(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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/s^{2}

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.

3. 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/s^{2}

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/s^{2}

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. :)

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# Homework Help: Tension homework help

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