SOLUTION
The vertical component of motion is only considered since that is the component that changes due to the action of gravity.
The equation of motion in the vertical \(y\) direction is given by \(F=ma\). Hence\begin{align*} my^{\prime \prime } & =-mg\\ y^{\prime \prime } & =-g \end{align*}
Integrating once gives\[ y^{\prime }-y^{\prime }\left ( 0\right ) =-gt \] Where \(y^{\prime }\left ( 0\right ) =v_{0}\sin (\alpha )\). The time for the particle to reach a final velocity of zero in the vertical direction is now find by solving the above for \(t\) \[ y_{\text{final}}^{\prime }=y^{\prime }\left ( 0\right ) -gt \] Where \(y_{\text{final}}^{\prime }=0\). Solving the above for the time \(t\) gives \begin{align*} 0 & =v_{0}\sin \alpha -gt\\ t & =\frac{v_{0}\sin \alpha }{g} \end{align*}
Hence the total time to reach back to its starting point is twice the above time, which is\[ \fbox{total time $=2\left ( \frac{v_0\sin \alpha }{g}\right ) $}\] For \(\alpha =20\) degree and \(v_{0}=2.4 \frac{m}{s}\) and \(g=9.81 \frac{m}{s^2}\), the total time is found from \begin{align*} \text{total time} & =2\left ( \frac{2.4\sin 20^o }{9.81}\right ) \\ & =\boxed{0.167 \hspace{2pt}\text{second}} \end{align*}
SOLUTION
The free body diagram is shown below for each mass.
The acceleration of each body is the same. Let this acceleration be \(a\). From the above free body diagram of the \(2m\) body the equation of motion is now derived (using positive direction as show)\begin{align*}{\displaystyle \sum } F & =2ma\\ 2mg\sin \theta -\mu _{k}F_{N}-T & =2ma \end{align*}
Using \(F_{N}=2mg\cos \theta \) the above becomes\begin{equation} 2mg\sin \theta -\mu _{k}2mg\cos \theta -T=2ma \tag{1} \end{equation} The tension \(T\) in the string is found from the free body diagram of the smaller hanging mass since the tension \(T\) is same. From the free body diagram of the small mass the equation of motion is\begin{align*}{\displaystyle \sum } F & =ma\\ -mg+T & =ma \end{align*}
Hence \[ T=m\left ( a+g\right ) \] Substituting \(T\) in (1) gives\begin{align*} 2mg\sin \theta -\mu _{k}2mg\cos \theta -m\left ( a+g\right ) & =\left ( 2m\right ) a\\ 2g\sin \theta -\mu _{k}2g\cos \theta -g & =3a \end{align*}
Therefore\[ a=\frac{2}{3}\left ( g\sin \theta -\mu _{k}g\cos \theta -\frac{1}{2}g\right ) \] For constant speed, \(a=0\) at some angle \(\theta _{c}\). The above reduces to \begin{align} \frac{2}{3}\left ( g\sin \theta _{c}-\mu _{k}g\cos \theta _{c}-\frac{1}{2}g\right ) & =0\nonumber \\ \sin \theta _{c}-\mu _{k}\cos \theta _{c}-\frac{1}{2} & =0\nonumber \\ \sin \theta _{c}-\mu _{k}\cos \theta _{c} & =\frac{1}{2} \tag{2} \end{align}
To solve this, the following identity is used\begin{equation} R\sin \left ( \theta _{c}+\alpha \right ) =R\left ( \sin \theta _{c}\cos \alpha +\cos \theta _{c}\sin \alpha \right ) \tag{3} \end{equation} Comparing the RHS of (3) with the LHS of (2) gives\begin{align} R\cos \alpha & =1\tag{4}\\ R\sin \alpha & =-\mu _{k} \tag{5} \end{align}
Dividing (5) by (4) gives \(\tan \alpha =\) \(-\mu _{k}\) or \[ \alpha =\tan ^{-1}\left ( -\mu _{k}\right ) =-\tan ^{-1}\left ( \mu _{k}\right ) \] Squaring (4) and (5) and adding gives\begin{align*} R^{2}\cos ^{2}\alpha +R^{2}\sin ^{2}\alpha & =1+\mu _{k}^{2}\\ R & =\sqrt{1+\mu _{k}^{2}} \end{align*}
