Mechanics problem

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December 19th 2010, 08:54 AM
FGT12

Mechanics problem

A 2kg textbook rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is 0.15m, to a hanging book with mass 3kg. The system is released from rest, and the books are observed to move 1.2m in 0.8s.

My solution so far:
using constant acceleration formula
$s = ut+\frac{1}{2}at^2$
$1.2 = 0 +\frac{1}{2}(a)(0.8)^2$
$a = 3.75$

For the hanging part:
$3g - T_1 = ma$
$T_1 = 3g-ma$
$T_1 = 3g-3(3.75) = 18.2$ (rounded) This agrees with my textbook

however for the horizontal tension in the cord
I make it
$T_1 - T_2 = ma$
$T_2 = T_1 - ma$
$T_2 = 18.2 - 2(3.75)$
$T_2 = 10.7N$

however the answer in my textbook is 7.5 which would suggest
$T_2=2(3.75)$

where has the $T_1$ force gone?
Have I missed out a force
Also how can you have different tensions in the cord surely it would snap?
December 19th 2010, 08:59 AM
snowtea

If you draw the free body diagram for the 2kg book. $T_1 = (2kg)a$

The equation $T_1 - T_2 = ma$ is incorrect.
If your pulley has no mass or is frictionless, then $T_1 = T_2$ .
However, this is not the case, so there is some inertia in the pulley.
$(T_2 - T_1)(0.15m/2)$ is the torque on the pulley.
December 19th 2010, 09:17 AM
FGT12

why wouldnt the cord snap if it has different tensions in it?
December 19th 2010, 09:27 AM
snowtea

Tension is just a force, it can be balanced by other forces such as friction.

Think about dragging a really long heavy chain on the ground.
The front link of the chain has the most tension and friction.
The back link of the chain may have only a little tension and a little friction.
However, you are right, the net acceleration at each link must be the same (if it has fixed length).