# Solving Force with 3 Masses Problem: Acceleration of m2

• StephenDoty
In summary, the system consists of three masses, m1=2.5kg, m2=1.5kg, and m3=4.5kg, connected by two strings (with uniform tension throughout) and initially at rest on a horizontal table with coefficients of static friction us=.3 and kinetic friction uk=.25. The net force equations for m1, m2, and m3 are T1-m1g=m1a, T3-fk-T1=m2a, and m3g-T3=m3a, respectively. To find the acceleration of m2, these three equations can be solved simultaneously, keeping in mind that the tension forces will change as the masses accelerate.
StephenDoty
A mass m2= 1.5kg rests on a horizontal table. us= .3 and uk = .25. M2 is attached by strings to m1= 2.5kg hanging over oneside of the table and m3= 4.5kg hanging over the other side. The system is initially at rest. Once released what is the acceleration of m2?

Ok for the force involving m1: m1 < m3 so m1 will be going up, so F1= T-m1g
For the force involving m3: m3 is going down so F3=m3g - T

I do not know what to do from here.

Any help would be appreciated. Thank you.

Stephen

StephenDoty said:
Ok for the force involving m1: m1 < m3 so m1 will be going up, so F1= T-m1g
For the force involving m3: m3 is going down so F3=m3g - T
The tensions in the strings are different, so label them T1 & T3 (or something).

You found the net force on m1 & m3, so apply Newton's 2nd law to each.

And don't forget m2. What forces act on it? Apply Newton's 2nd law to it as well.

You'll get three equations and three unknowns.

Ok for the force involving m1: m1 < m3 so m1 will be going up, so F1= T1-m1g=m1a
For the force involving m3: m3 is going down so F3=m3g - T4=m3a

F2=T3-(fk+T2)=m3a?

I do not know what to do from here.
What do I solve for to find the acceleration of m2?
Any help would be appreciated. Thank you.

Stephen

First you should calculate the static friction- the coefficient of static friction, $\mu_S$ times the weight of M2. (I guessed that us and ks were the coefficients of static and kinetic friction- it would have been a good idea to say that.) None of the masses will move until the net force on M2 is greater than the static friction.

The net force on M2 is, of course, the difference between the weights of the two other masses.

If M3 and M1 are such that the net force on M2 is greater than the static friction, you must calculate the kinetic friction and subtract that from the net force to find the total force accelerating M2.

There are only two strings, so there are only two tensions. Let's call them T1 (for the string attached to m1) and T3 (for the string attached to m3).
StephenDoty said:
Ok for the force involving m1: m1 < m3 so m1 will be going up, so F1= T1-m1g=m1a
Good.
For the force involving m3: m3 is going down so F3=m3g - T4=m3a
Good, but let's rewrite it using T3: m3g - T3 = m3a

F2=T3-(fk+T2)=m3a?
I assume this is supposed to be the force equation for m2, so it should be:
T3 -fk -T1 = m2a

What's fk?

To find the acceleration (for all the masses, since they are connected) solve those three equations together. Hint: See if you can combine the equations to eliminate the tensions.

HallsofIvy said:
The net force on M2 is, of course, the difference between the weights of the two other masses.
That's only true for the static case; if it accelerates, the tensions will no longer equal the weights of the hanging masses.

there are 2 strings?
one from the first hanging mass to the mass2 then from mass2 to mass three?

So F for mass 1= T1-mg
F for mass 3= mg-T2

and for mass 2 F= T2-(k friction) - (T1)

But if there are 2 strings with four sections of tension: one from first hanging mass to pulley, next from pulley to mass 2, then from mass 2 to other pulley, and lastly from pulley to hanging mass 3, then my origional 3 equations were right.

Which set of equations do I use??

And are we not assuming that after they are released the k friction is greater than the s friction? If not the s friction = 4.45N.
What do I do with this information?

How do I find the acceleration of m2?? And what do I solve for to help me find the acceleration??

thank you

Stephen

Last edited:
StephenDoty said:
there are 2 strings?
one from the first hanging mass to the mass2 then from mass2 to mass three?
Correct.

So F for mass 1= T1-mg
F for mass 3= mg-T2

and for mass 2 F= T2-(k friction) - (T1)

But if there are 2 strings with four sections of tension: one from first hanging mass to pulley, next from pulley to mass 2, then from mass 2 to other pulley, and lastly from pulley to hanging mass 3, then my origional 3 equations were right.
The tension is uniform throughout the string. There are two strings and thus two tensions. String 1 exerts the same tension force on m1 that it does on m2.

Which set of equations do I use??

How do I find the acceleration of m2?? And what do I solve for to help me find the acceleration??
See post #5.

(Note: I assume that the strings are massless, and that any pulleys are massless and frictionless.)

## 1. How do you calculate the acceleration of m2 in a force problem with three masses?

The acceleration of m2 can be calculated using the formula a = F/m2, where F is the net force acting on m2 and m2 is the mass of m2.

## 2. What is the relationship between the acceleration of m2 and the net force in this problem?

According to Newton's Second Law of Motion, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. So, in this problem, the acceleration of m2 will increase as the net force acting on it increases.

## 3. Can you use this formula to solve for the acceleration of any of the other masses in the problem?

Yes, this formula can be used to calculate the acceleration of any of the masses in the problem. However, you will need to use the net force and mass of the specific object you are trying to find the acceleration for.

## 4. How do you determine the direction of the acceleration in this problem?

In this problem, the direction of the acceleration will depend on the direction of the net force acting on m2. If the net force is in the same direction as the acceleration, then the acceleration will be positive. If the net force is in the opposite direction, then the acceleration will be negative.

## 5. What other variables should be considered when solving this force problem with three masses?

When solving this problem, it is important to also consider the initial velocity of each mass, as well as any external forces acting on the system. These variables can impact the overall net force and therefore affect the acceleration of m2.

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