0
$\begingroup$

In my book (Halliday Resnick Walker) there is a solved example which is as follows

There is a uniform disk with mass $M$ = 2.5kg and radius $R$ = 20cm, mounted on a fixed horizontal axle. A block with mass $m$ = 1.2kg hangs from a massless cord that is wrapping around the rim of the disk. Find the acceleration of the falling block, the angular acceleration of the disk, and the tension in the cord. The cord does not slip and there is no friction at the axle.

We need to find the acceleration of the block. So the equation to relate the forces on block to acceleration of body is given in the solution as $T - mg = ma$

but I think it should be $mg - T = ma$ as the block is moving in downward direction $mg$ should be greater than $T$ which is tension. Why is it given on the book as $T - mg = ma$?

Also can anyone help me how to find the tension forces acting on a string in complex system while making their free body diagram?

$\endgroup$
  • $\begingroup$ The tension in the string is also responsible for the Torque which makes the disk rotate about its axis. this might help $\endgroup$ – Hritik Narayan Jan 17 '15 at 12:50
  • $\begingroup$ You need not worry about it. If you put $T-mg$ instead of $mg-T$, the answer will only come as negative due to it being in the opposite direction. $\endgroup$ – AvZ Jan 17 '15 at 14:29
1
$\begingroup$

When solving problems of this kind, you shouldn't worry if you subtract larger quantities from smaller ones. The quantities such as $T$, $mg$, etc. are vectors, and the +/- sign only denotes the direction they are pointed towards. In this case $T$ and $mg$ are in opposite directions so either one could have the minus sign, as you've pointed out in the example. It would make no difference in the outcome of the problem if you were to interchange the signs. It only depends on the coordinate system you use. i.e. which direction is positive, and which negative, in this case.

After making Free Body Diagrams, apply all possible equations such as $F_{net}=ma$, $\tau_{net}=I\alpha$, Conservation of Energy.etc as applicable in the problem.

$\endgroup$
1
$\begingroup$

but I think it should be $mg - T = ma$ as the block is moving in downward direction $mg$ should be greater than $T$ which is tension. Why is it given on the book as $T - mg = ma$?

You are arguing about a sign convention, and by doing so, you are missing the point.

Newton's second law is $\vec F = m\vec a$, not $F=ma$. Force and acceleration are vector quantities. The vectorial nature of Newton's second law can be reduced to a scalar problem when all of the forces and motion are collinear. This conversion from a vector problem to a scalar problem involves a completely arbitrary selection of which direction is positive and which direction is negative.

Your textbook uses the widespread convention that up is positive, down is negative. Your $mg-T=ma$ is implicitly using the opposite convention. There's nothing wrong with either convention. Done correctly, both will yield the exact same answer.

$\endgroup$
0
$\begingroup$

One thing that I tell people on every single force problem is to draw a free body diagram to help you envision the situation. Now in the scenario you quote there are two objects to consider, but for the specific question you pose to this site, you only need to look at the block's motion. Draw your free body diagram, then keep reading.

One thing you'll notice when you look the free body diagram, which you have already figured out is that there are two blocks that act on the block, the tension $T$ and the weight $mg$. Because these forces act in opposite directions you are right to subtract them. Either your order or the textbooks order could be correct depending on your choice of coordinate system. Your equation of $mg-T=ma$ will give the same magnitude of the net force and acceleration as the textbook, but will differ by a minus sign.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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