Mechanics Problem involving connected particles on inclined plane

[James10492]

Hello good math people,

I am having a bit of difficulty with this problem which I have spent a long time looking at, I thought I had come up with the solution, and then I checked my answer and it turns out it is wrong. I feel like I need another set of eyes at this point to check my working.

Here is a quick sketch diagram of the problem from my notebook:



"Two boxes, P and Q, of masses 3m kg and 2m kg respectively, are attached to the ends of a light inextensible string. Box P lies on a rough surface that is inclined at 30 degrees to the horizontal. The coefficient of friction (myu) between the box and the surface is .1"

The system is released from rest with the strings taut.

Work out the time taken for P to reach a speed of 10m/s".

(You need to formulate equations of motions to work out the acceleration of the particles. Through a bit of trial and error I worked out that the system is accelerating towards the right-hand side of the diagram):

[math]2mg - T = 2ma \\ T - 3mg\sin 30 - F = 3ma[/math]
F = R x myu = 3mgcos30 x myu, so [math]T - 3mg\sin30 - .1(3mg\cos30) = 3ma \\ T - \frac{1}{2}mg(3-.3\sqrt3) = 3ma \\ T-1.24 mg = 3ma[/math]
Now you can add the simplified equations of motion to determine the acceleration of the system:

[math]3ma=T-1.24mg\\ 2ma=2mg-T\\ 5ma=.76mg \\ a = .152g[/math]
.152g = .152 x 9.8 = 7.5m/s^2

Now P started at rest, so the initial velocity is 0,

final velocity is 10,

10 = 7.5 x t

t = 1.33 (wrong answer)

What am I doing wrong?

 

[topsquark]

Hello good math people,

I am having a bit of difficulty with this problem which I have spent a long time looking at, I thought I had come up with the solution, and then I checked my answer and it turns out it is wrong. I feel like I need another set of eyes at this point to check my working.

Here is a quick sketch diagram of the problem from my notebook:

View attachment 32598

"Two boxes, P and Q, of masses 3m kg and 2m kg respectively, are attached to the ends of a light inextensible string. Box P lies on a rough surface that is inclined at 30 degrees to the horizontal. The coefficient of friction (myu) between the box and the surface is .1"

The system is released from rest with the strings taut.

Work out the time taken for P to reach a speed of 10m/s".

(You need to formulate equations of motions to work out the acceleration of the particles. Through a bit of trial and error I worked out that the system is accelerating towards the right-hand side of the diagram):

[math]2mg - T = 2ma \\ T - 3mg\sin 30 - F = 3ma[/math]

You put a huge amount of work into this. I'm sorry to say you made a sign error almost at the start. Note that the acceleration of the 2m mass is downward so we have T - 2mg = - 2ma.

-Dan

 

[James10492]

You put a huge amount of work into this. I'm sorry to say you made a sign error almost at the start. Note that the acceleration of the 2m mass is downward so we have T - 2mg = - 2ma.

-Dan

I'm still not getting it; [math]3ma = T-1.24mg \\ -2ma = 2mg - T \\ ma = .76mg \\ a = .76g = 7.5m/s^2[/math]
I thought that because the particles move in the same (clockwise) direction they should have the same sign? Now that I think about it it does make sense that you should still consider the x and y direction. But then why is it still wrong with your adjustment?

(I see I made a mistake in my original working with .152*9.8 = 7.5 sorry I am tired)

 

[James10492]

The given solution by the way is 2.12s (so acceleration would be 4.71)

 

[topsquark]

Okay, I made a goof on the coordinate system. Sorry about that!

Looking more carefully, you wrote the two following lines:
[imath]T - 3mg ~ sin(30) - 0.1(3mg ~ cos(30)) = 3ma[/imath]

[imath]T - \dfrac{1}{2} (mg(3 - 0.3 \sqrt{3} )) = 3ma[/imath]

Note that there is a sign error in the [imath]0.3 \sqrt{3}[/imath] term.

However I'm getting an acceleration of 0.471 m/s^2, not 4.71 m/s^2.

-Dan

 

[skeeter]

[imath]2mg - T = 2ma[/imath]

[imath]T - 3mg\sin{\theta} - \mu \cdot 3mg \cos{\theta} = 3ma[/imath]

sum the equations …

[imath]2mg -3mg(\sin{\theta}+\mu \cos{\theta}) = 5ma[/imath]

[imath]\dfrac{g[2-3(\sin{\theta} + \mu \cos{\theta})]}{5} = a[/imath]

I get an acceleration that’s 1/10 of your 4.71, which makes the time 10 times longer than the 2.12 sec

 

[James10492]

[imath]2mg - T = 2ma[/imath]

[imath]T - 3mg\sin{\theta} - \mu \cdot 3mg \cos{\theta} = 3ma[/imath]

sum the equations …

[imath]2mg -3mg(\sin{\theta}+\mu \cos{\theta}) = 5ma[/imath]

[imath]\dfrac{g[2-3(\sin{\theta} + \mu \cos{\theta})]}{5} = a[/imath]

I get an acceleration that’s 1/10 of your 4.71, which makes the time 10 times longer than the 2.12 sec

Thanks for your help; I arrived at that solution now. Your method is clearer than mine. I will approach these problems in the same way from now on. I will assume the solution in the book is a typo.

Regarding the acceleration: I remembered that you are supposed to give both particles the same acceleration when you are given the modelling assumption that they are connected by a rope/cable that is inextensible.