Q3.4-18E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

When an object slides on a surface, it encounters a resistance force called friction. This force has a magnitude of $\mathbf{\mu N}$ , where $\mathbf{\mu }$  the coefficient of kinetic friction and N is the magnitude of the normal force that the surface applies to the object. Suppose an object of mass  30 kg is released from the top of an inclined plane that is inclined  $\mathbf{30}\mathbf{°}$ to the horizontal (see Figure 3.11). Assume the gravitational force is constant, air resistance is negligible, and the coefficient of kinetic friction $\mathbf{\mu }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2}$ . Determine the equation of motion for the object as it slides down the plane. If the top surface of the plane is 5 m long, what is the velocity of the object when it reaches the bottom?

Step-by-Step Solution

Answer

The velocity of the object when it reaches the bottom is $V\left(t\right)=5.66m/\sec$  and $x\left(t\right)=5.66t+c$ .

Step1: Find the value of velocity

There are two forces are written as,

 $$\begin{gathered} F_1=mg \sin {30}^o \\ {F}_2=-\mu mg \cos {30}^o \end{gathered}$$

 

Now put the given values then;

$$\begin{gathered} m\frac{dv}{dt}=mgsin{30}^o-\mu mgcos{30}^o \\ \frac{dv}{dt}=gsin{30}^o-\mu gcos{30}^o \left(\mathrm{Puttingthevaluesof} \mathrm{g}=0.2\right) \\ \frac{dv}{dt}=3.207 \end{gathered}$$

Since $v\left(t\right)=V\left(x\left(t\right)\right)$ ,

 

So, the values are written as;

$$\begin{gathered} \frac{dv}{dt}=V\frac{dV}{dx} \\ V\frac{dV}{dx}=3.207 \end{gathered}$$

 

 

Step 2: Find the value of velocity by limits.

Now, find the value of velocity then,

$$\begin{gathered} \int VdV=\underset{0}{\overset{5}{\int }}3.207dx \\ \frac{V^2}{2}-{\left.3.207x\right|}_0^5=0 \left(\mathbf{Integratewithlimits}\mathbf{0}\mathbf{to}\mathbf{5}\right) \\ {V}^2=32.07 \\ V\left(t\right)=5.66m/\sec \end{gathered}$$

Step 3: Evaluate the equation of motion.

Now, for the value of  x(t),

 $$\begin{gathered} v=5.66 \\ \frac{dx}{dt}=5.66 \\ x \left(t\right)=5.66 t+c \end{gathered}$$

 

 

Therefore, the results are  $x\left(t\right)=5.66 t+c$ and  $v\left(t\right)=5.66 m/\sec$.