Updated spacing between @ latex expressions

astrosander edited
View solution History
revision #10309 parent #10078 GitHub c702ebe ← older newer →
en/2.1.49.md +19 −19
@@ -77,16 +77,16 @@
</center>
<p>
Newton's second law of motion:
− $$OX: \;T - T \sin \alpha = m_0 a_{0x} $$
− $$ T \sin \alpha = m_{a1x} $$
+ $$OX: \;T - T \sin\alpha = m_0 a_{0x} $$
+ $$ T \sin\alpha = m_{a1x} $$
− $$OY: \;N - mg - T \cos \alpha = m_0 a_{0y} = 0 $$
− $$ T \cos \alpha - mg = m_{a1y} $$
− $$ \tan \alpha = \frac{x_0 - x_1}{y_0 - y_1} $$
+ $$OY: \;N - mg - T \cos\alpha = m_0 a_{0y} = 0 $$
+ $$ T \cos\alpha - mg = m_{a1y} $$
+ $$ \tan\alpha = \frac{x_0 - x_1}{y_0 - y_1} $$
Since $\alpha = \text{const}$
<div class="scroll-wrapper">
$$
− \frac{d}{dt} \left( \tan \alpha \right) = \frac{\left( \frac{dx_0}{dt} - \frac{dx_1}{dt} \right)(y_0 - y_1) - \left( \frac{dy_0}{dt} - \frac{dy_1}{dt} \right)(x_0 - x_1)}{(y_0 - y_1)^2} = 0
+ \frac{d}{dt} \left( \tan\alpha\right) = \frac{\left( \frac{dx_0}{dt} - \frac{dx_1}{dt} \right)(y_0 - y_1) - \left( \frac{dy_0}{dt} - \frac{dy_1}{dt} \right)(x_0 - x_1)}{(y_0 - y_1)^2} = 0
$$</div>
<div class="scroll-wrapper">
@@ -103,37 +103,37 @@
$$
(a_{0x} - a_{1x})(y_0 - y_1) = (a_{0y} - a_{1y})(x_0 - x_1)
$$
− $$ \tan \alpha (a_{0y} - a_{1y}) = a_{0x} - a_{1x}$$
+ $$ \tan\alpha (a_{0y} - a_{1y}) = a_{0x} - a_{1x}$$
From (1)
$$a_{0y} = 0$$
− $$ -\tan \alpha\, a_{1y} = a_{0x} - a_{1x} $$
+ $$ -\tan\alpha\, a_{1y} = a_{0x} - a_{1x} $$
Length of the rope:
− $$ L = x_w - x_0 + \frac{y_0 - y_1}{\cos \alpha} $$
+ $$ L = x_w - x_0 + \frac{y_0 - y_1}{\cos\alpha} $$
Since the rope is inextensible
<div class="scroll-wrapper">
− $$ \frac{d^2 L}{dt^2} = -\frac{d^2 x_0}{dt^2} + \frac{1}{\cos \alpha} \left( \frac{d^2 y_0}{dt^2} - \frac{d^2 y_1}{dt^2} \right) = 0 $$
+ $$ \frac{d^2 L}{dt^2} = -\frac{d^2 x_0}{dt^2} + \frac{1}{\cos\alpha} \left( \frac{d^2 y_0}{dt^2} - \frac{d^2 y_1}{dt^2} \right) = 0 $$
</div>
<div class="scroll-wrapper">
$$ \left( \frac{d^2 x_0}{dt^2} = a_{0x}, \frac{d^2 y_0}{dt^2} = a_{0y}, \frac{d^2 y_1}{dt^2} = a_{1y} \right) $$</div>
− $$ a_{0x} \cdot \cos \alpha = a_{0y} - a_{1y}$$
+ $$ a_{0x} \cdot \cos\alpha = a_{0y} - a_{1y}$$
From $(1)$: $a_{0y} = 0$
− $$ a_{0x} \cos \alpha = -a_{1y} $$
+ $$ a_{0x} \cos\alpha = -a_{1y} $$
By dividing the equations from (1):
<div class="scroll-wrapper">
− $$ \frac{T \sin \alpha}{T \cos \alpha} = \frac{m a_{1x}}{m (g + a_{1y})} \to \tan \alpha (g + a_{1y}) = a_{1x} $$</div>
− $$ \tan \alpha a_{1y} = a_{1x} - g \tan \alpha$$
+ $$ \frac{T \sin\alpha}{T \cos\alpha} = \frac{m a_{1x}}{m (g + a_{1y})} \to \tan\alpha (g + a_{1y}) = a_{1x} $$</div>
+ $$ \tan\alpha a_{1y} = a_{1x} - g \tan\alpha$$
and using (4)
− $$\boxed{a_{0x} = g \tan \alpha }$$
+ $$\boxed{a_{0x} = g \tan\alpha}$$
By dividing the equations from (1):
$$
− \frac{T(1 - \sin \alpha)}{T \cos \alpha} = \frac{M}{m} \cdot \frac{a_{0x}}{g + a_{1y}}$$
+ \frac{T(1 - \sin\alpha )}{T \cos\alpha} = \frac{M}{m} \cdot \frac{a_{0x}}{g + a_{1y}}$$
Note that:
− $$a_{1y} = -g \sin \alpha$$
+ $$a_{1y} = -g \sin\alpha$$
− $$ \frac{1 - \sin \alpha}{\cos \alpha} \frac{M}{m} = \frac{g \tan \alpha}{g - g \sin \alpha} \to m = \frac{M \sin \alpha}{(1 - \sin \alpha)^2} $$
− $$ \boxed{m = \frac{M \sin \alpha}{(1 - \sin \alpha)^2}} $$
+ $$ \frac{1 - \sin\alpha}{\cos\alpha} \frac{M}{m} = \frac{g \tan\alpha}{g - g \sin\alpha} \to m = \frac{M \sin\alpha}{(1 - \sin\alpha )^2} $$
+ $$ \boxed{m = \frac{M \sin\alpha}{(1 - \sin\alpha )^2}} $$
</p>
<p style="text-align: right; font-style: italic; font-size: 16;">physicshub</p>
unchanged lines 15