Problem 7.1.
Step 1: We parametrize the position of the body by defining the angle \(\theta\) (to define the position of \(B\) whose motion is circular) and the angle \(\phi\) (to define the orientation of the body). We define the fixed basis \((\boldsymbol{\hat{\imath}},\boldsymbol{\hat{\jmath}},\boldsymbol{\hat{k}})\) and the moving bases \((\boldsymbol{\hat{u}},\boldsymbol{\hat{v}})\) (\(d\boldsymbol{\hat{u}}/dt =\dot{\theta}\boldsymbol{\hat{v}}\)) and \((\boldsymbol{\hat{u}}_1, \boldsymbol{\hat{v}}_1)\) (\(d\boldsymbol{\hat{u}}_1/dt =\dot{\phi}\boldsymbol{\hat{v}}_1\)). See Figure 1.
Step 2: We determine the kinematics of the body, starting with velocities: \[ \boldsymbol{\omega}= \dot{\phi}\boldsymbol{\hat{k}}, \qquad \boldsymbol{v}_B = L \dot{\theta}\boldsymbol{\hat{v}}, \qquad \boldsymbol{v}_G = L \dot{\theta}\boldsymbol{\hat{v}}+ l \dot{\phi}\boldsymbol{\hat{v}}_1 \] Then we find accelerations \[ \boldsymbol{\alpha}= \ddot{\phi}\boldsymbol{\hat{k}}, \qquad \boldsymbol{a}_B = L \ddot{\theta}\boldsymbol{\hat{v}}- L\dot{\theta}^2 \boldsymbol{\hat{u}}, \qquad \boldsymbol{a}_G = L \ddot{\theta}\boldsymbol{\hat{v}}- L\dot{\theta}^2 \boldsymbol{\hat{u}}+ l \ddot{\phi}\boldsymbol{\hat{v}}_1 - l\dot{\phi}^2 \boldsymbol{\hat{u}}_1 \]
Step 3: We sketch a FBD. The body is subject to the string’s tension \(-T \boldsymbol{\hat{u}}\) and gravity \(mg \boldsymbol{\hat{\jmath}}\).
Step 4: We apply Euler’s first and second principles. Equation \(m\boldsymbol{a}_G = {\bf F}_{ext\to rod}\) gives 2 equations by projection on basis \((\boldsymbol{\hat{u}},\boldsymbol{\hat{v}})\) \[ m\big(- L\dot{\theta}^2 + l \ddot{\phi}\sin (\theta-\phi) -l \dot{\phi}^2 \cos(\theta-\phi) \big) = -T + mg \sin\theta\qquad{(1)} \] \[ m \big(L \ddot{\theta}+ l \ddot{\phi}\cos(\theta-\phi) +l \dot{\phi}^2 \sin (\theta-\phi) \big) = m g \cos\theta\qquad{(2)} \] Then equation \(d{\bf H}_G /dt = {\bf M}_{G, ext\to rod}\) gives (on \(\boldsymbol{\hat{k}}\)) \[ \frac{1}{3}ml^2 \ddot{\phi}= l T \sin (\theta-\phi) \qquad{(3)} \] where we used angular momentum \({\bf H}_G = I_{Gzz}\omega\boldsymbol{\hat{k}}\).
Equation (2) is an equation of motion. By combining (1) and (3) (eliminating tension \(T\)), we can obtain a second equation of motion. These 2 ODEs can be integrated with the initial conditions \(\theta=\theta_0\), \(\phi = 0\), \(\dot{\theta}=\dot{\phi}=0\). First integrals of motion are not readily obtained. (They do exist!)
Problem 7.2.
Choose Cartesian coordinate axes \(Oxyz\) with \(Oz\) normal to the plane directed upward, \(Ox\) along the steepest incline directed downward. See Figure 2. We use polar coordinates \((r=l, \theta)\) to define the position of \(P\) in plane \(Oxy\). The velocity of \(P\) is then \(\boldsymbol{v}_P = l\dot{\theta}\boldsymbol{\hat{e}}_\theta\) and its angular momentum about \(O\) is \({\bf H}_O = l \boldsymbol{\hat{e}}_r \times ml\dot{\theta}\boldsymbol{\hat{e}}_\theta= ml^2 \dot{\theta}\boldsymbol{\hat{k}}\).