Therefore the equation \(R\sin \left ( \theta +\alpha \right ) =\frac{1}{2}\) becomes\begin{align*} \sqrt{1+\mu _{k}^{2}}\sin \left ( \theta _{c}+\alpha \right ) & =\frac{1}{2}\\ \sin \left ( \theta _{c}-\tan ^{-1}\left ( \mu _{k}\right ) \right ) & =\frac{1}{2\sqrt{1+\mu _{k}^{2}}}\\ \theta _{c}-\tan ^{-1}\left ( \mu _{k}\right ) & =\sin ^{-1}\left ( \frac{1}{2\sqrt{1+\mu _{k}^{2}}}\right ) \end{align*}
Therefore
\[ \theta _{c}=\sin ^{-1}\left ( \frac{1}{2\sqrt{1+\mu _{k}^{2}}}\right ) +\tan ^{-1}\left ( \mu _{k}\right ) \] For the case of no friction, where \(\mu _{k}=0\) the above gives \(\theta _{c}=\sin ^{-1}\left ( \frac{1}{2}\right ) =30^{0}\). As \(\mu _{k}\) increases, the angle \(\theta _{c}\) will increase. (in the limit, as \(\mu _{k}\rightarrow \infty ,\theta _{c}\rightarrow 90^{0}\)). This is a plot showing how the angle changes as \(\mu _{k}\) increases.
SOLUTION
Looking at the case where the small mass \(m\) is not moving (not sliding down the side), and considering both \(M+m\) as one body. Let the horizontal acceleration of both bodies be \(a\)
\begin{align} \sum F_{x} & =(M+m)a\nonumber \\ F-\mu _{k}(M+m)g & =(M+m)a\nonumber \\ a & =\frac{F-\mu _{k}(M+m)g}{(M+m)} \tag{1} \end{align}
The small mass \(m\) is now considered. The static friction force between \(m\) and \(M\) has to be larger than the weight \(mg\) so that \(m\) does not move and fall. This implies \(f_{s_{\max }}=\mu _{s}N\) must be larger than the weight \(mg\)
This implies the following condition is required \begin{equation} \mu _{s}N\geq mg \tag{2} \end{equation} Where \(N\) is the normal force on \(m\). But \[ ma=N \] From (1) we find \[ N=m\left ( \frac{F-\mu _{k}(M+m)g}{(M+m)}\right ) \] Therfore (2) becomes\[ \mu _{s}m\left ( \frac{F-\mu _{k}(M+m)g}{(M+m)}\right ) \geq mg \] Hence\begin{align*} F-\mu _{k}(M+m)g & \geq \frac{g}{\mu _{s}}(M+m)\\ F & \geq \frac{g}{\mu _{s}}(M+m)+\mu _{k}(M+m)g\\ & \geq (M+m)g\left ( \frac{1}{\mu _{s}}+\mu _{k}\right ) \end{align*}
Hence\[ \fbox{$F\geq (M+m)g\left ( \frac{1+\mu _s\mu _k}{\mu _s}\right ) $}\]
SOLUTION
From the free body diagram
The equation of motion is \begin{align*} \sum F_{y} & =my^{\prime \prime }\\ mg-(c_{1}y^{\prime }+c_{2}(y^{\prime 2}) & =my^{\prime \prime } \end{align*}
At the terminal velocity the body is not accelerating. Setting \(y^{\prime \prime }=0\) in the above gives an equation to solve for the terminal velocity (where now \(y^{\prime }\) is written as \(v_{t}\)) \begin{align*} mg-(c_{1}v_{t}+c_{2}v_{t}^{2}) & =0\\ c_{2}v_{t}^{2}+c_{1}v_{t}-mg & =0 \end{align*}
This is a quadratic equation in \(v_{t}\), hence the roots are given by \begin{align*} v_{t} & =\frac{-b}{2a}\pm \frac{\sqrt{b^{2}-4ac}}{2a}\\ & =\frac{-c_{1}}{2c_{2}}\pm \frac{\sqrt{c_{1}^{2}+4c_{2}mg}}{2c_{2}}\\ & =\frac{-c_{1}}{2c_{2}}\pm \sqrt{\left ( \frac{c_{1}}{2c_{2}}\right ) ^{2}+\frac{mg}{c_{2}}} \end{align*}
Since the terminal velocity \(v_{t}\) has to be positive as indicated in the diagram above, then the solution is the positive root given by\[ \fbox{$v_t=\frac{-c_1}{2c_2}+\sqrt{\left ( \frac{c_1}{2c_2}\right ) ^2+\frac{mg}{c_2}}$}\]
SOLUTION
The following is the free body diagram used to solve this problem.
