(adsbygoogle = window.adsbygoogle || []).push({}); Multiple Pulleys and Masses with Friction (Picture Included)

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 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}

3. 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.

**Physics Forums - The Fusion of Science and Community**

# Multiple Pulleys and Masses with Friction

Have something to add?

- Similar discussions for: Multiple Pulleys and Masses with Friction

Loading...

**Physics Forums - The Fusion of Science and Community**