The forces acting on \(P\) are:
the string’s tension \(-T\boldsymbol{\hat{e}}_r\),
the support normal reaction force \(N\boldsymbol{\hat{k}}\),
the gravitational force \(mg (\cos\alpha\boldsymbol{\hat{\imath}}-\sin\alpha\boldsymbol{\hat{k}})\).
The moment about \(O\) of the forces acting on \(P\) is given by \[ {\bf M}_O = \boldsymbol{r}_{OP}\times N\boldsymbol{\hat{k}}+ \boldsymbol{r}_{OP} \times m{\bf g} = Nl (-\cos\theta\boldsymbol{\hat{\jmath}}+\sin\theta\boldsymbol{\hat{\jmath}}) + mgl (-\cos\alpha\sin\theta\boldsymbol{\hat{\imath}}+ \cos\alpha\cos\theta\boldsymbol{\hat{\jmath}}- \sin\alpha\sin\theta\boldsymbol{\hat{k}}) \] We apply the moment form of Newton second law on \(\boldsymbol{\hat{k}}\) (the other components involve \(N\)): \[ \underbrace{ml^2 \ddot{\theta}}_{\boldsymbol{\hat{k}}\cdot\frac{d}{dt}{\bf H}_O} = \underbrace{-mgl \sin\alpha\sin\theta}_{\boldsymbol{\hat{k}}\cdot{\bf M}_O} \] which gives the ODE \[ \boxed{ \ddot{\theta}+ \frac{g}{l} \sin\alpha\sin\theta=0 } \] For small angles, we can use \(\sin\theta\approx \theta\) to get the linear ODE \[ \boxed{ \ddot{\theta}+ (\frac{g}{l} \sin\alpha) \theta=0 } \] corresponding to the period \[ \boxed{ T = 2\pi \sqrt{ \frac{l}{g\sin\alpha}} } \] Numerically, we obtain $T=2.4 $ s.
Problem 7.3.
We adopt a basis \((\boldsymbol{\hat{u}},\boldsymbol{\hat{v}},\boldsymbol{\hat{k}})\) as shown in Figure 3. We can then express the position and velocity of \(P_1\) and \(P_2\) as follows \[ \boldsymbol{r}_{OP_1} = l \boldsymbol{\hat{u}}, \qquad \boldsymbol{v}_{P_1} = l \dot{\theta}\boldsymbol{\hat{v}} \] \[ \boldsymbol{r}_{OP_2} = 2l \boldsymbol{\hat{u}}, \qquad \boldsymbol{v}_{P_2} = 2l \dot{\theta}\boldsymbol{\hat{v}} \]
Method a. We apply Euler’s 2nd principle about \(O\) by finding the angular momentum of the system \(\Sigma = \{ P_1, rod, P_2\}\) as the sum (the rod is massless) \[ {\bf H}_{O,\Sigma}= \boldsymbol{r}_{OP_1} \times m \boldsymbol{v}_{P_1} + \boldsymbol{r}_{OP_2} × m\boldsymbol{v}_{P_2} = ml^2 \boldsymbol{\hat{u}}\times \dot{\theta}\boldsymbol{\hat{v}}+ m4l^2 \boldsymbol{\hat{u}}\times \dot{\theta}\boldsymbol{\hat{v}} = 5ml^2 \dot{\theta}\boldsymbol{\hat{k}} \] The system is subjected to the following external forces:
the reaction at \(O\) (whose moment about \(O\) is zero)
the gravitational forces acting at \(P_1\) and \(P_2\):
This gives the total moment about \(O\) \[ {\bf M}_O = \boldsymbol{r}_{OP_1} \times m {\bf g} + \boldsymbol{r}_{OP_2} × m{\bf g} =- 3l mg \sin\theta\boldsymbol{\hat{k}} \] Euler’s 2nd principle \(d{\bf H}_{O,\Sigma}/dt ={\bf M}_{O,ext\to \Sigma}\) then gives \[ 5ml^2 \ddot{\theta}= -3l m g \sin\theta\qquad{(1)} \] For small angle \(\theta\), equation (1) is approximated as \[ \ddot{\theta}+\frac{3g}{5 l} \theta=0 \] using \(\sin\theta\approx \theta\). The corresponding period of (small amplitude) oscillations is then given by \[ \boxed{ T = 2\pi \sqrt{\frac{5 l}{3g} } } \] (Recall that the solutions of \(\ddot{\theta}+ \omega^2 \theta=0\) are given by \(\theta= A \cos\omega t + B \sin\omega t\): it is periodic, of period \(2\pi/\omega\).)
