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459
2. Gokul43201

11,141
Staff Emeritus
Looks good to me! Cleared your cache yet?

3. kirbykirbykirby

20
I found this on archive. I just post for 3 seconds.
sorry...

$$Hmm--I prefer to keep the static friction coefficients as \mu _{S,1} and \mu _{S,2} until the end, where \left\{ \begin{gathered} \mu _{S,1} = 0.60 \hfill \\ \mu _{S,2} = 0.20 \hfill \\ \end{gathered} \right\} and F_{net,1} is the net force for the top block, F_{net,2} is the net force for the bottom block. Here's my solution: \left\{ \begin{gathered} F_{net,1} = a \cdot 4.0kg = F_T - 39N\mu _{S,1} \hfill \\ F_{net,2} = a \cdot 3.0kg = 39N\mu _{S,1} - 69N\mu _{S,2} \hfill \\ \end{gathered} \right\} \Rightarrow a = \frac{{F_T - 39N\mu _{S,1} }}{{4.0kg}} \Rightarrow 3.0kg\frac{{F_T - 39N\mu _{S,1} }}{{4.0kg}} = 39N\mu _{S,1} - 69N\mu _{S,2} \Rightarrow \frac{{3.0}}{{4.0}}\left( {F_T - 23N} \right) = 9.6N \Rightarrow F_T = 36N \therefore a = \frac{{36N - 23N}}{{4.0kg}} = 3.3\frac{m}{{s^2 }} *Because v_0 = 0 \frac{m}{s} , and I like to set x_0 = 0 m , \therefore v = \left( {3.3\frac{m}{{s^2 }}} \right)t \Rightarrow x = 5.0m = \left( {3.3\frac{m}{{s^2 }}} \right)\frac{{t^2 }}{2} \Rightarrow t = 1.7s$$

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