How can you stop it from descending without friction?

[fl00f]

The particular situation is below:

"For the system shown in the attached image, find the force required to prevent the mass m2 from descending. Use the values of m1 = 5.25 kg, m2 = 6.50 kg and m3 = 2.500 kg. Assume that all surface are frictionless and that the ropes do not stretch. (Also, in the figure, assume that mass m2 is flush against m1.) "My first approach is to solve it via centripetal force by assuming that the rope will spin the system in a circle. This will result in a frictional force acting on m2 from m1 upwards against gravity. But how is that possible if all surfaces are frictionless as the question states?

Thank you for your help!

 

[gneill]

A frictional force, to exist, requires a force acting in the opposite direction (static friction) or relative motion between surfaces (kinetic friction). Neither of these situations apply to the situation here.

Think of inertial forces. Draw the FBD diagrams for the bodies assuming that the force applied via the rope is accelerating the system linearly.

 

[fl00f]

In drawing a FBD, I see that the Force applied is at 90° to the force of gravity pulling m2 down. How do perpendicular forces cancel each other? Or is that not the case...

 

[gneill]

In drawing a FBD, I see that the Force applied is at 90° to the force of gravity pulling m2 down. How do perpendicular forces cancel each other? Or is that not the case...

It is not the case. Perpendicular forces cannot cancel So the force to restrain the falling mass has to come from m3 (hint, hint). Again, think of inertial forces. What happens when the system is pulled to the right by the rope?

 

[fl00f]

When the system is pulled to the right, m3 resists (I presume that this is the inertial force? ..I'm not quite sure of what inertial forces are). If this resisting force is equal to the force of gravity pulling m2 and m3 downwards [(m2+m3)g] then m2 will not fall?

However, I don't believe I understand the situation completely b/c I can't see how m1 should be involved?

 

[gneill]

When the system is pulled to the right, m3 resists (I presume that this is the inertial force? ..I'm not quite sure of what inertial forces are). If this resisting force is equal to the force of gravity pulling m2 and m3 downwards [(m2+m3)g] then m2 will not fall?

However, I don't believe I understand the situation completely b/c I can't see how m1 should be involved?

Inertial forces are the forces that resist changes in motion. Masses have inertia -- they "want" to keep moving in whatever direction and at whatever velocity they have. It requires external force to be applied to change that motion, and they resist change. Newton's laws describe this -- 1st law and 3rd law.

Gravity is pulling down on all three blocks, but only block 2 is able to move vertically. Let's isolate our attention on blocks 1 and 2 from a moment. If block one is being held in place somehow (don't ask how yet!), then what will be the tension in the rope connecting blocks 1 and 2?

 

[fl00f]

If by block 1 you mean m3 in the image then the tension would be equal to (m2)g in magnitude

So it would be erroneous to consider the force of gravity acting on the system to equal (m3+m2)g, b/c it would only be (m2)g, as m3 does not act in the vertical direction?

 

[gneill]

If by block 1 you mean m3 in the image then the tension would be equal to (m2)g in magnitude

So it would be erroneous to consider the force of gravity acting on the system to equal (m3+m2)g, b/c it would only be (m2)g, as m3 does not act in the vertical direction?

Yes, sorry, I'd mixed up the block numbers. I should have consulted the diagram to refresh my memory.

Carrying on... Yes, only block 2 will have gravity contributing to the tension in the rope. Block 3 simply presses down vertically on block 1.

So, if the rope is pulling on block 3 with rope tension m2*g what force would be available to counteract that?

 

[fl00f]

The inertial force of block 3 which acts leftward and is equal in magnitude to the force causing the system to move rightward

 

[gneill]

The inertial force of block 3 which acts leftward and is equal in magnitude to the force causing the system to move rightward

Oooh, so close! Yes, it'll be the inertial reaction force of block 3 which does the trick. But it won't equal in magnitude the force causing the system to move rightward -- that force has to contend with moving all three blocks, not just block 3.

However, you've got the basic idea. The first thing to determine is the acceleration that block 3 has to experience in order to provide the required force to counter m2's rope tension.

 

[fl00f]

Ohh hmm! I solved the question using your suggestions:

1. m3*a = m2*g
a = m2*g / m3

2. F applied = (m1 + m2 + m3) a
= (m1 + m2 + m3) * (m2*g / m3)
= 363 N

which is the correct answer.

It now makes sense to me that the force that needs to be resisted is the downward force of m2*g and that the resisting force is by m3. The next logical step would be to equate it to m3*a as you hinted. The acceleration necessary would be provided for by the applied force. This applied force must accelerate all three blocks and be thus equal to (m1 +m2 +m3)*a.

What still perplexes me is the "inertial reaction force" or the force by m3 that resists m2...is this a real force or the resultant of other forces? (like centrifugal force) It confuses me that a mass can resist a force when there is no friction present. Or is this characteristic explained in Newton's 3rd law, when it states that all forces have a reaction force ?

Much thanks by the way for the frequent and very helpful answers!

 

[gneill]

Ohh hmm! I solved the question using your suggestions:

1. m3*a = m2*g
a = m2*g / m3

2. F applied = (m1 + m2 + m3) a
= (m1 + m2 + m3) * (m2*g / m3)
= 363 N

which is the correct answer.

Yup! Well done.

It now makes sense to me that the force that needs to be resisted is the downward force of m2*g and that the resisting force is by m3. The next logical step would be to equate it to m3*a as you hinted. The acceleration necessary would be provided for by the applied force. This applied force must accelerate all three blocks and be thus equal to (m1 +m2 +m3)*a.

What still perplexes me is the "inertial reaction force" or the force by m3 that resists m2...is this a real force or the resultant of other forces? (like centrifugal force) It confuses me that a mass can resist a force when there is no friction present. Or is this characteristic explained in Newton's 3rd law, when it states that all forces have a reaction force ?

Inertial forces are the reaction forces that you get from mass when acceleration takes place. It's mass "wanting" to move at a constant speed in a straight line. As you surmise, it's a result of Newton #1 and Newton #3. The magnitude of the reaction force in response to acceleration is proportional to what is called the "inertial mass" of the object. F = M*A. It is observed that in all cases, so far as we know, inertial mass is equal to gravitational mass for all things. Inertial force, the result of accelerating inertial mass, is the reason why all masses fall at the same rate in a gravitational field.

Inertial force is a form of reaction force -- inertial forces show up whenever you have masses being accelerated, or if you happen to choose a non-inertial reference frame for your coordinate system (such as a coordinate system associated with an accelerating train. Or a coordinate system that rotates). Centrifugal force is an inertial effect.

Inertial forces are sometimes called pseudo-forces when they are the result of a choice of reference frame.

 

[fl00f]

That's a new concept I wasn't familiar with. I'll read more into it. Thanks so much for your help! You're very good at explaining things :)