# Thread: pulley system on slope

1. ## pulley system on slope

the 4 equations i came up with to solve this was t-4g-4g=8a 2s-t=2a 5g-s=5(b+a) s-3g=3(b-a). i got theses equations by looking at each of the particles. if yu dnt understand how i got one of the equations i can explain it to you. how ever when i slove these i dont get the right answer

2. ## Re: pulley system on slope

Originally Posted by markosheehan

the 4 equations i came up with to solve this was t-4g-4g=8a 2s-t=2a 5g-s=5(b+a) s-3g=3(b-a). i got theses equations by looking at each of the particles. if yu dnt understand how i got one of the equations i can explain it to you. how ever when i slove these i dont get the right answer
Your equations are correct, though personally I would have chosen a different set of letters. For instance I would have called all the accelerations using the letter a: A, ap, a1, a2. But as long as they are labeled clearly on your diagram you are fine. In the future, though, it would be nice to have a larger and better labeled picture. Remember if your teacher can't read it they are not likely to give you full value even for a correct answer.

The good news is that I disagree with your text as well. I'm getting a = (2/5)g.

-Dan

3. ## Re: pulley system on slope

Hmmm ... I get the acceleration of the 8kg block to be $\dfrac{3g}{20}$ up the incline.

4. ## Re: pulley system on slope

Okay, let's do this the way I wanted to originally. All we need are FBDs on the 8 kg mass and the total weight acting on the pulley.

8 kg mass: +x up the slope, +y in the direction of the normal force.

$\sum F_x = t - f_k - w_x = 8 a$

etc.
$t - 4 g - 4 g = 8 a$

Pulley: +x downward.

The mass of the "pulley" is 2 g + 3 g + 5 g.

$\sum F_x = -t + 10 g = 10 a$

Two equations, two unknowns. I get a = g/9 and t = 80g/9

Clearly the book is wrong but there must also something wrong in the equations originally given by markosheehan, though I didn't find anything.

-Dan

5. ## Re: pulley system on slope

Rechecked my work and arrived at the book solution of $a=\dfrac{3g}{35}$

Since the lower pulley system is accelerating downward at $a$, the apparent weights of the 5kg and 3kg mass are $5(g-a)$ and $3(g-a)$ respectively. Let the tension in the lower pulley system string be $T_2$ and the acceleration of the 5kg and 3kg masses within their reference frame be $a'$.

$5(g-a)-T_2=5a'$
$T_2-3(g-a)=3a'$

$a'=\dfrac{g-a}{4} \implies T_2=\dfrac{15(g-a)}{4}$

Let the tension in the string attached to the sliding 8kg mass be $T_1$

Force equations for the 8kg mass and 2kg pulley are ...

$T_1 - 8g[\mu \cos(30) + \sin (30)] = 8a \implies T_1 - 8g = 8a$

$2T_2 + 2g - T_1 = 2a \implies \dfrac{15(g-a)}{2} + 2g - T_1 = 2a \implies \dfrac{19g}{2} -T_1 = \dfrac{19a}{2}$

adding the last two equations yields ...

$\dfrac{3g}{2}=\dfrac{35a}{2} \implies a= \dfrac{3g}{35}$

6. ## Re: pulley system on slope

There we go! Good job skeeter!

-Dan

7. ## Re: pulley system on slope

$5(g-a)-T_2=5a'$
$T_2-3(g-a)=3a'$

do the equations for the lower pulley 5g-t2=5(a+b) and t2-3g=(b-a) work?. in your answer why dont you account for the acceleration of both pulleys you wrote ^. should it the acceleration in the equation not be 5(a'+a) and 3(a'-a)

8. ## Re: pulley system on slope

do the equations for the lower pulley 5g-t2=5(a+b) and t2-3g=(b-a) work?
rearranging the two scalar equations I posted ...

$5(g-a) - T_2 = 5a' \implies 5g - 5a - T_2 = 5a' \implies 5g - T_2 = 5(a+a')$

$T_2 - 3(g-a) = 3a' \implies T_2 - 3g + 3a = 3a' \implies T_2 - 3g = 3(a'-a)$

see if that translates to the variables you chose to use ...

Ok thanks