Question

Sliding Box-Continued: (a) In Problem 48 lets(t) be the distance measured down the inclined plane from the highest point. Use $\mathit{d}\mathit{s}\mathbf{/}\mathit{d}\mathit{t}\mathbf{=}\mathit{v}\mathbf{\left(}\mathit{t}\mathbf{\right)}$and the solution for each of the three cases in part (b) of Problem 48 to find the time that it takes the box to slide completely down the inclined plane. A root-finding application of a CAS may be useful here.

(b) In the case in which there is friction $\mathbf{(}\mathit{\mu }\mathbf{\ne }\mathbf{0}\mathbf{)}$but no air resistance, explain why the box will not slide down the plane starting from rest from the highest point above ground when the inclination angle $\mathit{\theta }$satisfies $\mathbf{tan}\mathit{\theta }\mathbf{\le }\mathit{\mu }$..

(c) The box will slide downward on the plane when $\mathbf{tan}\mathit{\theta }\mathbf{\le }\mathit{\mu }$if it is given an initial velocity $\mathit{v}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathit{v}}_{\mathbf{0}}\mathbf{>}\mathbf{0}$. Suppose that $\mathit{\mu }\mathbf{=}\raisebox{1ex}{$\sqrt{\mathbf{3}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{4}$}\right.$and $\mathit{\theta }\mathbf{=}\mathbf{23}\mathbf{°}$Verify that $\mathbf{tan}\mathit{\theta }\mathbf{\le }\mathit{\mu }$. How far will the box slide down the plane if ${\mathit{v}}_{\mathbf{0}}\mathbf{=}\mathbf{1}\mathbf{}\mathit{f}\mathit{t}\mathbf{/}\mathit{s}$?

(d) Using the values $\mathit{\mu }\mathbf{=}\raisebox{1ex}{$\sqrt{\mathbf{3}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{4}$}\right.$and $\mathit{\theta }\mathbf{=}\mathbf{23}\mathbf{°}$approximate the smallest initial velocity v0 that can be given to the box so that, starting at the highest point 50 ft above ground, it will slide completely down the inclined plane. Then find the corresponding time it takes to slide down the plane.

Short answer

(a) (i) $t=3.54$seconds. ii. $t=7.1$seconds, and iii. $t=10$seconds

(b) $v⩽0$, when $\tan \theta ⩽\mu$and the box will not slide down the plane.

(c) 1.992 feet

(d) $v_0=8.015ft/s$; $t=31.92$seconds.

Step-by-step solution

Find the distance measured down the inclined plane v(t)=g2t

a) i) From the solution of problem (38), for the case (1) we have

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

And we have the relation between the velocity of the box and the length of the inclined plane as

$\frac{ds}{dt}=v\left(t\right)$

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

$\begin{array}{rcl}\int ds& =& \int \frac{g}{2}tdt\\ s& =& \frac{g}{2}\times \frac{1}{2}{t}^2+k\\ s& =& \frac{32}{4}{t}^2+k\\ s& =& 8t^2+k......\left(1\right)\end{array}$

Now, to find the value of constant k, we have apply the point of condition$(s,t)=(0,0)$into equation (1), then we have,

$\begin{array}{rcl}0& =& 8\left(0\right)+k\\ k& =& 0\end{array}$

After that, substitute with the value of constant k into equation (1), then we have,

$s=8t^2$----------- (2)

Now, to find the time needed for the box to reach the ground, substitute with$s=50\sin 30$into equation (2), then we have,

$\begin{array}{rcl}\frac{50}{\sin 30}& =& 8t^2\\ \frac{50}{\frac{1}{2}}& =& 8t^2\\ 100& =& 8t^2\end{array}$

Then we have

$\begin{array}{rcl}t& =& \sqrt{\frac{100}{8}}\\ & \approx & 3.54\end{array}$

Seconds (required)

Find the distance measured down the inclined plane v(t)=g8t

ii. From the solution of problem (38), for the case (2) we have

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

And we have the relation between the velocity of the box and the length of the inclined plane as

$\frac{ds}{dt}=v\left(t\right)$

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

$\begin{array}{rcl}\int ds& =& \int \frac{g}{8}tdt\\ s& =& \frac{g}{8}\times \frac{1}{2}{t}^2+k\\ s& =& \frac{32}{16}{t}^2+k\\ s& =& 2t^2+k.......\left(3\right)\end{array}$

Now, to find the value of constant k, we have apply the point of condition$(s,t)=(0,0)$ into equation (1), then we have,

$\begin{array}{rcl}0& =& 2\left(0\right)+k\\ k& =& 0\end{array}$

After that, substitute with the value of constant k into equation (3), then we have,

$s=2t^2$------- (4)

Now, to find the time needed for the box to reach the ground, substitute with$s=50\sin 30$ into equation (4), then we have,

$\begin{array}{rcl}\frac{50}{\sin 30}& =& 8t^2\\ \frac{50}{\frac{1}{2}}& =& 8t^2\\ 100& =& 2t^2\end{array}$

Then we have

$\begin{array}{rcl}t& =& \sqrt{\frac{100}{2}}\\ & \approx & 7.1\end{array}$

Seconds (required)

Find the distance measured down the inclined plane v(t)=48g-48ge-1384t

iii. From the solution of problem (38), for the case (3) we have

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

And we have the relation between the velocity of the box and the length of the inclined plane as

