e.

$$\begin{gathered} F\left(t\right)=mx\text{'}\left(t\right) \\ F\left(t\right)=7.4[-154.2e^{-0.25t}\sin (4.7t-5.8)-16.5e^{-0.25t}\cos (4.7t-5.80] \end{gathered}$$

Part (c) Step 1. Calculation

Consider the acceleration $x\text{'}\text{'}\left(t\right)=154.2e^{-0.25t}\sin (4.7t-5.8)+16.5e^{-0.25t}\cos (4.7-5.8)$

This is positive towards the ground.

Now consider the upward force, which is negative towards the ground.

$F\left(t\right)=[154.2e^{-0.25t}\sin (4.7t-5.8)+16.5e^{-0.25t}\cos (4.7-5.8\left)\right]$

When $t>0,$

The first local maximum point of the acceleration is obtained by the point

$(t,x\text{'}\text{'}(t\left)\right)=(0.197,147.404)$

When $t>12$

The second local maximum point of the acceleration is obtained by the point 

$(t,x\text{'}\text{'}(t\left)\right)=(1.534,105.527)$

And so on.

Hence, the greatest upward force is obtained by the point where the acceleration is maximum for time $t>1.2 i.e. (1.534,105.527)$

$$\begin{gathered} F\left(t\right)=7.4[-154.2e^{-0.25t}\sin (4.7t-5.8)-16.5e^{-0.25t}\cos (4.7t-5.8\left)\right] \\ F\left(t\right)=-7.4(105.527) \\ F\left(t\right)=-780.9 \end{gathered}$$

Thus, Maximum upward force$F\left(t\right)=-780.9$ newton.