- #1

- 339

- 1

https://www.physicsforums.com/showthread.php?p=1130323&posted=1#post1130323

Test:

[tex]\frac {1}{2}[/tex]

EDIT:It now seems to be working...

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- LaTeX
- Thread starter scott1
- Start date

- #1

- 339

- 1

https://www.physicsforums.com/showthread.php?p=1130323&posted=1#post1130323

Test:

[tex]\frac {1}{2}[/tex]

EDIT:It now seems to be working...

- #2

Gokul43201

Staff Emeritus

Science Advisor

Gold Member

- 7,082

- 20

Looks good to me! Cleared your cache yet?

- #3

- 21

- 0

sorry...

[tex]

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

[/tex]

Share: