Q3.4-19E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

An object of mass 60 kg starts from rest at the top of a 45º inclined plane. Assume that the coefficient of kinetic friction is 0.05 (see Problem 18). If the force due to air resistance is proportional to the velocity of the object, say, -3, find the equation of motion of the object. How long will it take the object to reach the bottom of the inclined plane if the incline is 10 m long?

Step-by-Step Solution

Answer

• The equation of motion of an object is  $\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{131}\mathbf{.}\mathbf{8}\mathbf{(}\mathbf{t}\mathbf{+}\mathbf{20}{\mathbf{e}}^{\mathbf{-}\mathbf{1}\mathbf{/}\mathbf{20}}\mathbf{)}\mathbf{-}\mathbf{2636}$
• The object take to reach the bottom of inclined plan is 1.768 seconds, if the incline is 10 m 

Step1: Find the equation of motion of an object

There are two forces are 

 $$\begin{gathered} {\mathrm{F}}_1=\mathrm{mg} \sin {45}^{\mathrm{o}} \\ {\mathrm{F}}_2=-\mu \mathrm{mg} \cos {45}^{\mathrm{o}} \\ {\mathrm{F}}_3=-3v \end{gathered}$$

Now

$$\begin{gathered} \mathrm{m}\frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{mg} \sin {45}^{\mathrm{o}}-\mu \mathrm{mg} \cos {45}^{\mathrm{o}}-3v \\ \frac{\mathrm{dv}}{\mathrm{dt}}=6.57-\frac{v}{20} \\ \int \frac{20 \mathrm{dv}}{131.8-v}=\int \mathrm{dt} \\ \int \frac{20 \mathrm{dv}}{131.8-v}=\mathrm{t}+\mathrm{C} \end{gathered}$$

Step 2: Find ∫ 20   dv 131 . 8 - v .

Put $$\begin{gathered} \mathrm{|}\mathrm{131}\mathrm{.}\mathrm{8}\mathrm{-}\mathrm{v}\mathrm{=}\mathrm{t} \\ \mathrm{dv}\mathrm{=}\mathrm{-}\mathrm{dt}\mathrm{|} \end{gathered}$$ then

$$\begin{gathered} \int \frac{20 \mathrm{dv}}{131.8-v}=-20\int \frac{\mathrm{dt}}{\mathrm{t}} \\ =-20 \ln \left|\mathrm{t}\right| \\ =-20 \ln \left|131.8-v\right| \\ =-20 \ln \left|131.8-v\right|=\mathrm{t}+\mathrm{C} \\ =\ln \left|131.8-v\right|=-\frac{\mathrm{t}}{20}+\mathrm{C} \\ \mathrm{v}\left(\mathrm{t}\right)=131.8-{\mathrm{Ce}}^{-1/20} \end{gathered}$$

When v(t_o) = 0 then C = 131.8

$\mathrm{v}\left(\mathrm{t}\right)=131.8(1-{\mathrm{e}}^{\frac{-1}{20}})$

Step 3: Find the value of x(t)

$$\begin{gathered} \mathrm{x}\left(\mathrm{t}\right)=\int 131.8(1-{\mathrm{e}}^{\frac{-1}{20}})\mathrm{dt} \\ \mathrm{x}\left(\mathrm{t}\right)=131.8(\mathrm{t}+20{\mathrm{e}}^{\frac{-1}{20}})+\mathrm{C} \end{gathered}$$

When x(0) = 0 then C=-2636 so

$\mathrm{x}\left(\mathrm{t}\right)=131.8(\mathrm{t}+20{\mathrm{e}}^{\frac{-1}{20}})-2636$

Hence, the equation of motion of an object is $\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{131}\mathbf{.}\mathbf{8}\mathbf{(}\mathbf{t}\mathbf{+}\mathbf{20}{\mathbf{e}}^{\mathbf{-}\mathbf{1}\mathbf{/}\mathbf{20}}\mathbf{)}\mathbf{-}\mathbf{2636}$

Step 4: Find the value of time

When x(t) = 10 then $10=131.8(\mathrm{t}+20{\mathrm{e}}^{\frac{-1}{20}})-2636$

By solving it, we obtain t = 1.768 sec.

 

Hence, the object takes to reach the bottom of inclined plan is 1.768 seconds, if the incline is 10 m.