Calculating Tension in Ropes with Multiple Blocks and Friction | Homework Help"

[Ogakor]

Homework Statement

Block A, B and C are placed as in the figure below and connected in ropes of negligible mass. Both A and B weighs 25N each and the coefficient of kinetic friction between each block and the surface is 0.35. The weight of Block C is 30.75774639 and it descends with constant velocity.
a.) Find the tension in the rope connecting blocks A and B
b.) If the rope connecting A and B were cut, what would be the acceleration?

[PLAIN][URL]http://i735.photobucket.com/albums/ww359/anyone11/awewae.jpg[/PLAIN][/URL]

given:
w1 = 25
w2 = 25
w3 = 30.75774639
coefficient of friction (u) = 0.35
normal force (N) = 25 (for both A and B)
g = 9.8m/s²

I already have an answer but it's different from my friend's. He assumed acceleration to be 0 because velocity is constant. But I really feel confident about my solution.

The Attempt at a Solution

a.)
T1 - f1 = m1g
T1 - u(N) = (w / g) x a
T1 - 0.35(25N) = 25N / 9.8 x a
T1 - 8.75N = 2.551020408a

T2 - f2 - T1 - w2x = m2a
T2 - u(w2 sin theta) - T1 - w2 (cos theta) = (w/g) x a
T2 - 0.35(25N sin 36.9) - T1 - (25 cos 36.9) = 25 / 9.8 x a
T2 - 25.24579343 - T1 = 2.551020408a

w3 - T2 = w3/g a
30.75774639 - T2 = 3.13854555 / 9.8 a

add the three..

T1 - 8.75N = 2.551020408a
T2 - 25.24579343 - T1 = 2.551020408a
30.75774639 - T2 = 30.75774639 / 9.8 a

-3.23804704 = 8.24086366 a

divide -3.23... by 8.24...

a = -0.392938911 m/s²

T1 - 8.75N = 2.551020408a
T1 - 8.75N = 2.551020408(-0.392938911)
T1 = 7.747604819N

b.)
T2 - 25.24579343 - T1 = 2.551020408a
30.75774639 - T2 = 3.13854555/ 9.8 a

add -25.24... and 30.76... and 2.5... and 2.55... and 3.14...
then, divide...

a = 7.012084432m/s²

So, that's my solution. Who's correct, my solution or my friend's where he assumed acceleration to be 0. Please teach me the correct formula. :)

 

[city25]

Well, I think yours is wrong. Your answers are different from mine.
and f2 should be 0.35(25cos36.9).
you seems to use the wrong component.
and what is w2x? the horizontal component of w2?

BTW, I want to know your friend's work.

anyway, I'm not the assistant, please wait for their help.

 

[Ogakor]

Well, I think yours is wrong. Your answers are different from mine.
and f2 shold be 0.35(25cos36.9).
you seems to use the wrong component.
and what is w2x? the horizontal component of w2?

BTW, I want to know your friend's work.

anyway, I'm not the assistant, please wait for their help.

Sorry, I forgot to use another variable. w2x is w2 (cos theta).
I forgot how my friend did it but I think his answer to letter A is 8.75(not sure though).
His letter B is 6m/s² (forgot the exact amount. acceleration is 6 point something)

 

[city25]

Sorry, I forgot to use another variable. w2x is w2 (cos theta).
I forgot how my friend did it but I think his answer to letter A is 8.75(not sure though).
His letter B is 6m/s² (forgot the exact amount. acceleration is 6 point something)

I have the answer 8.75N in a) too.
But my ans in b) is 3 point something.

 

[rwisz]

I cannot say who is correct without seeing your friend's solution. However, your solution seems to be more accurate as it takes into account the friction and acceleration of the blocks. Your friend's assumption of zero acceleration may not have taken into account the forces acting on the blocks and may not be accurate. It is important to use the correct formulas and equations when solving physics problems to ensure accuracy.