2 blocks one over the edge with a pulley

[vivitribal]

Homework Statement

I have 2 blocks of mass M one on a frictionless horizontal surface the other hanging over the edge of the surface connected by a string of length l. The string goes over a pulley of mass m. Find the acceleration.

Homework Equations

PE = mgh
KE = 1/2mv^2

The Attempt at a Solution

I know how to do the problem without the pulley having any mass how do you factor in the pulley's mass?

 

[rsk]

Are you sure this is an energy question? These are usually to do with forces on the blocks.

 

[rsk]

And come to think of it, the mass of the pulley isn't usually given or needed - at least in the versions we do...

 

[berkeman]

He can do it with energy, but the two m values will be different.

Alternately, draw the FBD for the rope -- what is the tension?

 

[berkeman]

Oops, I missed the pulley mass in the question. So you need to factor in something about the energy in the pulley wheel's rotation as well. Let's see some more equations, vivitribal.

 

[marlon]

I know how to do the problem without the pulley having any mass how do you factor in the pulley's mass?

HINT : Torque and the rotational inertia of the pulley (ie a circle)

marlon

 

[vivitribal]

Ok so the force down from the hanging block is M*g which has to pull the pulley and the other block which would be (m+M)*a? or would it be m*alpha +M*a?

 

[marlon]

You need three things here :

1) Newton's second law for the mass on the table (tension on mass1 T1)
2) Newton's second law for the mass hanging down (tension on mass2 T2 + gravity)
3) Newton's second law in angular form for the pulley : F' R = I (F' is the tangential force on the pulley, which is the tension in the wire ofcourse. Keep in mind that you have TWO forces acting on the pulley). To get rid of the alpha, you can use : = a/R

For 3) you need the rotational inertia I = 1/2(mR^2)of the pulley, which is a solid disk (m is the mass of the pulley, r is the radius).


ATTENTION : keep an eye on the signs of the force vectors here. i suggest you take the positive direction along the x and y-axis to be the ones in which the blocks move (x : to the left, y to the right)

marlon

 

[berkeman]

i suggest you take the positive direction along the x and y-axis to be the ones in which the blocks move (x : to the left, y to the right)

I think maybe that's a small typo? If the pulley is on the left edge of the table (say), maybe you meant take +x aimed to the left and +y aimed down?

 

[vivitribal]

Let's try this again sum of the forces = m*a so M1*T1+M2*g+I*(T2-T1)-M2*T2= (M1+M2+I)*a and solve for a. Is that right or am I still missing something

 

[berkeman]

No, you are mixing unlike quantities. Tension is a force, not an acceleration. And the moment of inertia I is not the same as a mass.

Remember F=ma, and review how to use the moment of inertia...

 

[vivitribal]

OK so ignoring the pulley for a second the equation would be T1-T2+M2*g=(M1+M2)*a is that correct at least?

 

[berkeman]

Which is the hanging mass (I'm assuming M2)? And if the pully wheel is massless, what can you say about T1-T2?

 

[vivitribal]

* M1-------o * ^ * +y
* ____T1-> | * | * -x +x
* _________ | * T2 * -y
* _________ |
* _________ M2

Wouldn't T1-T2 be the acceleration?

 

[franniemeow07]

So I did this question and one thing that is EXTREMELY tricky (i.e. I kept bangning my head against the wall, trying to do this while I was sleepy...), is that on the FBD for the blocks, you have T1 going to the right and T3 going up. For the FBD for the pulley, these are OPPOSITE in direction; i.e. T1 on the pulley is going to the left and T3 on the pulley is going down.

 

[vivitribal]

Yes I realized that about the tensions and the pullies which you can use to find the pulley's angular accel which has to be the same as the systems accel but how do you find tensions?

 

[vivitribal]

Can someone help please?

 

[OlderDan]

Can someone help please?

There is no solution to this problem as stated. Some people have mentioned the moment of inertia of the pulley, and that is needed to solve the problem. However, the problem statement gives you no way to find it.

Homework Statement

I have 2 blocks of mass M one on a frictionless horizontal surface the other hanging over the edge of the surface connected by a string of length l. The string goes over a pulley of mass m. Find the acceleration.

