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Brent0423

a pully has a mass of 2kg one one side and 4kg on the other side. calculate the acceleration of the system The net force on each block The tension in the string connecting the blocks

  • 2 years ago
  • 2 years ago

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  1. elica85
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    for massless pulley... \[F=ma\] \[mg-T=ma, T _{1}=m _{1} g-m _{1} a\] \[T-mg=ma, T _{2}=m _{2}a+m _{2}g\] acceleration is the same for both side, tensions will cancel

    • 2 years ago
  2. Brent0423
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    please solve the net force

    • 2 years ago
  3. Brent0423
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    please solve for the tension.. im really confused

    • 2 years ago
  4. elica85
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    \[m _{1}(g-a)=m _{2}(g+a)\] 4(g-a)=2(g+a) 39.2-4a=19.6+2a 19.6=6a a=3.3 m/s^2

    • 2 years ago
  5. elica85
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    net force: if you look at the side with the heavier mass, you can see that the force acting on the mass is gravity and tension. look at the lighter side, same thing. the tension is the same on both sides so they cancel. now the driving force of the system is the 4kg mass. the resistive force is the 2kg mass. so 4g-2g... 4g-2g=19.6N

    • 2 years ago
  6. elica85
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    plug in numbers in the first set of equations i gave you to solve for tension. it is already set up to solve for T and it will be the same for both sides

    • 2 years ago
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