# Thread: Forces in Equilibrium (Vectors) II

1. ## Forces in Equilibrium (Vectors) II

Here is another question I am suck on, it seems simple but I cant get it.

Maybe i am drawing the free body diagram wrong. Does the tension in the rope still need to be include since its the same on both sides? (since there is one rope). Thanks again!

2. Originally Posted by Oblivionwarrior
Here is another question I am suck on, it seems simple but I cant get it.

Maybe i am drawing the free body diagram wrong. Does the tension in the rope still need to be include since its the same on both sides? (since there is one rope). Thanks again!
The tension in the rope must be the same at the end of both boxes.

Let the blue box have a mass of $m_1$ and let the orange box have a mass of $m_2$

Doing a FB analysis on the blue box [the x direction is what matters] , we see that

$\sum F_x=0$:

$-m_1g\cos\alpha+T=0\implies T=m_1g\cos\alpha$

Doing a FB analysis on the orange box [the x direction is what matters] , we see that

$\sum F_x=0$:

$m_2g\cos(30^{\circ})-T=0\implies T=m_2g\cos(30)$

Therefore, since the tensions are the same, we see that $m_2g\cos(30)=m_1g\cos(66)$

Can you take it from here?

--Chris

3. Ooh that makes sense now, I was adding the normal force into the mix before. It was just making it complicated. So you don't need the normal force?

4. Sorry, shouldn't the force due to gravity along the incline be $mg\sin \theta$? Otherwise it'd be the rest of the analysis follows: $m_{1} \sin 66^{\circ} = m_{2} \sin 30^{\circ}$

unless I've thoroughly confused myself...

5. Originally Posted by o_O
Sorry, shouldn't the force due to gravity along the incline be $mg\sin \theta$? Otherwise it'd be the rest of the analysis follows: $m_{1} \sin 66^{\circ} = m_{2} \sin 30^{\circ}$

unless I've thoroughly confused myself...
I'm the one that confused myself!

--Chris

6. Originally Posted by Oblivionwarrior
Ooh that makes sense now, I was adding the normal force into the mix before. It was just making it complicated. So you don't need the normal force?
No. It just tells us that $N_1=m_1g$ and $N_2=m_2g$. It wouldn't tell us much, because the masses aren't known. Since their is tension in the string and they are inclined at different angles, it would have been best to look at the forces along the incline, rather than perpendicular to the incline.

--Chris

7. Are you sure its ? I got the same thing but with Cos.

8. sine, not cosine.

9. Would the ratio be 1.83:1?

10. Originally Posted by Oblivionwarrior
Would the ratio be 1.83:1?
Yes!

$1.83:1$ would signify that $m_2$ is $1.83$ times larger than $m_1$.

--Chris

11. Originally Posted by Chris L T521
Yes!

$1.83:1$ would signify that $m_2$ is $1.83$ times larger than $m_1$.

--Chris
I put 1.83:1 as the answer and it was wrong, did I screw up the ratio? Should It be 0.55:1 or 1:0.55 or 1:1.83

12. an "online" response, no doubt. the ratio for $\frac{M}{m}$ is correct ... how many decimal places do you need?

$\frac{\sin(66)}{\sin(30)} \approx 1.82709091529$

13. Originally Posted by skeeter
an "online" response, no doubt. the ratio for $\frac{M}{m}$ is correct ... how many decimal places do you need?

$\frac{\sin(66)}{\sin(30)} \approx 1.82709091529$
I got it to the right amount of decimals, but after reading the questions again " Determine the ratio of the mass of the right box to the mass of the left box" would the ratio be reversed? 0.55:1?

14. left side ... right side

$mg\sin(66) = Mg\sin(30)$

$\frac{\sin(66)}{\sin(30)} = \frac{M}{m}$

what else can I say?

15. Originally Posted by skeeter
left side ... right side

$mg\sin(66) = Mg\sin(30)$

$\frac{\sin(66)}{\sin(30)} = \frac{M}{m}$

what else can I say?
Hmmm, maybe I should just enter just the 1.827 instead of 1.827:1?