Conceptual Help on Force F Applied to Block & Bracket

[Nat3]

My question is a little different in that I don't need help with the calculations for my problem, but rather I just need help understanding a concept. Take a look at the situation below:

[PLAIN]http://img546.imageshack.us/img546/3459/phys.png

The problem is then, "What is the maximum force F that can be applied if the block is not to slide on the bracket?"

I tried doing the problem and ended up getting the wrong answer. Upon asking my professor what I did wrong, I was told that the force acting on the bracket in the +x direction (to the right) is $2\vec F$. I've tried and tried to understand this, and maybe I'm just stupid, but I simply don't get it. No matter how I think about it, it seems to me that the force acting on the bracket should be $\vec F$, not $2\vec F$.

Could anyone try to explain this to me?

I really appreciate your help. Thanks.

 

[gash789]

Imagine if the fource was equal to the weight of the mass, such that it was in equilibrium. Then the weight is F, you can then draw a force diagram for the pulley, this will therefore have two strings with a force F meaning 2F in the positive x (for the pulley).

Does this help?

 

[grzz]

Clearly mass M does not move to the left by itself. It is being pulled in that direction. Now if mass M is being pulled to the left by the rope then the rope feels itself being pulled to the right by M.(remember Newton's 3rd law?). Hence the lower rope is pulling on the bracket (grey system) to the right.
But then there is also force F pulling on the bracket to the right.
Assuming there is no friction between pulley and rope, then BOTH sections of rope are pulling by force F on bracket to the right.

 

[Nat3]

So if I'm pulling on the rope with a force of 10N, the bracket experiences a force of 20N? That doesn't make any sense to me since it's like energy is being created out of nowhere. It doesn't seem like looping the rope around the pulley and attaching it to a mass should make the bracket twice as easy to pull?

Thanks for your help :shy:

 

[PhanthomJay]

So if I'm pulling on the rope with a force of 10N, the bracket experiences a force of 20N? That doesn't make any sense to me since it's like energy is being created out of nowhere. It doesn't seem like looping the rope around the pulley and attaching it to a mass should make the bracket twice as easy to pull?

Thanks for your help :shy:

Actually in this example, it is not twice as easy to pull...you apply the same force to move the bracket and mass system with or without the pulley, and you use the same amount of energy (work) even though the force on the bracket increases (but not the force on the bracket-mass system).
There are many cases, however, where you can use less force to move an object, using multiple pulleys (see 'mechanical advantage'), but even in these cases, using less force requires applying that force over a greater distance to achieve the desired displacemnt, meaning you still the need the same energy (force times distance) to accomplish the task. You can't get 'something for nothing'.

 

[SammyS]

Draw a Free Body Diagram for each object, the bracket and the block.

Indeed, the rope exerts a force of 2F on the bracket, each of the lengths exerting a force of F. This force is to the right (+x direction).

The block exerts a force on the bracket by way of friction. This force is to the left.

Of course, the net force on the bracket is to the right and is less than 2F.

There are other forces involved, but those are the horizontal forces exerted on the bracket.

 

[Nat3]

Actually in this example, it is not twice as easy to pull...you apply the same force to move the bracket and mass system with or without the pulley, and you use the same amount of energy (work) even though the force on the bracket increases (but not the force on the bracket-mass system).

...You can't get 'something for nothing'.

I think that's part of what's so confusing to me. We say that we're pulling on the rope with a force of F, but that the bracket experiences a force of 2F. Isn't that saying that the force is being doubled?

Draw a Free Body Diagram for each object, the bracket and the block.

Here is my attempt:

[PLAIN]http://img13.imageshack.us/img13/5706/fbdw.png

The block exerts a force on the bracket by way of friction. This force is to the left.

Of course, the net force on the bracket is to the right and is less than 2F.

I'm kind of confused on this point. Could you elaborate a bit?

Thanks.

 

[SammyS]

Draw a Free-Body-Diagram for the block also.

Since the block moves (accelerates) to the right, there must be a force exerted upon it that's to the right. That force must be friction. The only other horizontal force acting on the block is the force exerted by the rope, and that's to the left.

 

[PhanthomJay]

I think that's part of what's so confusing to me. We say that we're pulling on the rope with a force of F, but that the bracket experiences a force of 2F. Isn't that saying that the force is being doubled?

Suppose you have a bracket attached to the ceiling, you fasten a pulley to it, and wrap a rope around it. Now hang a 100 N block on one free end of the rope, and pull on the other free end with a force of 100 N. The system is in equilibrium. You are pulling with a force of 100 N. The bracket is being pulled on with a force of 200 N, or double what you are pulling at. It's the tension force from the hanging weight that provides the other 100 N. Do you have a problem with that?

 

[Nat3]

Here is my attempt:

[PLAIN]http://img13.imageshack.us/img13/5706/fbdw.png

Draw a Free-Body-Diagram for the block also.

[PLAIN]http://img268.imageshack.us/img268/8513/fbd2.png

Since the block moves (accelerates) to the right, there must be a force exerted upon it that's to the right. That force must be friction. The only other horizontal force acting on the block is the force exerted by the rope, and that's to the left.

I'll try to process this :shy:

Suppose you have a bracket attached to the ceiling, you fasten a pulley to it, and wrap a rope around it. Now hang a 100 N block on one free end of the rope, and pull on the other free end with a force of 100 N. The system is in equilibrium. You are pulling with a force of 100 N. The bracket is being pulled on with a force of 200 N, or double what you are pulling at. It's the tension force from the hanging weight that provides the other 100 N. Do you have a problem with that?

No, that makes sense..

 

[SammyS]

Horizontal Forces:

If the bracket exerts a force (to the right) f_s on the block, then the block exerts an equal (in magnitude) and opposite (in direction) force on the bracket. So there is an additional force, f_s that's exert (to the left) on the bracket.

BTW: In addition to what you have, the rope exerts a force, F, on the block which is to the left.

Include both of these additional forces in your Free-Body-Diagrams.

Notice that both objects have the same acceleration, a, which is to the right.​

Vertical Forces:

What are you using for the normal force, FN ?

What you have drawn for the bracket isn't correct, but probably won't cause you to get the wrong numerical answer. However, some graders (like me) would deduct points for the error.​

 

[Nat3]

Is this correct, then?

[PLAIN]http://img11.imageshack.us/img11/1087/fbd3.png

Since the block is not moving and the force F is pulling the block to the left, f_s = F ?

Since the bracket exerts force f_s on the block to the right, then the block exerts force f_s to the left on the bracket.

Does this mean that the net force acting on the bracket is 2F-f_s = 2F-F = F ?

 

[SammyS]

Is this correct, then?

[PLAIN]http://img11.imageshack.us/img11/1087/fbd3.png

Since the block is not moving and the force F is pulling the block to the left, f_s = F ?

No, the block moves with the bracket, so its acceleration is the same as that of the bracket. I would call it a & assume it's to the right.

Since the bracket exerts force f_s on the block to the right, then the block exerts force f_s to the left on the bracket.

Does this mean that the net force acting on the bracket is 2F-f_s = 2F-F = F ?

No, it's simply 2F-f_s, which you can't evaluate until you solve the problem

Where is any ma ?