Method b. The mass center \(G\) of the system is given by \[ 2m \boldsymbol{r}_{OG} = m \boldsymbol{r}_{OP_1} + m \boldsymbol{r}_{OP_2} = m l \boldsymbol{\hat{u}}+ m 2l \boldsymbol{\hat{u}} \] which gives \[ \boldsymbol{r}_{OG}= \frac{3l}{2}\boldsymbol{\hat{u}} \] Then we apply Euler’s 2nd principle about \(O\) by considering that the system is equivalent to the system \(\Sigma'= \{rod, G\}\) with mass \(2m\) (as we must do with Euler’s 1st principle): \[ \frac{d{\bf H}_{O,\Sigma'}}{dt} = {\bf M}_{O, ext\to \Sigma' } \] with \[ {\bf H}_{O, \Sigma'} = \boldsymbol{r}_{OG} \times 2m \boldsymbol{v}_G = 2m \frac{9l^2}{4} \dot{\theta}\boldsymbol{\hat{k}} \] and where the external moment \[ {\bf M}_{O, ext\to \Sigma' } =\boldsymbol{r}_{OG}\times 2m{\bf g}= - 2mg \frac{3l}{2} \sin\theta\boldsymbol{\hat{k}} \] is found as the moment about \(O\) of force \(2m {\bf g}\) applied at \(G\). This then gives \[ m \frac{9l^2}{2} \ddot{\theta}= - 3lmg \sin\theta\qquad{(2)} \] This leads to the linearized ODE: \[ \ddot{\theta}+\frac{2g}{3 l} \theta=0 \] and the period \[ \boxed{ T = 2\pi \sqrt{\frac{3 l}{2g} } } \]
Conclusion: The results are not identical. One of the two methods is incorrect. The 2nd principle is properly applied in Method a). However, we cannot treat the system as if equivalent to a particle \(G\) (the mass center) as must be done in Euler’s 1st principle for a system of particles. Method b) must not be used.
Problem 7.4.
As the rod is released in the vertical position, it first rotates about cornerpoint \(O\) up to an inclination \(\theta_1\). We must show that when \(\theta=\theta_1\) is reached, it slips along axis \(Ox\) and leaves the vertical wall.
In the rotation phase, the position of the rod is completely defined by the angle \(\theta\). The value \(\theta_1\) is found by determining the reaction force \({\bf R}\) of the walls at the endpoint \(O\) of the rod. When the horizontal component \({\bf R}\cdot \boldsymbol{\hat{\imath}}= R_x\) of \(\bf R\) becomes zero, the rod must lose contact with the vertical wall.
a. The kinematics of the rod is defined by its angular velocity \(\boldsymbol{\omega}= -\dot{\theta}\boldsymbol{\hat{k}}\), and the velocity \(\boldsymbol{v}_G = l \dot{\theta}\boldsymbol{\hat{v}}\) of its mass center (see Figure 4). Note that \(\dot{\theta}> 0\).
The external forces acting on the rod are (i) the gravitational force \(m {\bf g} = - m g \boldsymbol{\hat{\jmath}}\) and (ii) the reaction \({\bf R} = R_x \boldsymbol{\hat{\imath}}+ R_y \boldsymbol{\hat{\jmath}}\) of the horizontal support and the vertical wall on the extremity \(O\). The rotation of the rod about \(O\) lasts as long as \(R_x \geq 0\). Force \(\bf R\) is found by applying Euler’s first principle \(m \boldsymbol{a}_G = m {\bf g} + {\bf R}\): \[ R_x = m l ( \ddot{\theta}\cos\theta- \dot{\theta}^2 \sin\theta) , \qquad R_y = mg + m l (- \ddot{\theta}\sin\theta- \dot{\theta}^2 \cos\theta) \qquad{(1-2)} \] since \(\boldsymbol{a}_G = l \ddot{\theta}\boldsymbol{\hat{v}}- l\dot{\theta}^2 \boldsymbol{\hat{u}}\).
