LaTeX Latex Equation Troubleshooting: Fixing the Fraction

  • Thread starter Thread starter scott1
  • Start date Start date
  • Tags Tags
    Latex
AI Thread Summary
The discussion revolves around troubleshooting an equation related to static friction and net forces acting on two blocks. Initially, there was confusion regarding the formatting of a fraction, which was later resolved by clearing the cache. The user prefers to keep the static friction coefficients, denoted as μS, until the end of the calculations. The equations presented detail the net forces acting on the top and bottom blocks, leading to the calculation of tension (FT) and acceleration (a). The final results indicate that the tension is 36N, with an acceleration of 3.3 m/s², and the time taken for the blocks to move a distance of 5.0m is calculated to be approximately 1.7 seconds.
Physics news on Phys.org
Looks good to me! Cleared your cache yet?
 
I found this on archive. I just post for 3 seconds.
sorry...

<br /> Hmm--I prefer to keep the static friction coefficients as \mu _{S,1} and \mu _{S,2} until the end, where <br /> \left\{ \begin{gathered}<br /> \mu _{S,1} = 0.60 \hfill \\<br /> \mu _{S,2} = 0.20 \hfill \\ <br /> \end{gathered} \right\} <br /> and F_{net,1} is the net force for the top block, F_{net,2} is the net force for the bottom block. Here&#039;s my solution:<br /> \left\{ \begin{gathered}<br /> F_{net,1} = a \cdot 4.0kg = F_T - 39N\mu _{S,1} \hfill \\<br /> F_{net,2} = a \cdot 3.0kg = 39N\mu _{S,1} - 69N\mu _{S,2} \hfill \\ <br /> \end{gathered} \right\} \Rightarrow <br /> a = \frac{{F_T - 39N\mu _{S,1} }}{{4.0kg}} \Rightarrow <br /> <br /> 3.0kg\frac{{F_T - 39N\mu _{S,1} }}{{4.0kg}} = 39N\mu _{S,1} - 69N\mu _{S,2} \Rightarrow <br /> \frac{{3.0}}{{4.0}}\left( {F_T - 23N} \right) = 9.6N \Rightarrow F_T = 36N <br /> \therefore a = \frac{{36N - 23N}}{{4.0kg}} = 3.3\frac{m}{{s^2 }} <br /> <br /> *Because v_0 = 0 \frac{m}{s} , and I like to set x_0 = 0 m ,<br /> \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 <br />
 

Similar threads

Replies
12
Views
3K
Replies
8
Views
3K
Replies
1
Views
2K
Replies
11
Views
5K
Replies
8
Views
2K
Replies
5
Views
3K
Replies
16
Views
3K
Back
Top