In the vertical direction, with positive taken upwards as shown in the diagram, the equation of motion is given by\begin{align}{\displaystyle \sum } F_{y} & =my^{\prime \prime }\nonumber \\ -mg-kmy^{\prime } & =my^{\prime \prime }\nonumber \\ y^{\prime \prime }+ky^{\prime } & =-g \tag{1} \end{align}
In the horizontal direction, the equation of motion is\begin{align}{\displaystyle \sum } F_{x} & =mx^{\prime \prime }\nonumber \\ -kmx^{\prime } & =mx^{\prime \prime } \tag{2} \end{align}
The initial conditions for equation of motion in the vertical direction are \(y\left ( 0\right ) =0,y^{\prime }\left ( 0\right ) =v_{0}\sin \alpha \) and the initial conditions for the equation of motion in the horizontal direction are \(x\left ( 0\right ) =0,x^{\prime }\left ( 0\right ) =v_{0}\cos \alpha \).
Equation (1) is now solved. The characteristic equation is \(\lambda ^{2}+k\lambda =0\) or \(\lambda \left ( \lambda +k\right ) =0\), hence the roots are \(\lambda =0,\lambda =-k\), and therefore the homogeneous solution is\[ y_{h}\left ( t\right ) =A+Be^{-kt}\] The particular solution is now found. Let \(y_{p}\left ( t\right ) =ct\) where \(c\) is some constant. Substituting this into (1) gives\begin{align*} kc & =-g\\ c & =\frac{-g}{k} \end{align*}
Hence the particular solution is \(y_{p}\left ( t\right ) =-\frac{g}{k}t\,\), and the complete solution in the \(y\) direction is\begin{align*} y\left ( t\right ) & =y_{h}\left ( t\right ) +y_{p}\left ( t\right ) \\ & =\left ( A+Be^{-kt}\right ) -\frac{g}{k}t \end{align*}
The initial conditions are now applied to determine the constants \(A,B\). (Initial conditions must be used in the complete solution and not the homogeneous solution). When \(t=0\), \(y\left ( 0\right ) =0\) and the above gives\[ A=-B \] Since \(y^{\prime }\left ( t\right ) =-Bke^{-kt}-\frac{g}{k}\) and since \(y^{\prime }\left ( 0\right ) =v_{0}\sin \alpha \), then at \(t=0\)\begin{align*} v_{0}\sin \alpha & =-Bk-\frac{g}{k}\\ B & =-\left ( \frac{g}{k^{2}}+\frac{v_{0}\sin \alpha }{k}\right ) \end{align*}
Using values for the constants \(A,B\), the complete solution for equation of motion in the vertical direction becomes\begin{align} y\left ( t\right ) & =\left ( A+Be^{-kt}\right ) -\frac{g}{k}t\nonumber \\ & =\left ( \frac{g}{k^{2}}+\frac{v_{0}\sin \alpha }{k}\right ) -\left ( \frac{g}{k^{2}}+\frac{v_{0}\sin \alpha }{k}\right ) e^{-kt}-\frac{g}{k}t\nonumber \end{align}
Hence\begin{equation} y\left ( t\right ) =\left ( \frac{g+kv_{0}\sin \alpha }{k^{2}}\right ) \left ( 1-e^{-kt}\right ) -\frac{g}{k}t \tag{3} \end{equation} The duration time \(T\) is now found by solving for \(y=0\) from (3). Hence\begin{align} 0 & =\left ( \frac{g+kv_{0}\sin \alpha }{k^{2}}\right ) \left ( 1-e^{-kT}\right ) -\frac{g}{k}T\nonumber \\ T & =\left ( \frac{g+kv_{0}\sin \alpha }{gk}\right ) \left ( 1-e^{-kT}\right ) \tag{4} \end{align}
An analytical solution based on perturbation method for this is given in the text book at page 67 as\[ \fbox{$T\simeq \frac{2v_0\sin \alpha }{g}\left ( 1-\frac{kv_0\sin \alpha }{3g}\right ) $}\] However in this solution equation (4) was solved numerically instead for \(T\) for the numerical values given in this problem, and the results are summarized on the following table
| \(k\) | \(T\) (sec) |
| \(0.001\) | \(50.5427\) |
| \(0.01\) | \(47.2597\) |
| \(0.1\) | \(34.3395\) |
The equation of motion in the \(x\) direction is now solved. This equation is given above in (2) as \(x^{\prime \prime }+kx^{\prime }=0\). The characteristic equation is \(\lambda ^{2}+k\lambda =0\) or \(\lambda \left ( \lambda +k\right ) =0\), hence the roots are \(\lambda =0,\lambda =-k\), and therefore, the homogeneous solution is\[ x_{h}\left ( t\right ) =A+Be^{-kt}\] Since there is no forcing function, the complete solution is the same\begin{equation} x\left ( t\right ) =A+Be^{-kt} \tag{5} \end{equation} The constants are found from the initial conditions. At \(t=0\)\begin{align*} 0 & =A+B\\ A & =-B \end{align*}
Since \(x^{\prime }\left ( t\right ) =-Bke^{-kt}\), then at \(t=0\)\begin{align*} v_{0}\cos \alpha & =-Bk\\ B & =\frac{-v_{0}\cos \alpha }{k} \end{align*}
Substituting the above values for \(A,B\) into (4) gives the solution for the motion in the horizontal direction\begin{equation} x\left ( t\right ) =\frac{v_{0}\cos \alpha }{k}\left ( 1-e^{-kt}\right ) \tag{6} \end{equation}
The following shows the projectile path for each different \(k\) value.
From the above, an estimate of the range for each \(k\) is given in the following table
| \(k\) | range (meters) |
| \(0.001\) | \(21500\) |
| \(0.01\) | \(16500\) |
| \(0.1\) | \(4100\) |
SOLUTION
\(f\left ( x\right ) =\cos \left ( x\right ) \) about \(x=0\)\begin{align*} f\left ( x\right ) & \simeq f\left ( 0\right ) +hf^{\prime }\left ( 0\right ) +\frac{1}{2}h^{2}f^{\prime \prime }\left ( 0\right ) +\frac{1}{3!}h^{3}f^{\prime \prime \prime }\left ( 0\right ) +\frac{1}{4!}h^{4}f^{\left ( 4\right ) }\left ( 0\right ) +\cdots \\ & =\cos \left ( 0\right ) +x\left ( -\sin \left ( 0\right ) \right ) +\frac{1}{2}x^{2}\left ( -\cos \left ( 0\right ) \right ) +\frac{1}{6}x^{3}\sin \left ( 0\right ) +\frac{1}{24}x^{4}\cos \left ( 0\right ) +\cdots \\ & =1-\frac{1}{2}x^{2}+\frac{1}{24}x^{4}-\cdots \end{align*}
\(f\left ( x\right ) =\cosh \left ( x\right ) \) about \(x=0\)\begin{align*} f\left ( x\right ) & \simeq f\left ( 0\right ) +hf^{\prime }\left ( 0\right ) +\frac{1}{2}h^{2}f^{\prime \prime }\left ( 0\right ) +\frac{1}{3!}h^{3}f^{\prime \prime \prime }\left ( 0\right ) +\frac{1}{4!}h^{4}f^{\left ( 4\right ) }\left ( 0\right ) +\cdots \\ & =\cosh \left ( 0\right ) +x\left ( \sinh \left ( 0\right ) \right ) +\frac{1}{2}x^{2}\left ( \cosh \left ( 0\right ) \right ) +\frac{1}{6}x^{3}\sinh \left ( 0\right ) +\frac{1}{24}x^{4}\cosh \left ( 0\right ) +\cdots \\ & =1+\frac{1}{2}x^{2}+\frac{1}{24}x^{4}+\cdots \end{align*}
