- #1
Wara
- 9
- 0
Multiple Pulleys and Masses with Friction (Picture Included)
http://content.screencast.com/users/Waraa/folders/Snagit/media/2d699ba1-32e0-4d9c-b79f-1a1d85dccd12/03.19.2012-15.49.46.png
I have solved the question but I am still iffy as to whether my solution is correct or not.
Let m1 = 10kg.
Let m2 = 20kg.
(1) m1a1 = T1 - μm1g
(2) m2aa = m2g - T2
(3) 2T1 = T2
(4) 2a2 = a1
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.
Homework Statement
http://content.screencast.com/users/Waraa/folders/Snagit/media/2d699ba1-32e0-4d9c-b79f-1a1d85dccd12/03.19.2012-15.49.46.png
I have solved the question but I am still iffy as to whether my solution is correct or not.
Homework Equations
Let m1 = 10kg.
Let m2 = 20kg.
(1) m1a1 = T1 - μm1g
(2) m2aa = m2g - T2
(3) 2T1 = T2
(4) 2a2 = a1
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:
