Question

Sliding Box: (a) A box of mass m slides down an inclined plane that makes an angle $\mathit{\theta }$with the horizontal as shown in Figure 3.1.16. Find a differential equation for the velocity$\mathit{v}\mathbf{\left(}\mathit{t}\mathbf{\right)}$of the box at time t in each of the following three cases:

(i) No sliding friction and no air resistance

(ii) With sliding friction and no air resistance

(iii) With sliding friction and air resistance

In cases (ii) and (iii), use the fact that the force of friction opposing the motion of the box is $\mathit{\mu }\mathit{N}$ where $\mathit{\mu }$ the coefficient of sliding friction is and N is the normal component of the weight of the box. In case (iii) assume that air resistance is proportional to the instantaneous velocity.

(b) In part (a), suppose that the box weighs 96 pounds, that the angle of inclination of the plane is $\mathit{\theta }\mathbf{=}\mathbf{30}\mathbf{°}$, that the coefficient of sliding friction is $\mathit{\mu }\mathbf{=}\frac{\sqrt{\mathbf{3}}}{\mathbf{4}}$, and that the additional retarding force due to air resistance is numerically equal to $\frac{\mathbf{1}}{\mathbf{4}}\mathit{v}$. Solve the differential equation in each of the three cases, assuming that the box starts from rest from the highest point 50 ft above ground.

Short answer

a) i) $\frac{dv}{dt}=g\sin \theta$

ii) $\frac{dv}{dt}=g(in\theta -\mu \cos \theta )$

iii) $\frac{dv}{dt}=g(\sin \theta -\mu \cos \theta )-\frac{k}{m}v$

b) i) $v\left(t\right)=\frac{g}{2}t$

ii) $v\left(t\right)=\frac{g}{8}t$

iii)$v\left(t\right)=48g-48g{e}^{-\frac{1}{384}t}$

Step-by-step solution

No sliding friction and no air resistance

a) We have a box of mass m is sliding down on an inclined plane with an angle $\theta$with the horizontal zone as shown in the figure, and we have to write the differential equation for the velocity$v\left(t\right)$ for the given cases according to Newton’s law as

i) With no sliding friction and no air resistance, we have

$m\frac{dv}{dt}=mg\sin \theta$

$\frac{dv}{dt}=g\sin \theta$------------- (1)

With sliding friction and no air resistance

ii) With sliding friction and no air resistance, we have

$m\frac{dv}{dt}=mg\sin \theta -\mu N$

Which is,

$\begin{array}{rcl}m\frac{dv}{dt}& =& mg\sin \theta -\mu \left(mg\cos \theta \right)\\ \frac{dv}{dt}& =& g\sin \theta -\mu g\cos \theta \\ \frac{dv}{dt}& =& g(in\theta -\mu \cos \theta )......\left(2\right)\end{array}$

With sliding friction and with air resistance

iii) With sliding friction and air resistance, where air resistance is proportional velocity we have

$m\frac{dv}{dt}=mg\sin \theta -\mu N-kv$

Which is

$\begin{array}{rcl}m\frac{dv}{dt}& =& mg\sin \theta -\mu \left(mg\cos \theta \right)-kv\\ m\frac{dv}{dt}& =& mg\sin \theta -\mu mg\cos \theta -kv\\ m\frac{dv}{dt}& =& mg(\sin \theta -\mu \cos \theta )-kv\\ \frac{dv}{dt}& =& g(\sin \theta -\mu \cos \theta )-\frac{k}{m}v......\left(3\right)\end{array}$

Assume the value θ=30° in No sliding friction and no air resistance and find the velocity 

b) i) Since we have $\theta =30°$, then by substituting into equation(1), we have

$$\begin{gathered} \frac{dv}{dt}=g\sin 30° \\ \frac{dv}{dt}=\frac{g}{2} \end{gathered}$$

This is a first order linear and separable differential equation, and we can solve it as

$\begin{array}{rcl}dv& =& \frac{g}{2}dt\\ \int dv& =& \frac{g}{2}\int dt\\ v\left(t\right)& =& \frac{g}{2}t+c\end{array}$

After that, to find the value of constant c, we have apply the point of condition $(v,t)=(0,t)$into equation (a), then we have,

$0=\frac{g}{2}\left(0\right)+c$

Then we have

$c=0$

After that, substitute with the value of constant into equation (a), then we have,

$v\left(t\right)=\frac{g}{2}t$, is the velocity of the box with no sliding friction and no air resistance at time t.

Assign the value θ=30°  and μ=34 in With sliding friction and no air resistance and find the velocity 

ii) Since we have $\theta =30°$and $\mu =\frac{\sqrt{3}}{4}$, then by substituting into equation (2), we have

$$\begin{gathered} \frac{dv}{dt}=g(\sin 30-\frac{\sqrt{3}}{4}\cos 30) \\ \frac{dv}{dt}=g\left(\frac{1}{2}-\frac{\sqrt{3}}{4}\times \frac{\sqrt{3}}{2}\right) \\ \frac{dv}{dt}=g\left(\frac{4}{8}-\frac{3}{8}\right) \\ \frac{dv}{dt}=\frac{g}{8} \end{gathered}$$

This is a first order linear and separable differential equation, and we can solve it as

$\begin{array}{rcl}dv& =& \frac{g}{8}dt\\ \int dv& =& \frac{g}{8}\int dt\\ v\left(t\right)& =& \frac{g}{8}t+h\end{array}$

After that, to find the value of constant h, we have apply the point of condition $(v,t)=(0,t)$into equation (a), then we have,

$0=\frac{g}{2}\left(0\right)+h$

Then we have

$h=0$

After that, substitute with the value of constant h into equation (a), then we have,

$v\left(t\right)=\frac{g}{8}t$, is the velocity of the box with sliding friction at time t.

Assign the valuem=96Ib,θ=30°and μ=34 in With sliding friction and no air resistance and find the velocity

iii) Since we have $m=96Ib,\theta =30°,\mu =\frac{\sqrt{3}}{4}$and $k=\frac{1}{4}$, then by substituting into equation (3), we have

$$\begin{gathered} \frac{dv}{dt}=g(\sin 30-\frac{\sqrt{3}}{4}\cos 30)-\frac{1}{4\times 96}v \\ \frac{dv}{dt}=g\left(\frac{1}{2}-\frac{\sqrt{3}}{4}\times \frac{\sqrt{3}}{2}\right)-\frac{1}{384}v \\ \frac{dv}{dt}=g\left(\frac{4}{8}-\frac{3}{8}\right)-\frac{1}{384}v \\ \frac{dv}{dt}=\frac{g}{8}-\frac{1}{384}v \end{gathered}$$

This is a first order linear and separable differential equation, and we can solve it as

$\begin{array}{rcl}\frac{dv}{\frac{g}{8}-\frac{1}{384}v}& =& dt\\ -384\int \frac{\frac{1}{384}}{\frac{g}{8}-\frac{1}{384}v}dv& =& \int dt\\ -384In\left(\frac{g}{8}-\frac{1}{384}v\right)& =& t+h_1\\ In\left(\frac{g}{8}-\frac{1}{384}v\right)& =& -\frac{1}{384}t+h_2\\ e^{In\left(\frac{g}{8}-\frac{1}{384}v\right)}& =& e^{-\frac{1}{384}t+h_2}\\ \frac{g}{8}-\frac{1}{384}v& =& h{e}^{-\frac{1}{384}t}\\ \frac{1}{384}v& =& \frac{g}{8}-h{e}^{-\frac{1}{384}t}\end{array}$

Then we have

$\begin{array}{rcl}v\left(t\right)& =& 384\times \frac{g}{8}-384h{e}^{-\frac{1}{384}t}\\ & =& 48g-c{e}^{-\frac{1}{384}t}\end{array}$

After that, to find the value of constant c, we have apply the point of condition $(v,t)=(0,t)$into equation (c), then we have,

$0=48g-c{e}^0$

Then we have

$c=48g$

After that, substitute with the value of constant into equation (b), then we have,

$v\left(t\right)=48g-48g{e}^{-\frac{1}{384}t}$, is the velocity of the box with sliding friction and air resistance at time t.