How Is Tension Calculated in a Pulley System with Multiple Masses?

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To calculate the minimum tension in a pulley system with multiple masses, the equation 2T - mg = ma can be applied, where T is tension, m is mass, and g is gravitational acceleration. In this scenario, the forces acting on the crate include its weight (W = mg) and the tension in the rope. When the system is at equilibrium, the net force is zero, meaning the upward tension equals the downward weight. Therefore, the relationship simplifies to 2T = W, indicating that the total tension must counterbalance the weight of the crate. This approach ensures the crate can be raised slowly with constant velocity.
mparsons06
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The angle is θ = 55.0°. The masses are, for the small pulley, m1 = 3.1 kg, for the traveling pulley, M2 = 5.7 kg, and for the crate, MC = 41.0 kg. What is the minimum tension with which the operator must pull on the cable (assume the cable is of negligible mass) in order to slowly raise the crate.

Please see the figure: http://i92.photobucket.com/albums/l18/bonbons06/prob93_upmass2pulleys.gif


My book shows an example of the equation:

2 x F(T) - mg = ma


I'm not sure if that's the equation to use and if it is, how do I apply it?
 
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Do a Force FBD.

Tension is the only force holding up the Box. The Weight force is pushing the box down.
When these two forces are equal the net force is zero and it is the minimum force to raise the crate. (force can be zero, but still have constant velocity to raise the box)

Weight for up is mg.

right above the box, the rope has equal tension...this tension is equal to the tension that the guy must pull. The tension throughout the rope is equal.

2T - W = Fnet ..If Fnet is zero
2T= W
 
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