- #1

Wara

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**Multiple Pulleys and Masses with Friction (Picture Included)**

## 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 m

_{1}= 10kg.

Let m

_{2}= 20kg.

(1) m

_{1}a

_{1}= T

_{1}- μm

_{1}g

(2) m

_{2}a

_{a}= m

_{2}g - T

_{2}

(3) 2T

_{1}= T

_{2}

(4) 2a

_{2}= a

_{1}

## The Attempt at a Solution

T

_{2}= m

_{2}g

T

_{2}= (20kg)(9.8m/s

^{2})

T

_{2}= 196N

T

_{1}= [itex]\frac{T_{2} + μm_{1}g}{2}[/itex]

T

_{1}= [itex]\frac{196N + (0.32)(10kg)(9.8m/s^{2})}{2}[/itex]

T

_{1}= 113.68N

__Sub (3) and (4) into (2)__

m

_{2}a

_{2}= m

_{2}g - T

_{2}

m

_{2}([itex]\frac{a_{1}}{2}[/itex]) = m

_{2}g - 2T

_{1}

__Called this new equation (5).__

(5) m

_{2}([itex]\frac{a_{1}}{2}[/itex]) = m

_{2}g - 2T

_{1}

(1) m

_{1}a

_{1}= T

_{1}- μm

_{1}g

(5) m

_{2}([itex]\frac{a_{1}}{2}[/itex]) = m

_{2}g - 2T

_{1}

__(1) + (5)__

m

_{1}a

_{1}+ m

_{2}([itex]\frac{a_{1}}{2}[/itex]) = T

_{1}- μm

_{1}g + m

_{2}g - 2T

_{1}

a

_{1}(m

_{1}+ m

_{2}([itex]\frac{1}{2}[/itex]) = -μm

_{1}g + m

_{2}g - T

_{1}

a

_{1}(10kg + 20kg([itex]\frac{1}{2}[/itex]) = -(0.32)(10kg)(9.8m/s

^{2})+(20kg)(9.8m/s

^{2})-113.68N

a

_{1}= 2.5448 m/s

^{2}

__from (4)__

2a

_{2}= a

_{1}

a

_{2}= [itex]\frac{a_{1}}{2}[/itex]

a

_{2}= [itex]\frac{2.548m/s^{2}}{2}[/itex]

a

_{2}= 1.274 m/s

^{2}

**Therefore, T**

_{1}= 113.68N, T_{2}= 196N, a_{1}= 2.548 m/s^{2}, and a_{2}= 1.274 m/s^{2}.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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