If you assume a radius R for the pulley, and make the usual assumption that the pulley is a disk, then you can solve it. Since you were not asked about the tensions in the string, you could solve this with conservation of energy. If you prefer, you can use the approach outlined by marlon in #8, except for the typo. I = ½mR², not ½m².

If you use energy, only one object in the problem is changing its gravitational potential energy (GPE). The loss of GPE of that one object is transformed into KE of all three objects. If you use marlon's outline, you should write three separate equations for the three objects before trying to lump everything into one equation. If you write those separate equations, it will be easy for others to check what you have done.

 

[marlon]

I think maybe that's a small typo? If the pulley is on the left edge of the table (say), maybe you meant take +x aimed to the left and +y aimed down?

yep you are right

marlon

 

[marlon]

OK so ignoring the pulley for a second the equation would be T1-T2+M2*g=(M1+M2)*a is that correct at least?

How did you acquire that ?

Listen,you ALWAYS need to split up the problem in several distinct directions (here x and y): THIS IS THE VERY FUNDAMENT OF NEWTONIAN PHYSICS. After having drawn a FBD, you need to apply Newton's second law in each direction separately. Just calculate each vector along x and y directions, beware of the signs !

1) mass 1 : only a horizontal tension force T1 : m1a = ?
2) mass2 : vertical tension force T2 and gravity : m2a = ?
3) Pulley, TWO FORCES T1 and T2, calculate the torque for both forces

WATCH THE SIGNS OF THE VECTORS (many students make mistakes with these). In the case of the sign of the torque and alpha : use the right hand rule. Keep in mind that the positive z axis is POINTING DOWN, so clockwise rotation is positive here !

The answer has nearly been given now !

 

[vivitribal]

Ok I think I have it.

T1=M1*a
T2=M2*a+M2*g
T2-T1=1/2*m*r^2*alpha=1/2*m*r*a
combining we get
M2*a+M2*g-M1*a=1/2*m*r*a
solving for a we get
a=2*M2*g/(m*r)

Is that right or am I still missing something?

 

[vivitribal]

Can someone confirm or deny this?

 

[Doc Al]

T1=M1*a

OK.

T2=M2*a+M2*g

How do the accelerations of M1 and M2 relate to each other? Not just in magnitude, but in sign.

T2-T1=1/2*m*r^2*alpha=1/2*m*r*a

That should be torque = I alpha, not force.

 

[vivitribal]

How do the accelerations of M1 and M2 relate to each other? Not just in magnitude, but in sign.

That should be torque = I alpha, not force.

In reference to the first one T2=-M2*a-M2*g thanks for pointing that out.

For the second one are you saying that (T2-T1)*r=I*alpha?

 

[OlderDan]

Ok I think I have it.

T1=M1*a <== OK for mass M1 on table
T2=M2*a+M2*g <== inconsistent sign. This a is - the a above
T2-T1=1/2*m*r^2*alpha=1/2*m*r*a <== Dimensionally inconsistent. Something is missing.
combining we get
M2*a+M2*g-M1*a=1/2*m*r*a
solving for a we get
a=2*M2*g/(m*r)

Is that right or am I still missing something?

See anotations above. You have a sign problem and something is missing in your third equation.

 

[Doc Al]

In reference to the first one T2=-M2*a-M2*g thanks for pointing that out.

The signs are still not quite right. Only the sign of "a" was wrong.

For the second one are you saying that (T2-T1)*r=I*alpha?

Yes.

 

[vivitribal]

Here we go again

T1=M1*a
T2=-M2*a+M2*g
(T2-T1)*r=1/2*m*r^2*alpha
T2-T1=1/2*m*a <--r^2 cancels from alpha=a/r and for the r on the other side
substituting we get
-M2*a+M2*g-M1*a=1/2*m*a
M2*g=a*(1/2*m+M2+M1)
a=M2*g/(1/2*m+M2+M1)

So hopefully for the last time is this correct?

 

[Doc Al]

Looks good to me.

 

[vivitribal]

Finally thank you everyone for your help.