To obtaine \(R_x\) vs \(\theta\), we need to find \(\ddot{\theta}\) and \(\dot{\theta}^2\) as functions of angle \(\theta\). The evolution of angle \(\theta\) can be found by applying Euler’s second principle about point \(O\): \[ \frac{d{\bf H}_O}{dt} = {\bf M}_O \] where
\({\bf H}_O = I_{Ozz} \boldsymbol{\omega}= - \frac{4}{3} m l^2 \dot{\theta}\boldsymbol{\hat{k}}\) is the rod’s angular momentum about \(O\),
\({\bf M}_O = \boldsymbol{r}_{OG}\times m {\bf g} =- m g l \sin\theta\boldsymbol{\hat{k}}\) is the moment about \(O\) of the external forces.
This yields an ODE governing angle \(\theta\) only \[ \ddot{\theta}= \frac{3g}{4 l} \sin\theta\qquad{(3)} \] which by integration (pre-multiply by \(\dot{\theta}\)) gives \[ \dot{\theta}^2 = \frac{3g}{2 l} (1 - \cos\theta) \qquad{(4)} \] using the initial conditions \(\theta= 0\) and \(\dot{\theta}= 0\) at \(t=0\).
Note that it is much faster to apply Euler’s 2nd principle about \(O\) rather than \(G\), since this eliminates the contribution of the reaction force. Since \(O\) is a fixed point of the body, Euler’s 2nd principle about \(O\) is just as easy to apply as it would be about the mass center.
Now we can express \(R_x\) and \(R_y\) as functions of \(\theta\) by using equations (3) and (4). We find (after some algebraic work) \[ \boxed{ R_x = \frac{3}{4} mg \sin\theta(3 \cos\theta-2 ) , \quad R_y = \frac{1}{4} mg (1- 3 \cos\theta)^2 } \] It is possible to plot \(R_x /mg\) and \(R_y/mg\) versus \(\theta\). We see that the horizontal reaction initially increases from the value \(R_x =0\) at \(\theta= 0\), then decreases to again reach the zero value at the angle \(\boxed{\theta_1 = \cos^{-1} (2/3) = 48.2^o}\) At angle \(\theta= \theta_1\), the rod must lose contact with the vertical wall. Note that \(R_y \geq 0\) but is never negative.
b. When \(\theta=\theta_1\) is reached, the angular velocity is found from equation (4): \[ \boxed{ \dot{\theta}_1 = \sqrt{\frac{g}{2l}} } \]
Problem 7.5.
4.14. With the choice of basis \((\boldsymbol{\hat{\imath}},\boldsymbol{\hat{\jmath}},\boldsymbol{\hat{k}})\) shown in Figure 5, the slipping motion of the crate on plane \(Oxz\) corresponds to a translation along axis \(Ox\). Its kinematics is defined by the velocity of its mass center \(G\) given by \(\boldsymbol{v}_G = v_G \boldsymbol{\hat{\imath}}\) and angular velocity \(\boldsymbol{\omega}=\boldsymbol{0}\).
The crate is subject to
gravity \(m {\bf g} = mg (\sin\theta\boldsymbol{\hat{\imath}}- \cos\theta\boldsymbol{\hat{\jmath}})\) whose moment is zero at point \(G\),
the reaction force \({\bf R} = -F \boldsymbol{\hat{\imath}}+ N \boldsymbol{\hat{\jmath}}\) of the support, written as a sum of a normal and tangential components. We assume that the line of action of this force passes through a point \(I\) located in the vertical plane \(Gxy\) (by symmetry) on the segment \(AB\) in contact with \(Ox\). The motion being assumed planar, all forces acting on the cube must lie in the same plane. We define the position of \(I\) by introducing the unknown coordinate \(b\) such that \(\boldsymbol{r}_{GI} = -l \boldsymbol{\hat{\jmath}}+ b \boldsymbol{\hat{\imath}}\). The motion being downward (in the direction of \(Ox\)), the friction force must be pointing opposite \(\boldsymbol{\hat{\imath}}\).
