Tension, Net Force, and Acceleration relationship?

  • Thread starter Riman643
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Yes, because the net forces on each of the objects are pointing in different directions, and force is a vector. So if someone taught it to you in this way, they used bad judgment.
Thanks. I think I understand now where my confusion was coming from.
 
  • #27
kuruman
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$$ma=T-mg\sin{8}-mg(\cos{8})f$$
Easy to miss, but the angle of the upper incline relative to the horizontal is ##\theta_2=20^o+8^o## which affects the components of the weight.

On edit: Wait a minute, what happened to OP's drawing where both inclines were on the same side with the pulley where the slope changes? I am confused.
 
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  • #28
haruspex
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it looks like you have written separate expressions for the net force acting on each object.
Please use the reply and quote features so that readers know to which text your comment is directed.
 
  • #29
haruspex
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Problem Statement: What is the relationship between Tension, Net Force, and Acceleration?

The tension force on the first object be equal to the tension force on the second object?
Not always.

In pulley problems, the usual assumption is that the string does not slip on the pulley. If the question statement refers to a frictionless pulley it means the axle, not the contact with the string.

If the pulley is massless and has no axial friction then the tension must be the same on each side of the pulley; otherwise there would be a net nonzero torque on the pulley, leading to infinite angular acceleration.

If the system is static and there is no axial friction then, again, the two tensions must be equal, even if the pulley has mass.

In all other cases you need to allow two different tensions and should draw a separate FBD for the pulley.

Finally, there is the rope/string. If that has mass and the system is accelerating then the two ends of the same section of rope will have different tensions. Draw an FBD for each section.
 

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