# Thread: pulley system on block

1. ## pulley system on block

Im stuck on question 22 the first part
i resolved the forces paralel and horizontal to the slope.
i got the 2 equations T-√3 mg/2=ma using F=ma and kgm/2-T=kma
substituting these into each other I get a=kg+
√3 g /2k+2

From here i can not see how to prove k>
√3 if the weight of mass m is to accelerate towards P
Ive done most of the work but i am missing something small. maybe some one can tell me?

2. ## Re: pulley system on block

Parentheses would help. Do you mean kg+(√3 g/2k)+ 2 or (kg+ √3g)/(2k+ 2)?

3. ## Re: pulley system on block

(kg+ √3g)/(2k+ 2) is what I mean. Sorry I should of used brackets

4. ## Re: pulley system on block

You calculated $a$ wrong.

$T-\dfrac{mg\sqrt{3}}{2}=ma$
$\dfrac{kmg}{2}-T = kma$

So,
$\dfrac{kmg}{2}-T = k\left(T-\dfrac{mg\sqrt{3}}{2}\right)$
implies
$T = \dfrac{kmg(1+\sqrt{3})}{2(k+1)}$

Plugging that back in, you have:

$\dfrac{k\cancel{m}g}{2}-\dfrac{k\cancel{m}g(1+\sqrt{3})}{2(k+1)} = \cancel{m}a$

Thus, after simplification, you get:

$a = \dfrac{kg(k-\sqrt{3})}{2(k+1)}$

Now, the particle is moving towards P if $a>0$, so:

$0 < \dfrac{kg(k-\sqrt{3})}{2(k+1)} \Longrightarrow \sqrt{3}<k$

5. ## Re: pulley system on block

Thanks for the answer.
I'm not sure how you go from
Kmg/2 -kmg(1+✓3)/2(k+1)=ma
To a=kg(k-✓3)/2(k+1)

6. ## Re: pulley system on block

$T-\dfrac{\sqrt{3}}{2}mg = ma$

$\dfrac{1}{2}kmg - T = kma$

summing the two equations to eliminate $T$ yields ...

$\dfrac{mg}{2} \left(k-\sqrt{3}\right) = ma(k+1)$

solving for $a$ ...

$a=\dfrac{g(k-\sqrt{3})}{2(k+1)}$

Since the initial net force equations were formed under the assumption that $a > 0$, then $k > \sqrt{3}$