Euler’s principles then give \[\begin{align*} m \dot{v}_G = -F + mg \sin\theta\\ N - mg \cos\theta=0 \\ -l F + b N = 0 \end{align*}\] where \(b\) is unknown. An additional equation is needed and is found by imposing \(|F|= F = \mu N\) since the crate is slipping. With \(N = mg \cos\theta\), we find that \(\dot{v}_G\) is given by \[ \dot{v}_G = g ( \sin\theta- \mu \cos\theta), \] leading then to \[ F = mg \mu \cos\theta, \qquad \frac{b}{l} = \mu \] We must impose two conditions for the motion to be realizable:
the acceleration \(\dot{v}_G\) must be positive for a downward motion: \(\tan\theta> \mu\),
point \(I\) must be located between \(A\) and \(B\), that is, \(b\leq l\): this gives the condition \(\mu \leq 1\).
In conclusion, the slipping motion of the crate is realizable if the following conditions
are guaranteed:
\[
\boxed{
\mu \leq 1 , \quad {\rm and}
\qquad \tan\theta> \mu
}
\]
- 4.15. The tipping motion of the cube about lower edge \(A\) corresponds to a pure rotation about fixed point \(A\). See Figure 6.
At time \(t=0+\) (immediately after its release), the motion of the cube is characterized by the angular velocity \(\boldsymbol{\omega}_0 =0\) and the angular acceleration \(\boldsymbol{\alpha}_0 = - \alpha_0 \boldsymbol{\hat{k}}\). The acceleration of its mass center \(G\) is given by \[ \boldsymbol{a}_G = \boldsymbol{\alpha}_0 \times \boldsymbol{r}_{AG} = l \alpha_0 (\boldsymbol{\hat{\imath}}+ \boldsymbol{\hat{\jmath}}) \] Note that the rotation about \(A\) is feasible only if \(\alpha_0 > 0\).
The forces acting on the body are
due to gravity, of resultant \(m {\bf g} = mg (\sin\theta\boldsymbol{\hat{\imath}}- \cos\theta\boldsymbol{\hat{\jmath}})\) and zero moment about \(G\),
due to contact, of resultant \({\bf R} = F \boldsymbol{\hat{\imath}}+ N \boldsymbol{\hat{\jmath}}\) and zero moment about \(A\). The components \(F\) and \(N\) must satisfy the condition \(|F| \leq \mu N\) (otherwise the cube slips).
The equations governing the motion at \(t=0^+\) are found by applying Euler’s first and second principles \[\begin{eqnarray*} m l \alpha_0 = F + mg \sin\theta\\ m l \alpha_0 = N - mg \cos\theta\\ - \frac{2}{3} ml^2 \alpha_0 = lF + l N \end{eqnarray*}\] where the quantity ${2 } ml^2 $ is the cube’s moment of inertia about axis \(Gz\). We solve these equations and find \[\begin{eqnarray*} \boxed{\alpha_0 = {3g \over 8l} (\sin\theta- \cos\theta)} \\ N = {1 \over 8} mg (3 \sin\theta+ 5 \cos\theta) \\ F = - {1 \over 8} mg (5 \sin\theta+ 3 \cos\theta) \end{eqnarray*}\]
Other method: apply Euler’s 2nd principle about \(A\). It is easy because \(A\) is a fixed point of the body. This also eliminates \(F\) and \(N\): \[ I_{Azz} \alpha_0 = mg l(\sin\theta-\cos\theta) \] with \(I_{Azz}= I_{Gzz}+ 2ml^2\). We find the same result.
