Question 8db2f

Nov 26, 2017

See below.

Explanation:

For rod $\overline{A B}$

${\vec{f}}_{A} + {\vec{f}}_{B} + {\vec{p}}_{A B} + {\vec{p}}_{B} = \vec{0}$

${\vec{m}}_{A} = \left({\vec{f}}_{B} + {\vec{p}}_{B}\right) \times \left(B - A\right) + {\vec{p}}_{A B} \times \frac{B - A}{2} = \vec{0}$

for rod $\overline{B C}$

$- {\vec{f}}_{B} + {\vec{p}}_{B C} + {\vec{f}}_{C} = \vec{0}$

${\vec{m}}_{C} = - {f}_{B} \times \left(B - C\right) + {\vec{p}}_{B C} \times \frac{B - C}{2} = \vec{0}$

Here

${\vec{f}}_{A} = \left({h}_{A} , {v}_{A}\right)$
${\vec{f}}_{B} = \left({h}_{B} , {v}_{B}\right)$
${\vec{f}}_{C} = \left({h}_{C} , {v}_{C}\right)$
${\vec{p}}_{A B} = \left(0 , - k m g\right)$
${\vec{p}}_{B C} = \left(0 , - k m g\right)$
${\vec{p}}_{B} = \left(0 , - m g\right)$

$A = \left(0 , 2 a \cos \theta\right)$
$B = \left(2 a , 2 a \cos \theta\right)$
$C = \left(2 a + 2 a \sin \theta , 0\right)$

and the condition

${h}_{C} = \mu {v}_{C}$

Those equations give

{(h_A + h_B = 0), (v_A + v_B = (k+1)mg), (h_B - h_C = 0), (v_C -v_B= k m g), (2 a g m + a g k m - 2 a v_B = 0), (2 a h_B Cos theta - a g k m Sin theta - 2 a v_B Sin theta = 0), (h_C =mu v_C):}

Solving for ${h}_{A} , {v}_{A} , {h}_{B} , {v}_{B} , {h}_{C} . {v}_{C} , \mu$ we get at

{(),(h_A = -g (1 + k) m Tan theta), (v_A = (g k m)/2), (h_B = g (1 + k) m Tan theta), (v_B = 1/2 g (2 + k) m), (hc = g (1 + k) m Tan theta), (vc = 1/2 g (2 + 3 k) m), (mu = (2 Sec theta (Sin theta + k Sin theta))/(2 + 3 k)):}

Assuming $\theta = \frac{\pi}{6}$ and $k = 3$ we have

$\mu = \frac{8}{11 \sqrt{3}}$

Nov 27, 2017

Free Body diagram of the situation ...!

Explanation:

We are going to apply the static equilibrium conditions
(translational equilibrium + rotational equilibrium) to both bars separately.

Translational Equilibrium Condition: $\setminus {\sum}_{k} {\vec{F}}_{k} = \vec{0}$
Rotational Equilibrium Condition: $\setminus {\sum}_{k} \vec{\tau} = \vec{0}$

Symbols Used:
$\left(1\right) \setminus \quad {N}_{A}$ - Normal force at A ; $\setminus q \quad \left(2\right) {N}_{B} \setminus \quad$ - Normal force at B;
$\left(3\right) \setminus \quad {N}_{C}$ - Normal force at C; $\setminus q \quad \left(4\right) \setminus \quad {F}_{f}$ - Static friction at C;
$\left(5\right) \setminus \quad {M}_{2} g$ - Weight of bar BC; $\setminus q \quad \left(6\right) \setminus \quad {M}_{1} g$ - Weight of bar AB;

$L$ - Length of the bars AB and BC;

Given : \quad L = 2a; \qquad M_1 = k.m; \qquad M_2 = m

Situation: The two bars, AB and BC, are hinged at B where their interaction produces a normal force ${N}_{B}$ that makes an angle $\setminus \phi$ with respect to the horizontal.

The component of ${N}_{B}$ responsible for generating a torque on bar AB is : \quadN_{B_|_} = N_B\sin\phi;\quad (labeled 2a)

The component of ${N}_{B}$ responsible for generating a torque on bar BC is: \quadN_B^{BC} = N_B\cos(\phi-\theta);\quad (labeled 2b)

[1] Static Equilibrium on Bar AB: Forces acting on this bar are ${\vec{N}}_{A} , \setminus \quad {\vec{N}}_{B}$ and ${M}_{1} \vec{g}$

[1a] Rotational Equilibrium Condition: Calculating the torques about point A -
$\vec{\setminus} {\tau}_{\text{net}} = \vec{\setminus} {\tau}_{A} + \vec{\setminus} {\tau}_{g} + \vec{\setminus} {\tau}_{B}$
$\vec{\setminus} {\tau}_{\text{net}} = + {N}_{A} .0 - {M}_{1} g \left(\frac{L}{2}\right) + {N}_{B \bot} . L = \vec{0}$

${N}_{B \bot} = \frac{{M}_{1.} g}{2}$ - Thus we get our first unknown.
But, N_{B_|_} = N_B\sin\phi; \qquad \rightarrow N_B = (M_1.g)/(2\sin\phi) ...... (EQ1)
[1b] Translational Equilibrium Condition: Vector sum of all the forces acting on bar AB must vanish.
vec F_{"net"} = vec F_A + vec F_g + vec F_B = vec 0;
Vertical Component: $\setminus q \quad {N}_{A} - {M}_{1} g + {N}_{B \bot} = 0$

${N}_{A} = {M}_{1.} g - {N}_{B \bot} = {M}_{1.} g - \frac{{M}_{1.} g}{2} = \frac{{M}_{1.} g}{2}$

[2] Static Equilibrium Condition on Bar BC: Forces acting on this bar are ${\vec{N}}_{B} , \setminus \quad {\vec{N}}_{C} , \setminus \quad {M}_{2.} \vec{g} ,$ and ${\vec{F}}_{f}$.