$\frac{ds}{dt}=v\left(t\right)$

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

$\begin{array}{rcl}\int ds& =& \int \left(48g-48g{e}^{-\frac{1}{384}t}\right)dt\\ s& =& 48g\int dt-48g\int e^{-\frac{1}{384}t}dt\\ s& =& 48\times 32t-48\times 32\times \left(-384e^{-\frac{1}{384}t}\right)\\ s& =& 1,536t+589,824e^{-\frac{1}{384}t}+k......\left(5\right)\end{array}$

Now, to find the value of constant k, we have apply the point of condition $(s,t)=(0,0)$into equation (5), then we have,

$\begin{array}{rcl}0& =& 1,536\left(0\right)+589,824e^0+k\\ k& =& -589,824\end{array}$

After that, substitute with the value of constant into equation (5), then we have,

$s=1,536t+589,824e^{-\frac{1}{384}t}-589,824$------------ (6)

Now, to find the time needed for the box to reach the ground, substitute with $s=50\sin 30$into equation (6), then we have,

$\begin{array}{rcl}\frac{50}{\sin 30}& =& 1,536t+589,824e^{-\frac{1}{384}t}-589,824\\ 100& =& 1,536t+589,824e^{-\frac{1}{384}t}-589,824\\ 1,536t+589,824e^{-\frac{1}{384}t}-589,924& =& 0\\ t+384e^{-\frac{1}{384}t}-384.06& =& 0\end{array}$

Then we have

$t\approx 10$ seconds (required)

Explain why the box will not slide down the plane starting from rest from the highest point

b) For the second case from solution of problem (48) we have that D.E

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

With the initial conditions

$v\left(0\right)=0$

And we can apply why the box will not slide down the plane starting from rest for this case as the following:

We can solve and the simplify our given differential as

$\begin{array}{rcl}v& =& g\cos \theta \left(\frac{\sin \theta }{\cos \theta }-\mu \right)t\\ & =& g\cos \theta (\tan \theta -\mu )\end{array}$

Now if we substitute with $\mu =\tan \theta$into equation (7), we can have

$\begin{array}{rcl}v& =& g\cos \theta (\tan \theta -\tan \theta )t\\ & =& 0\end{array}$

And if we substitute with $\mu >\tan \theta$into equation (7), we can have

$\begin{array}{rcl}v& =& g\cos \theta (\tan \theta -(>\tan \theta \left)\right)t\\ & <& 0\end{array}$

Then, since we have the velocity of the box $v⩽0$, then the box will not slide down the plane starting from rest from the highest.

Verify that tanθ≤μ

c) First, since we have $\theta =23$and $\frac{\sqrt{3}}{4}$then we have to verify that $\tan \theta <\mu$as

$\tan 23=0.422447$------------ (a)

$\mu =\frac{\sqrt{3}}{4}=0.433$----------- (b)

Then from (a) and (b), we find that

$\tan 23<\mu$

After that,for the second case from solution of problem (48) we have that D.E

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

With the initial conditions

$v\left(0\right)=v_0=1ft/s$

And we can apply why the box will not slide down the plane starting from rest for this case as the following:

We can solve and the simplify our given differential as

$\begin{array}{rcl}v& =& g\cos \theta \left(\frac{\sin \theta }{\cos \theta }-\mu \right)t+v_0\\ & =& 32\times \cos 23\left(\tan 23-\frac{\sqrt{3}}{4}\right)t+1\\ & =& 29.46(0.42447-0.433)t+1\\ & =& -0.2511t+1......\left(8\right)\end{array}$

After that, we have to obtain the time at which the box reaches the ground by substitute with $v=0$into equation (8), we can have

$\begin{array}{rcl}0& =& -0.2511t+1\\ 0.2511t& =& 1\end{array}$

Then we have

$\begin{array}{rcl}t& =& \frac{1}{0.2511}\\ & =& 3.982\text{sec}\end{array}$

Find the far the box slide down the plane if v0=1 ft/s

Now, since we have the relation between the distance that the box slides and velocity of the box as

$\frac{ds}{dt}=v$

Then we have

$\begin{array}{rcl}s& =& \int (-0.2511t+1)dt\\ & =& -0.2511\times \frac{1}{2}{t}^2+t\\ & =& -0.1256t^2+t......\left(9\right)\end{array}$

After that, substitute with$t=3.982$ seconds into equation (9), then we have,

$\begin{array}{rcl}s& =& -0.1256(3.982{)}^2+3.982\\ & =& 1.992ft\end{array}$

(Required), is the distance that the box slides down on the plane for this case.

Find the corresponding time it takes to slide down the plane

d) Since we have from the previous point, we have the functions of the velocity and distance as

$v=-0.2511t\times v_{0}s=-0.1256t^2+v_{0}t$

Which are

$0=0.2511t+v_0$------- (a)

And

$\frac{50}{\sin 23}=-0.1256t^2+v_{0}t$

$127.97=-0.1256t^2+v_{0}t$------- (b)

Then from (a) and (b), we find that

$\begin{array}{rcl}1.99{v_0}^2& =& 127.97\\ {v_0}^2& =& 64.27\end{array}$

Then we have

$\begin{array}{rcl}{v}_0& =& \sqrt{64.27}\\ & =& 8.015ft/s\end{array}$, is the initial velocity of the box.

After that, substitute with the value of $v_0$into equation (a), then we have,

$\begin{array}{rcl}0& =& -0.2511t+8.015\\ 0.2511t& =& 8.015\end{array}$

Then we have

$t=\frac{8.015}{0.2511}$

$t=31.92$ seconds (required), is the time needed for the box to reach the ground.