7
$\begingroup$

If we consider a simple pulley system with two masses hanging on each end of a MASSLESS and INEXTENSIBLE string around a MASSLESS and FRICTIONLESS pulley, how then can one reason that the tension at each end of the string must be the same?

My own reasoning:

MASSLESS ROPE means that for any segment of the rope with tension $T_1$ and $T_2$ we have that $\sum F = T_ 2 - T_1 = 0$ (since $m = 0$) and thus the tensions must be the same, on a non curved rope at least!

INEXTENSIBLE means that no energy can be stored in the string, however I fail to see how this is a neccesary condition (for equal tension)

MASSLESS PULLEY means that no rotational inertia exists, and thus no force can alter the tension of the string (?)

FRICTIONLESS PULLEY is hard for me to figure.

Needless to say, I feel quite at a loss conceptually!

Improve this question
$\endgroup$
1

3 Answers 3

Reset to default
6
$\begingroup$

What we mean by a frictionless pulley is that the friction in the bearings of the pulley is negligible, and the pulley is free to rotate without any resistance. We don't mean that the friction between the string and the pulley surface is negligible. In fact, we assume that there is enough static friction between the string and pulley surface to prevent the string from slipping. But, in this case, if you do a moment balance on the pulley, you must then conclude that the tensions in the string on either side are equal (even if the pulley has angular acceleration), since the moment of inertia of the pulley is assumed to be zero.

Improve this answer
$\endgroup$
3
$\begingroup$

In general case such system moves with acceleration... If rope is extensible we cannot assume that the magnitudes of accelerations of the two masses are the same( they will NOT move "jointly"). If the pulley has a mass then tensions on left and right of the pulley must differ to ensure the pulley's rotational acceleration. If pulley is not frictionless the friction force must be accounted for when writing 2-nd law of Newton.

Improve this answer
$\endgroup$
1
$\begingroup$

Since the rope is massless, and since two identical masses are attached to rope-ends, then as far as forces on rope are concerned, the problem has left-right symmetry. This symmetry itself assures you that tension in both sides of the rope must be equal. This is true whether or not the pulley is frictionless.

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.