Two trucks are around the pulley

[heartOFphysic]

(see image)

How do I go about this?

since a1 is larger, and the two trucks are around the pulley, then a2 is opposing (to some extent) so i thought the answer would be a1-a2? but I checked the answer and it's a1+a2/2

please help

thank you :)

 

[Tea Jay]

View attachment 42314
(see image)

How do I go about this?

since a1 is larger, and the two trucks are around the pulley, then a2 is opposing (to some extent) so i thought the answer would be a1-a2? but I checked the answer and it's a1+a2/2

please help

thank you :)

Well, both trucks are moving, but a pulley would split the force, so the total pull divided by 2 would be right.

On the other hand, in the drawing...it just looks to me like the pulley is not used as a pulley per se...but, that could be an artifact of the schematic.If you think of one (a) truck as stationary, and a pulley generally being used to double force while halving speed...the moving truck is essentially hoisting the third truck at half its speed.

If both are moving, they are both doing that, and both pulling at the same ratio, so the total is divided by two.If both trucks are going the same speed, but as the rigging is shown, it looks like they are just dragging the pulley along at that speed, and the third truck with it.If one truck is faster, essentially, the pulley is being dragged along at their average speed, or the two speeds divided by 2.

 

[cupid.callin]

View attachment 42314
(see image)

How do I go about this?

since a1 is larger, and the two trucks are around the pulley, then a2 is opposing (to some extent) so i thought the answer would be a1-a2? but I checked the answer and it's a1+a2/2

please help

thank you :)

Approach this problem like this:

let the masses be m1, m2, m3
Let in frame of pulley, truck 1 and 2 have acc. a in direction left and right
And let whole system has acc. a' in ground frame

And you know their individual acc. in ground frame is a1 and a2

so just write Newton eqn's and solve

 

[cupid.callin]

On the other hand, in the drawing...it just looks to me like the pulley is not used as a pulley per se...but, that could be an artifact of the schematic.

You will see the pulley working perfectly like a pulley if you were sitting on it ... its not looking like a motion due to pulley because the whole system is moving
Its like putting a system of 1 pulley and 2 masses in an elevator accelerating and watching from ground frame

 

[Tea Jay]

You will see the pulley working perfectly like a pulley if you were sitting on it ... its not looking like a motion due to pulley because the whole system is moving
Its like putting a system of 1 pulley and 2 masses in an elevator accelerating and watching from ground frame

I meant that its not really doubling the forces because the two ends are not pulling the third truck...its not anchored to itself for mechanical advantage.

If used the way a pulley is used for advantage, to simplify, one truck could drive 10 mph, pulling the third truck at 5 mph, with double the power. The pulley could be for example set with one rope to the bumper to anchor it, and the other to the pulley block.The way it looks on my admittedly small screen is that if we made the SHOWN pulley a simple harness that didn't slide/just held the two ropes to the third truck's rope...and drove forward with one truck, at 10 mph...we'd get a pull at the one truck's 10 mph speed.

If the same harness was used by both pulling trucks, and they went forward at the same 10 mph speed, even if one went slower, we'd get the 10 mph speed (Unless the slower truck was so slow it induced drag, etc...)

If we put the pulley back in, all it would do is allow slippage between the two truck's speeds, so the drag would be at the average speed...but the pulley is not acting mechanically to increase the speed or force of the pull in any scenario.This is a logic problem, not a math problem. The answer is obvious if you simply look at it without doing any math at all.

 

[heartOFphysic]

Approach this problem like this:

let the masses be m1, m2, m3
Let in frame of pulley, truck 1 and 2 have acc. a in direction left and right
And let whole system has acc. a' in ground frame

And you know their individual acc. in ground frame is a1 and a2

so just write Newton eqn's and solve

LOL, that went slightly over my head, is it possible to simplify (ie just explain a bit more) this solution. what's "ground frame" and what's "have acc. a in direction left and right"

Sorry for the inconvenience! :(

 

[PhanthomJay]

If the 2 trucks are accelerating at the same rate (a1 = a2), then it should be visualized that truck 3 must also be accelerating at that same rate and same speed , since it is solidly connected to the others thru the pulley. Only equation C gives that simple result of a3 = a1 = a2. Process of elimination.

 

[heartOFphysic]

If the 2 trucks are accelerating at the same rate (a1 = a2), then it should be visualized that truck 3 must also be accelerating at that same rate and same speed , since it is solidly connected to the others thru the pulley. Only equation C gives that simple result of a3 = a1 = a2. Process of elimination.

how do you know they are accelerating at the same rate?

 

[Tea Jay]

how do you know they are accelerating at the same rate?

It doesn't matter if they are or not.

He used that as an example to simply the thought process.

Forget its a pulley for a minute...and get rid of one truck.

OK, we have a truck dragging another truck along at a speed, say 10 mph.

The pulling truck is going 10 mph, so the truck being pulled is being pulled at 10 mph.

Now, tie a rope to a second truck's rear bumper, and run the other end to the first truck's rear bumper...then tie a new rope from the middle of the rope between them, and run that end to the third truck...tie it it its bumper.

