1
$\begingroup$

From my understanding, in this atwood machine, one mass is on a horizontal surface, and the other is hung off a pulley and left to freefall. Pictured below:

Modified Atwood

If only the hanging mass affects the acceleration of the entire system, why does the tension in m1 equal (m1*a)?

Trying to work through this myself: I may have misinterpreted that (m1*a) was mass times gravity, which isn't the case. However, that's still confusing. The acceleration is equal throughout the string, so the acceleration is equal to the force of gravity on m2. That makes sense to me algebraically and conceptually, but why does the TENSION equal (m1 * a)?

$\endgroup$
  • $\begingroup$ It might help to think of it as a problem where one force (the weight of the hanging mass) accelerates two the masses. $\endgroup$ – Farcher Nov 1 '18 at 8:50
1
$\begingroup$

Cut the string to the right of M1. You need to replace it with the force that the string exerted to the right on M1. That force is the tension in the string. Since there are no other forces acting on M1 horizontally (assumes frictionless table), the acceleration of M1 must be $\frac{T}{M_1}$.

Hope this helps.

$\endgroup$
0
$\begingroup$

Net Force for m1 = (m1 * a), but there's only one force acting in the x direction, being tension. You put in the tension for net force, and get T = (m1 * a).

$\endgroup$
0
$\begingroup$

I think it's easier to think about it as two different things, the force on a object is always $mass\times acceleration$

And in this case it's the $m2\times a2$ that gets transferred through the rope.

All of this is because the force on something is dependent from the mass of the object.

$\endgroup$
  • $\begingroup$ No, the force on something is NOT necessarily dependent on the mass of something. Maybe you meant acceleration is dependent on the mass? Consider pushing a block across a table with your finger. The force you exert depends on your pushing ability and is independent of the mass of the block. Or two charged objects interacting. The force of one on the other is independent of the masses of the objects. $\endgroup$ – Bill N Nov 2 '18 at 17:49
0
$\begingroup$

"If only the hanging mass affects the acceleration of the entire system,"

That's not true. Both masses affect the acceleration of the system. The weight of the hanging mass is the only external force acting on the system. If you change $m_1$ it's true you won't change the net force on the system, but you will definitely change (affect) the acceleration.

According the Newton's 2nd Law, the net force acting on a mass must be equal to the time derivative of the momentum. In constant mass systems, we often replace that time derivative with mass times the acceleration of that mass. This relationship holds regardless of the individual forces acting on the mass and whether they are internal or external to the system. In this case, there are 3 forces acting on $m_1$:

  1. the tension force of the string pulling parallel to the table top,
  2. the weight of $m_1$ acting down, and
  3. the normal (contact) force from the table top pushing upward.

The only acceleration observed is parallel to the table top, so the upward and downward forces must add to zero. Therefore, the tension force must be equal to $m_1\vec{a}$.

Notice that the tension force itself is NOT the same physical item as the product $m_1\vec{a}$. Mathematically, they are equal, but physically they represent different things. One is a force vector, and the other is an acceleration vector multiplied by a scalar. This is easier to understand if you reverse the traditional equation: $$\vec{a}=\frac{1}{m}\sum_j \vec{F}_j.$$ Here we are adding several force vectors, none of which are the same as an acceleration. After adding them and dividing by the mass we should get the same result as the single acceleration vector which we observe.

$\endgroup$

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.