# Pulley Force Vector: Check Friction on 10kg Weight

• inv
In summary: I'm guessing that the forces on the block counteract the tension in the string, giving the weight the advantage?In summary, the block of wood successfully acted as a lever directing the force perpendicular to the arm of the machine, as the red arrow indicates.
inv
I'd like to ask the question as above, I think it successfully acted as a lever and redirected the force vector perpendicular to the arm of the machine, but it does've > friction, reducing the force acting on the 10kg Weight, could anyone that knows pls help check for me?*Edit, ... I think it successfully acted as a *pulley ...

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Hello **-1,

Is this the complete text of the exercise ? What is the aim of the depicted contraption ?

What's **-1?

Yes, complete. There's no aim, it's just an exercise to test lever and force vector knowledge.

So the exercise is to compare the torque on the machine arm when the block is present with the torque when the block is absent ?
If so, the thing to do is calculate that torque in both cases !

No, my friction comment was just a trivia, the question is only as in the picture.

Ah, there is another way to read this: the machine is meant to lift the weight and you want to know if the weight is lifted more easily (with more accleration) when the block is present than with the block absent ?

In that case convert the torque from the machine (should be the same in both cases) to a tension in the cable for both situations.

(Can we assume the machine arm orientation is given as drawn ?)

No, the question is simply "Did the block of wood successfully act as a lever directing the force perpendicular to the arm of the machine, as the red arrow indicates?".

Yes, that's what it said. Is the weight a means to handle the arm or is the arm a means to lift he weight ? Not that it matters (symmetry).
inv said:
"Did the block of wood successfully act as a lever directing the force perpendicular to the arm of the machine, as the red arrow indicates?"
Why should it ? The wire also exerts an upward force on the block, counteracting the torque from the tangential component. Still need a calculation to show if one of the two changes wins.

I cannot see any mention of a pivot for the lever. Is it supposed to be somewhere, out of shot, at the other end of the 'arm'? Its position is important if you want to calculate the torque.
Presumably that circle on the left is a Pulley and not a "lever"?
If all of the stuff that has been added to the arm is rigid then the force marked in red is not relevant because the string tension where it joins the rigid arrangement is being shared by internal forces in the block etc.. The torque without the block is the string tension times the perpendicular distance from the pivot that we cannot see. The torque with the block is the string tension times the new perpendicular distance (the angle has changed). The details of the arm and block are not relevant; the arm could be any shape you like - it's just the direction of the string that counts and that affects the length of a perpendicular line, dropped from the pivot to the string.

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Sry typo, there're no levers in this question, all levers typed here were supposed to be pulleys, edited the picture w/ the question.

The force from the string at the red dot is 90° to the arm, if the string is 90° to the arm.

What is the block of wood attached to ?

This "arm" has to be a lever, doesn't it?

I think the short answer is no! The block does not achieve the desired effect if it is fixed to the arm.

As proof, consider what happens if the round pulley is moved up and to the right so that it's at the pivot point of the arm.

Merlin3189
BvU said:
What is the block of wood attached to ?
Another thinner block of wood screwed to the machine arm, as shown.
sophiecentaur said:
This "arm" has to be a lever, doesn't it?
It has to be attached to something at its end, so yes.
CWatters said:
I think the short answer is no! The block does not achieve the desired effect if it is fixed to the arm.

As proof, consider what happens if the round pulley is moved up and to the right so that it's at the pivot point of the arm.
I think there's no difference to the perpendicular force vector to the arm, what do you mean, maybe a picture can explain better?

I have to agree with Cwatters. I can't see why people are making such a meal of this.
IF the string and block are smooth (ie. frictionless) then the tension in the string at the red dot is the same as in the rest of the string. It will behave as if it were a pulley on a block attached to the arm.
BUT there will still be the additional forces on the block being transferred to the arm. So it will not behave like the *pulley on the left.
Practically it will not be frictionless, so it doesn't make sense to think of it as a pulley.
IMO, all the unnecessary calculations would yield the same results whether it was a rough block attached to the arm or a smooth pulley (of very small diameter!) attached to the arm.

Why don't you just get rid of the bits and pieces and tie the string to the arm, or to a nice solid extension if you need to offset the attachment?

sophiecentaur
The question has not been defined fully enough for a proper answer. That is one of the problems here; we have to make assumptions where the information has been omitted.
If you replace the corner of that block with a tiny pulley, it would make no difference to the force that the deflected string exerts on the 'arm'. The force on the arm is just the tension in the string times the sine of the angle between the string and the arm (the component of the tension at right angles to the arm). Whether you follow my original assumption of the arm being a lever or not, the final answer is the same.
It is important [NOT! Edit] to rely on any 'gut' reaction that one has, due to the way the picture has been drawn and to get to the basics of it - that is, a string pulling at an arm at an angle. I was assuming that the arm is a lever because it's more convenient and because it is very unlikely that the arm would be attached rigidly to 'something' without being intended to turn it. In any case, whether intentional or not, there will be a torque on the arm. Unless the object it is attached to is shaped so that its cm is actually behind the contact point of the string (i.e. somewhere outside the picture, to the right) then there is a torque, by definition.
It is rather disappointing that this question about such a concrete bit of Mechanics seems to be treated in such an arm waving way. Given the facts, there is only one solution.