Drive the two trucks forward at 10 mph, and it drags the third truck along at the same speed.Have one pulling truck slow down, and, until the end of the slack is used up, the towed truck is STILL being pulled along at 10 mph.Change the knot in the middle to a pullley, and now as the second truck slows, the pulley allows the slack to play through, so the pulley is being pulled forward more slowly...but, the SPEED is now the AVERAGE of the two pulling trucks.

The average speed is just the total divided by 2.

So, the only answer they list that looks like the average speed is the one that adds the two together, and then divides by 2.

 

[Tea Jay]

The way to look at the problem is to break it down into its parts, and see what makes sense to you.

IE: What in each part is doing what?

Each component is doing something that interacts with another or others. If I pull this rope, what is it pulling on? What should happen as a result?

Do I understand what I am looking at?

If you go to the math before you've established what the problem is really asking, you can have a mathematically correct answer to a question that was not really asked.In this case, it was simply looking for the end result of the towing arrangement on speed.

If you look at what pulls what, you see that the only role of the pulley is to allow a (temporary) speed difference between the two pulling rigs...not to provide a mechanical advantage.

Once that's observed, you can proceed to imagine what that would mean to a pulley's speed in that scenario.

That would then lead to the realization that the pulley will travel at whatever rate its being pulled at...and that if one truck was slower, the pulley would allow it to lag behind for a bit...which would slow the pulley's forward progress.

You'd see that the pulley's speed would be made faster by the faster truck, and slower by the slower truck.

As its a simple pulley, with no mechanical advantage, etc...the average of the two speeds will be the pulley's speed, as the faster and slower truck's speeds essentially proportionally cancel each other out.In this particular problem as there were no units, speeds, masses, etc to calculate, and you merely needed to "set-up" the equation...no math was required, just setting it up...and picking the multiple choice equation that looked like it did what you needed to be done.

So, logically, you see the pulley speed will be the average of the truck speeds, you see an equation that gives the average of the two trucks' speeds, and that's pretty much enough.I didn't bother to go through cupids math...on a cell phone, I was lucky to see the drawing. :D

:D

 

[Tea Jay]

Approach this problem like this:

let the masses be m1, m2, m3
Let in frame of pulley, truck 1 and 2 have acc. a in direction left and right
And let whole system has acc. a' in ground frame

And you know their individual acc. in ground frame is a1 and a2

so just write Newton eqn's and solve

Out of curiosity, could you post this all worked out?

 

[heartOFphysic]

Out of curiosity, could you post this all worked out?

YES! I too am interested in knowing how you can "mathematically" work it out!

 

[cupid.callin]

Out of curiosity, could you post this all worked out?

YES! I too am interested in knowing how you can "mathematically" work it out!

Hi ...

Well i redid my method and found a flaw ... so anyway i corrected it and here's how it should be done ...

black are acc. and red are forces

acceleration of system - a' - right

In ground frame: acc of m1 - a1 - right

acc of m2 - a2 - right​

In pulley frame: acc of m1 - a - right

acc of m2 - a - left​

Tension in wire = T & Forces F1 and F2

____________________
Now write force eqn's for m1 and m2 in ground frame (net acc will be a1 and a2)

(I'll be leaving this)

Eqn's in pulley frame ...

i'll just write one

T + m₂a' -F₂ = m₂a ... (I used pseudo force concept here)

Now put value of F1 and F2 from bottom 2 eqn in upper 2 ... then its just simple solving ...

EDIT: Sorry i can't post complete solution ... I don't want to get in some trouble :p

 

[Tea Jay]

Hi ...

Well i redid my method and found a flaw ... so anyway i corrected it and here's how it should be done ...

black are acc. and red are forces

acceleration of system - a' - right

In ground frame: acc of m1 - a1 - right

acc of m2 - a2 - right​

In pulley frame: acc of m1 - a - right

acc of m2 - a - left​

Tension in wire = T & Forces F1 and F2

____________________
Now write force eqn's for m1 and m2 in ground frame (net acc will be a1 and a2)

(I'll be leaving this)

Eqn's in pulley frame ...

i'll just write one

T + m₂a' -F₂ = m₂a ... (I used pseudo force concept here)

Now put value of F1 and F2 from bottom 2 eqn in upper 2 ... then its just simple solving ...

EDIT: Sorry i can't post complete solution ... I don't want to get in some trouble :p

So, do you see what I mean about the pulley not being used as a pulley would be used traditionally, IE: Its not used to increase mechanical advantage, merely to allow slippage between pulling truck speeds?Also - He knew the correct answer already, and why it was correct, so its not like there was a spoiler alert. :D

 

[cupid.callin]

So, do you see what I mean about the pulley not being used as a pulley would be used traditionally, IE: Its not used to increase mechanical advantage, merely to allow slippage between pulling truck speeds?Also - He knew the correct answer already, and why it was correct, so its not like there was a spoiler alert. :D

Well I would still comment that pulley IS working just like a pulley ...
Though yes, there no need and no benefit in using the pulley here.

PS: Happy New Year