Multiple Pulleys and Masses with Friction


by Wara
Tags: acceleration, force of friction, tension, two mass, two pulleys
Wara
Wara is offline
#1
Mar19-12, 03:12 PM
P: 9
1. The problem statement, all variables and given/known data


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

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

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


(1) m1a1 = T1 - μm1g
(5) m2([itex]\frac{a_{1}}{2}[/itex]) = m2g - 2T1

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

from (4)
2a2 = a1
a2 = [itex]\frac{a_{1}}{2}[/itex]
a2 = [itex]\frac{2.548m/s^{2}}{2}[/itex]
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.
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Wara
Wara is offline
#2
Mar19-12, 07:53 PM
P: 9
Please help me with this question. have i done this question correctly?


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