\(f\left ( x\right ) =\ln \left ( x\right ) \) about \(x=2\)\begin{align*} f\left ( x\right ) & \simeq f\left ( 2\right ) +hf^{\prime }\left ( 2\right ) +\frac{1}{2}h^{2}f^{\prime \prime }\left ( 2\right ) +\frac{1}{3!}h^{3}f^{\prime \prime \prime }\left ( 2\right ) +\frac{1}{4!}h^{4}f^{\left ( 4\right ) }\left ( 2\right ) +\cdots \\ & =\ln \left ( 2\right ) +\left ( x-2\right ) \left ( \frac{1}{x}\right ) _{x=2}+\frac{1}{2}\left ( x-2\right ) ^{2}\left ( \frac{-1}{x^{2}}\right ) _{x=2}+\frac{1}{6}\left ( x-2\right ) ^{3}\left ( \frac{2}{x^{3}}\right ) _{x=2}+\frac{1}{24}\left ( x-2\right ) ^{4}\left ( \frac{-6}{x^{4}}\right ) _{x=2}+\cdots \\ & =\ln \left ( 2\right ) +\frac{x-2}{2}-\frac{1}{2}\frac{\left ( x-2\right ) ^{2}}{4}+\frac{1}{3}\frac{\left ( x-2\right ) ^{3}}{8}-\frac{1}{4}\frac{\left ( x-2\right ) ^{4}}{16}+\cdots \\ & =\ln \left ( 2\right ) +\frac{x-2}{2}-\frac{\left ( x-2\right ) ^{2}}{8}+\frac{\left ( x-2\right ) ^{3}}{24}-\frac{\left ( x-2\right ) ^{4}}{64}+\cdots \end{align*}
\(f\left ( x\right ) =\frac{1}{x^{2}}\) about \(x=-1\)\begin{align*} f\left ( x\right ) & \simeq f\left ( -1\right ) +hf^{\prime }\left ( -1\right ) +\frac{1}{2}h^{2}f^{\prime \prime }\left ( -1\right ) +\frac{1}{3!}h^{3}f^{\prime \prime \prime }\left ( -1\right ) +\frac{1}{4!}h^{4}f^{\left ( 4\right ) }\left ( -1\right ) +\cdots \\ & =1+\left ( x+1\right ) \left ( \frac{-2}{x^{3}}\right ) _{x=-1}+\frac{1}{2}\left ( x+1\right ) ^{2}\left ( \frac{6}{x^{4}}\right ) _{x=-1}+\frac{1}{6}\left ( x+1\right ) ^{3}\left ( \frac{-24}{x^{5}}\right ) _{x=-1}+\frac{1}{24}\left ( x+1\right ) ^{4}\left ( \frac{120}{x^{6}}\right ) _{x=-1}+\cdots \\ & =1+\left ( x+1\right ) \left ( \frac{-2}{-1}\right ) +\frac{1}{2}\left ( x+1\right ) ^{2}\left ( \frac{6}{1}\right ) +\frac{1}{6}\left ( x+1\right ) ^{3}\left ( \frac{-24}{-1}\right ) +\frac{1}{24}\left ( x+1\right ) ^{4}\left ( \frac{120}{1}\right ) +\cdots \\ & =1+2\left ( x+1\right ) +3\left ( x+1\right ) ^{2}+4\left ( x+1\right ) ^{3}+5\left ( x+1\right ) ^{4}+\cdots \end{align*}
\(f\left ( x\right ) =\sqrt{1+x}\) about \(x=0\)\begin{align*} f\left ( x\right ) & \simeq f\left ( 0\right ) +hf^{\prime }\left ( 0\right ) +\frac{1}{2}h^{2}f^{\prime \prime }\left ( 0\right ) +\frac{1}{3!}h^{3}f^{\prime \prime \prime }\left ( 0\right ) +\frac{1}{4!}h^{4}f^{\left ( 4\right ) }\left ( 0\right ) +\cdots \\ & =1+x\left ( \frac{1}{2\sqrt{1+x}}\right ) _{x=0}+\frac{1}{2}x^{2}\left ( \frac{-1}{4\left ( 1+x\right ) ^{\frac{3}{2}}}\right ) _{x=0}+\frac{1}{6}x^{3}\left ( \frac{3}{8\left ( 1+x\right ) ^{\frac{5}{2}}}\right ) _{x=0}+\frac{1}{24}x^{4}\left ( \frac{-15}{16\left ( 1+x\right ) ^{\frac{7}{2}}}\right ) _{x=0}+\cdots \\ & =1+\frac{x}{2}-\frac{x^{2}}{8}+\frac{x^{3}}{16}-\frac{5}{128}x^{4}+\cdots \end{align*}