[2a] Rotational Equilibrium Condition: Calculating the torques about point C,
vec \tau_{"net"} = vec \tau_B + vec \tau_g + vec \tau_C + vec \tau_f = vec 0;
$\vec{\setminus} {\tau}_{\text{net}} = - {N}_{B}^{B C} . L + {M}_{2.} g \setminus \sin \setminus \theta \left(\frac{L}{2}\right) + \left({F}_{f} - {N}_{C \bot}\right) .0 = \vec{0}$
N_B^{BC} = (M_2.g\sin\theta)/2; \qquad \rightarrow N_B\cos(\phi-\theta) = (M_2.g\sin\theta)/2
${N}_{B} = \frac{{M}_{2.} g \setminus \sin \setminus \theta}{2 \setminus \cos \left(\setminus \phi - \setminus \theta\right)}$ ...... (EQ2)

Comparing (EQ1) and (EQ2) we can get,
$\frac{{M}_{1.} g}{2 \setminus \sin \setminus \phi} = \frac{{M}_{2.} g \setminus \sin \setminus \theta}{2 \setminus \cos \left(\setminus \phi - \setminus \theta\right)}$

${M}_{1} / {M}_{2} = \frac{\setminus \sin \setminus \theta \setminus \sin \setminus \phi}{\setminus \cos \left(\setminus \phi - \setminus \theta\right)} = \frac{\setminus \sin \setminus \theta \setminus \sin \setminus \phi}{\setminus \cos \setminus \theta \setminus \cos \setminus \phi + \setminus \sin \setminus \theta \setminus \sin \setminus \phi} = \frac{1}{1 + \setminus \cot \setminus \theta \setminus \cot \setminus \phi}$

\cot\phi = (M_2/M_1-1)\tan\theta = -2/(3\sqrt{3}); \qquad \phi = 68.9^o

Because, M_1 = k.m = 3m; \qquad M_2 = m; \qquad \theta = 30^o

[2b] Translational Equilibrium Condition: Vector sum of all the forces acting on bar BC must vanish.
vec F_{"net"} = vec F_B + vec F_g + vec F_C + vec F_f = vec 0;

Vertical Component: $\setminus q \quad {N}_{B \bot} - {M}_{2} g + {N}_{C} = 0$
${N}_{C} = {M}_{2.} g - {N}_{2 \bot} = {M}_{2.} g - {N}_{2} \setminus \sin \setminus \phi$
${N}_{C} = {M}_{2.} g - \frac{{M}_{1.} g}{2 \setminus \sin \setminus \phi} . \setminus \sin \setminus \phi = \left({M}_{2} - {M}_{1} / 2\right) g$ ...... (EQ3)

Calculating Friction Coefficient:
Now to find the frictional force apply the rotational equilibrium condition again, BUT this time calculating the torques about point B.
Rotational Equilibrium Condition (Again): Calculating the torques about point B,

vec \tau_{"net"} = vec \tau_B + vec \tau_g + vec \tau_C + vec \tau_f = vec 0;
$0 = {N}_{B \bot} .0 - {M}_{2.} g \setminus \sin \setminus \theta . \left(\frac{L}{2}\right) + \left({N}_{C} \setminus \sin \setminus \theta - {F}_{f} \setminus \cos \setminus \theta\right) . L$
$\frac{{M}_{2.} g \setminus \sin \setminus \theta}{2} = {N}_{C} \setminus \sin \setminus \theta - {F}_{f} \setminus \cos \setminus \theta = {N}_{C} \setminus \sin \setminus \theta - \setminus {\mu}_{s} {N}_{C} \setminus \cos \setminus \theta$
(M_2.g)/2 = N_C(1-\mu_s/\tan\theta); \qquad \mu_s = \tan\theta(1-(M_2.g)/(2N_C))

Using (EQ3) replace ${N}_{C}$ in the expression for $\setminus {\mu}_{s}$,

$\setminus {\mu}_{s} = \setminus \tan \setminus \theta \left(1 - \frac{{M}_{2.} g}{2 \left({M}_{2} - {M}_{1} / 2\right) g}\right) = \setminus \tan \setminus \theta \left(1 - \frac{{M}_{2}}{2 {M}_{2} - {M}_{1}}\right)$

M_1 = 3m; \qquad M_2 = m; \qquad \theta = 30^o#
$\setminus {\mu}_{s} = 2 \setminus \tan {30}^{o} = 2 \setminus \times \setminus \frac{\sqrt{3}}{2} = \setminus \sqrt{3}$