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BvU
inv said:
I think there's no difference to the perpendicular force vector to the arm, what do you mean, maybe a picture can explain better?

OK try this diagram. The rope is in red.

If you think the force on the arm is still to the left (causing it to rotate clockwise) then try this one. Which way will this one rotate?

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BvU and sophiecentaur
A.T. said:
The force from the string at the red dot is 90° to the arm, if the string is 90° to the arm.
The force on the string is at right angles but what about the force of the block on the arm, too? At the corner where the string bends around the block, there will be another force component that cannot be ignored. You cannot just pick a force that you fancy and use only that force in your conclusions.

CWatters
Merlin3189 said:
I have to agree with Cwatters. I can't see why people are making such a meal of this.
IF the string and block are smooth (ie. frictionless) then the tension in the string at the red dot is the same as in the rest of the string. It will behave as if it were a pulley on a block attached to the arm.
BUT there will still be the additional forces on the block being transferred to the arm. So it will not behave like the *pulley on the left.
Practically it will not be frictionless, so it doesn't make sense to think of it as a pulley.
IMO, all the unnecessary calculations would yield the same results whether it was a rough block attached to the arm or a smooth pulley (of very small diameter!) attached to the arm.

Why don't you just get rid of the bits and pieces and tie the string to the arm, or to a nice solid extension if you need to offset the attachment?

So the force vector on the block's perpendicular to the red dot?
sophiecentaur said:
The question has not been defined fully enough for a proper answer. That is one of the problems here; we have to make assumptions where the information has been omitted.
If you replace the corner of that block with a tiny pulley, it would make no difference to the force that the deflected string exerts on the 'arm'. The force on the arm is just the tension in the string times the sine of the angle between the string and the arm (the component of the tension at right angles to the arm). Whether you follow my original assumption of the arm being a lever or not, the final answer is the same.
It is important [NOT! Edit] to rely on any 'gut' reaction that one has, due to the way the picture has been drawn and to get to the basics of it - that is, a string pulling at an arm at an angle. I was assuming that the arm is a lever because it's more convenient and because it is very unlikely that the arm would be attached rigidly to 'something' without being intended to turn it. In any case, whether intentional or not, there will be a torque on the arm. Unless the object it is attached to is shaped so that its cm is actually behind the contact point of the string (i.e. somewhere outside the picture, to the right) then there is a torque, by definition.
It is rather disappointing that this question about such a concrete bit of Mechanics seems to be treated in such an arm waving way. Given the facts, there is only one solution.

If I'm correct, if a force is applied on an object's centre of gravity( generally center of mass), it will displace; Elsewhere, it'll turn.

Are you using the right angle triangle Sin Y°= Opposite/ Hypothenus formula?

But is the Force Vector at the point perpendicular to the arm?
CWatters said:
OK try this diagram. The rope is in red.
View attachment 99588
I think it'll pull up on the pivot and destroy it.

CWatters said:
If you think the force on the arm is still to the left (causing it to rotate clockwise) then try this one. Which way will this one rotate?
View attachment 99591
To the right. What's the reason for these 2 results?

sophiecentaur said:
The force on the string is at right angles but what about the force of the block on the arm, too? At the corner where the string bends around the block, there will be another force component that cannot be ignored. You cannot just pick a force that you fancy and use only that force in your conclusions.

Am I right to say you're saying the block's friction is causing the same force vector on it on the arm? If so, if the block's replaced with a pulley, it still causes the same "force vector" on the arm, unless it's separate from that arm, ie anywhere from its pivot, even while still on the machine?

inv said:
Am I right to say you're saying the block's friction is causing the same force vector on it on the arm? If so, if the block's replaced with a pulley, it still causes the same "force vector" on the arm, unless it's separate from that arm, ie anywhere from its pivot, even while still on the machine?
The corner (or the pulley you replace it with) is pushing the string to one side (to make it kinked). So there has to be an extra force on it, in addition to the 'tension' in the string from the corner to the red spot. That extra force is transmitted to the arm. The resultant of this force and the string tension will be the same as the force from the long piece of string (i.e. 100N) The direction will necessarily be in the direction of the string. Because the angle of the string is different (with the block in place) the force, normal to the arm, will be different and the torque will be different.

inv said:
But is the Force Vector at the point perpendicular to the arm?
Here we are again suffering from the fact that the OP was not precise enough. I am assuming we are trying to turn the arm so the relevant force is the force normal to the arm. Any other component is acting through the pivot and will have no effect. And... if we are interested in the torque, the relevant quantities are the Force (tension) and the Perpendicular distance between the pivot (hidden) and the string. If you are confused by the way the thread is going then don't blame me. Blame the OP. Perhaps we could benefit from a properly modified diagram?

