Multiple Pulleys and Masses with Friction

1. Mar 19, 2012

Wara

Multiple Pulleys and Masses with Friction (Picture Included)

1. The problem statement, all variables and given/known data
http://content.screencast.com/users/Waraa/folders/Snagit/media/2d699ba1-32e0-4d9c-b79f-1a1d85dccd12/03.19.2012-15.49.46.png [Broken]

I have solved the question but I am still iffy as to whether my solution is correct or not.

2. Relevant equations
Let m1 = 10kg.
Let m2 = 20kg.

(1) m1a1 = T1 - μm1g
(2) m2aa = m2g - T2
(3) 2T1 = T2
(4) 2a2 = a1

3. The attempt at a solution
T2 = m2g
T2 = (20kg)(9.8m/s2)
T2 = 196N

T1 = $\frac{T_{2} + μm_{1}g}{2}$
T1 = $\frac{196N + (0.32)(10kg)(9.8m/s^{2})}{2}$
T1 = 113.68N

Sub (3) and (4) into (2)
m2a2 = m2g - T2
m2($\frac{a_{1}}{2}$) = m2g - 2T1

Called this new equation (5).
(5) m2($\frac{a_{1}}{2}$) = m2g - 2T1

(1) m1a1 = T1 - μm1g
(5) m2($\frac{a_{1}}{2}$) = m2g - 2T1

(1) + (5)
m1a1 + m2($\frac{a_{1}}{2}$) = T1 - μm1g + m2g - 2T1
a1(m1 + m2($\frac{1}{2}$) = -μm1g + m2g - T1
a1(10kg + 20kg($\frac{1}{2}$) = -(0.32)(10kg)(9.8m/s2)+(20kg)(9.8m/s2)-113.68N
a1 = 2.5448 m/s2

from (4)
2a2 = a1
a2 = $\frac{a_{1}}{2}$
a2 = $\frac{2.548m/s^{2}}{2}$
a2 = 1.274 m/s2

Therefore, T1 = 113.68N, T2 = 196N, a1 = 2.548 m/s2, and a2 = 1.274 m/s2.

My question is if my solution is correct. Please check and reply back with suggestions or corrections. Thank you so much.

Last edited by a moderator: May 5, 2017
2. Mar 19, 2012