The tipping motion is realizable if \(\alpha_0 > 0\) and \(|F| \leq \mu N\). These conditions can be stated as \[ \boxed{ \qquad \tan\theta> 1 , \quad {\rm and} \qquad \mu \geq {5 \tan\theta+ 3 \over 3 \tan\theta+ 5 } } \]
Problem 7.6. (Optional. Good review problem)
a. Consider the system \(\Sigma = \{1,2\}\). As long as body 2 does not slip relative to body 1, \(\Sigma\) can be considered as a rigid body rotating about axis \(Oz\) with angular velocity \(\boldsymbol{\omega}=\dot{\theta}\boldsymbol{\hat{k}}\). See Figure 7 for notations. Its motion is predicted by applying Euler’s second principle: if the axle at \(O\) is frictionless, the forces it exerts on the plate must satisfy \({\bf M}_{O, axle \to 1}\cdot \boldsymbol{\hat{k}}=0\). So we apply the second principle in the form \[ \boldsymbol{\hat{k}}\cdot {\bf M}_{O, ext \to \Sigma} = \boldsymbol{\hat{k}}\cdot \frac{d{\bf H}_O}{dt} \qquad{(1)} \] with \[ \boldsymbol{\hat{k}}\cdot {\bf M}_{O, ext \to \Sigma} ={\bf M}_{O, axle \to 1}\cdot \boldsymbol{\hat{k}}+{\bf M}_{O, gravity \to \Sigma}\cdot \boldsymbol{\hat{k}} =\boldsymbol{\hat{k}}\cdot (\boldsymbol{r}_{OG}\times mg\boldsymbol{\hat{\jmath}}+\boldsymbol{r}_{OP} \times mg \boldsymbol{\hat{\jmath}}) = mg (l+x)\cos\theta \] and \[ {\bf H}_O = {\bf H}_{O,1}+ {\bf H}_{O,2}= \frac{4}{3}ml^2 \dot{\theta}\boldsymbol{\hat{k}}+ mx^2 \dot{\theta}\boldsymbol{\hat{k}} \] So equation (1) gives \[ \boxed{ (\frac{4}{3}l^2 +x^2) \ddot{\theta}= mg (l+x)\cos\theta } \qquad{(2)} \] This is an ODE with initial conditions \(\theta=0\) and \(\dot{\theta}=0\) at \(t=0\). It is integrated by pre-multiplying by \(\dot{\theta}\) and recognizing each term as \(d/dt\) of a quantity: \(\dot{\theta}\ddot{\theta}= \frac{d}{dt} (\frac{\dot{\theta}^2}{2})\) and \(\dot{\theta}\cos\theta= \frac{d}{dt} (\sin\theta)\) \[ (\frac{4}{3}l^2 +x^2) \frac{\dot{\theta}^2}{2} = mg (l+x)\sin\theta+ Cst \Longrightarrow \boxed{ \dot{\theta}^2 = 2 \frac{l+x}{\frac{4}{3}l^2 +x^2} g \sin\theta } \qquad{(3)} \] where we imposed \(\theta=0\) and \(\dot{\theta}=0\) at \(t=0\).
b. We sketch a FBD for body 2 and apply Newton’s second law to find \[ m x( \ddot{\theta}\boldsymbol{\hat{v}}- \dot{\theta}^2 \boldsymbol{\hat{u}}) = -N \boldsymbol{\hat{v}}-F \boldsymbol{\hat{u}}+mg (\sin\theta\boldsymbol{\hat{u}}+\cos\theta\boldsymbol{\hat{v}}) \] This gives two equations \[ -m x \dot{\theta}^2 = -F +mg \sin\theta, \qquad m x \ddot{\theta}= -N+ mg \cos\theta\qquad{(4-5)} \] Then we use the expressions of \(\ddot{\theta}\) and \(\dot{\theta}^2\) vs \(\theta\) given by equations (2-3): this gives \(N\) and \(F\) vs \(\theta\) \[ \boxed{ N= mg \left(1 -\frac{x(l+x)}{\frac{4}{3}l^2 +x^2} \right)\cos\theta , \qquad F= mg \left(1+ \frac{2x(l+x)}{\frac{4}{3}l^2 +x^2} \right)\sin\theta } \]
c. Body 2 will not lose contact with the plate as long as \(N>0\): this imposes the condition \[ 1 -\frac{x(l+x)}{\frac{4}{3}l^2 +x^2} >0 \Longrightarrow \frac{4}{3}l^2 +x^2 > x(l+x) \Longrightarrow \boxed{x< \frac{4}{3}l} \] If \(x \geq \frac{4}{3}l\), body 2 loses contact with the plate immediately upon release of the plate.
d. If \(x < \frac{4}{3}l\), body 2 will likely start to slip off the plate at some value of angle \(\theta=\theta_1\). Body 2 will not slip as long as the condition \(|F|\leq \mu_s N\) is satisfied: \[ \left(1+ \frac{2x(l+x)}{\frac{4}{3}l^2 +x^2} \right)\sin\theta\leq \mu_s \left(1 -\frac{x(l+x)}{\frac{4}{3}l^2 +x^2} \right)\cos\theta \] which gives \[ \boxed{ \tan\theta\leq \tan\theta_1 = \mu_s \frac{4l-3x}{4l^2 +6l x +9x^2} } \] For \(\mu_s = 0.4\), \(l=0.6\) m and \(x=0.4\) m, we find \(\boxed{\theta_1 = 15.5^o}\).