"So the force vector on the block <i>s perpendicular to the red dot?" No.
For a start, you can't be perpendicular to a dot. But if you mean, perpendicular to the arm at the dot, still no!
The string (provided it is thin and flexible) can only act along the line of the string. Putting a pulley in place which is attached to the arm makes no difference: you would need to have a pulley with an independant mounting (like the one on the left) to change the direction of the string before it touches the arm, any part of the arm, or anything fixed to the arm.

I have simplified your diagram. The string acts with force of magnitude T along the line and in the sense of the red arrow.
It could be considered to act at the red dot, or at the green dot, or anywhere along that line. At the moment when the arm and its attachments are in the position shown, that is how it acts. What you do with the string after it touches the corner (rough, smooth or pulley) makes no difference, so long as it ends up fixed to the arm. The net result of the force on the corner and any other forces it causes as you bend it round other parts of the arm and wherever you finally fix it to the arm will be exactly the same as if you just fixed it to the corner. There is no point in calculating them (if you could) because the simplest calculation is to imagine it attached at the red or green dot.
As to the effect of this force, as SophieCentaur says, we can only speculate: there is simply no information about the arm and other forces acting on it. Strictly, the whole system stays in its current state of rest or uniform motion in a straight line, unless acted on by an external force - none of which is shown. From the label "10kg weight" and the direction of the string between the pulley and the weight, I might infer the presence of a gravitational field acting down the page: then the whole system (as shown) would simply accelerate in the direction of that field.
At the moment there is no reason for T to be anything other than 0 and so it does not matter where it acts nor in what direction.

I don't want to join the speculators, because you asked if the corner could change the direction of action of the string tension and I've answered that - NO - and I've tried to give some explanation. But it might help if I suggest a hypothetical version of your arm and string, then say what would happen there.
Say the string was exerting tension T as shown, but forget the left pulley and weight, just say something is pulling the string in that direction. And if the arm is just a lump of wood or metal (as drawn) not attached to anything else. What will happen?
The only force acting is the tension in the string and it acts in the direction of the string. So the whole arm accelerates in that direction according to F=ma. It does not matter where the CG is - in line with the string, 10cm to the side, 50cm to the side, whatever. The only force is in that direction, so that's the direction you go.
But every system of forces reduces to a single force and a torque. If the CG happened to be in line with the string (impossible as drawn) the torque would be zero. Anywhere else just measure from the CG to the line of the string, perpendicular to the string and multiply by the tension in the string to get the torque. Then you can calculate the angular acceleration according to torque = moment of inertia x angular acceleration.
The point I'm trying to make (probably badly) is that what matters is the strength and direction of the tension in the string. That is what acts on your lump of arm. It does not matter how the string is attached, or what corners or pulleys on the arm it goes round. They are all part of the arm and forces between them are internal. The only force that has a net effect is the tension in the string away from the arm (including all attachments.)

sophiecentaur
I get the feeling that the OP is still wedded his original ideas and has not taken anything from what's been said on the thread.

BvU
@inv , could you respond instead of starting the identcal thread again ?

## 1. What is a pulley force vector?

A pulley force vector is a type of force that acts on an object due to the tension in a rope or cable that is wrapped around a pulley. It is typically used to change the direction of a force or to lift an object.

## 2. How does a pulley affect the force needed to lift a 10kg weight?

A pulley can reduce the amount of force needed to lift a 10kg weight by spreading the load over multiple ropes or cables. The mechanical advantage of the pulley system can also be increased by adding more pulleys, further reducing the force needed to lift the weight.

## 3. What is friction and how does it affect the pulley force vector?

Friction is a force that acts in the opposite direction of motion between two surfaces in contact. In the case of a pulley system, friction can occur between the rope or cable and the pulley. This friction can affect the amount of force needed to lift the 10kg weight, as some of the force will be lost to overcoming the friction.

## 4. How can we check the amount of friction on a 10kg weight in a pulley system?

To check the amount of friction on a 10kg weight in a pulley system, we can use a spring scale or force meter to measure the force needed to lift the weight. By comparing this force to the expected force based on the weight and mechanical advantage of the pulley system, we can determine the amount of friction present.

## 5. What factors can affect the amount of friction on a 10kg weight in a pulley system?

The amount of friction on a 10kg weight in a pulley system can be affected by several factors, including the type and condition of the rope or cable, the material and condition of the pulley, and the angle of the rope or cable as it wraps around the pulley. Additionally, external factors such as temperature and humidity can also affect the amount of friction present.

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