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Doubt: At pulley $A$, consider the free-body diagram. Why is the tension, $\vec{T}$, in both hands downwards? I know $\vec{T}_2$ is upwards since it is the force acting upwards in order to prevent breaking the rope. I don't understand the logic of $\vec T$ in the left and $\vec T$ in the right. Please help me.

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Tension in an ideal light string is a scalar, not a vector.

The tension is maintained by forces at the ends of the string equal in magnitude to the tension. If we are considering forces on pulley $A$, then the geometry of the attachment means that the forces from the string are downward.

If you were thinking about the forces on whatever is pulling down on the strings, then that force would be upward.

By declaring the force on the pulley to be $\vec T$, the problem is using shorthand to say that the force is equal in magnitude to the tension. For simple problems like this, it can be easy to miss the difference because the direction is pretty clear from examination.

I don't get your point. In system (a) Tension is pulling upward. Why there is a change in the system (b)?

"Tension" doesn't have a direction and doesn't pull in any direction. One end of the string (because of its orientation) pulls upward and the other end pulls downward. Both ends pull with the same magnitude. You have to define which end you are talking about to tell what direction the force from the string is oriented.

Because $A$ is above the string, it is pulled downward by the string. Because $B$ is below the string, it is pulled upward by the string.

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  • $\begingroup$ I don't get your point. In system (a) Tension is pulling upward. Why there is a change in the system (b)? $\endgroup$ – Unknown x Jul 17 '19 at 2:21
  • $\begingroup$ The difference is the orientation of the pulley and the string. I've added some to the answer. $\endgroup$ – BowlOfRed Jul 17 '19 at 4:18

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