Finding friction without coefficient?

[haruspex]

String, pulley, ticker timer, Gravity?

The string is internal to the system. We are only concerned with external forces here.
Yes, the pulley exerts an external force, but it does no work, so does not contribute to the system's acceleration. It also may have some friction at its axle. Strictly speaking that is a frictional torque, but for the purpose of the question we can just consider that part of the frictional force.
The ticker timer may slow things a bit, but that is just part of the total frictional force, so we do not need to separate that out either out.
That leaves gravity, and you know that force. So what equation do you get?

 

[David Lewis]

2.44/1.154 = 2.15m/s^2

The way I would write my scratch pad calculation is 2.44 N/1.154 kg = 2.15 m/s²
just to keep it all straight in my mind. Less chance of error.
And makes it easier for others to follow along.

 

[kelseybrahe]

[QUOTE="haruspex, post: 5539034, member: 334404]
That leaves gravity, and you know that force. So what equation do you get?[/QUOTE]

Will it be the sumF multiplied by gravity (9.8)?

 

[kelseybrahe]

The way I would write my scratch pad calculation is 2.44 N/1.154 kg = 2.15 m/s²
just to keep it all straight in my mind. Less chance of error.
And makes it easier for others to follow what you're doing.

Thank you, that makes sense!

 

[kelseybrahe]

[QUOTE="haruspex, post: 5539034, member: 334404]
That leaves gravity, and you know that force. So what equation do you get?

Will it be the sumF multiplied by gravity (9.8)?[/QUOTE]

No. I'll divide it by gravity?

 

[David Lewis]

You can't multiply or divide by gravity. Please append measurement units to your numbers (9.8 what)?

 

[haruspex]

Will it be the sumF multiplied by gravity (9.8)?

No. I'll divide it by gravity?[/QUOTE]
Gravity exerts a force on each of the two masses, but only one of those contributes to the motion of the system (the other being balanced by a normal force). What is the algebraic expression for that gravitational force?

 

[kelseybrahe]

No. I'll divide it by gravity?

Gravity exerts a force on each of the two masses, but only one of those contributes to the motion of the system (the other being balanced by a normal force). What is the algebraic expression for that gravitational force?[/QUOTE]

I don't understand, I'm sorry.
Is it w=mg?

 

[kelseybrahe]

You can't multiply or divide by gravity. Please append measurement units to your numbers (9.8 what)?

Sorry, 9.8 m/s^2.

 

[kelseybrahe]

Gravity exerts a force on each of the two masses, but only one of those contributes to the motion of the system (the other being balanced by a normal force). What is the algebraic expression for that gravitational force?

I don't understand, I'm sorry.
Is it w=mg?[/QUOTE]

Will it just be (sum) F = ma - g?

 

[haruspex]

Is it w=mg?

Yes.

 

[haruspex]

Will it just be (sum) F = ma - g?

No, that is not a sum of forces. What two forces have we narrowed it down to?

 

[kelseybrahe]

No, that is not a sum of forces. What two forces have we narrowed it down to?

Friction and gravity/weight

 

[haruspex]

Friction and gravity/weight

Right. So use those to write an equation in the form ΣF=ma. You already know the m and a to use, as I mentioned in post #27.

 

[kelseybrahe]

Right. So use those to write an equation in the form ΣF=ma. You already know the m and a to use, as I mentioned in post #27.

Friction + weight = total mass + observed acceleration

So friction = ma - mg?

 

[haruspex]

Friction + weight = total mass + observed acceleration

Nearly right. One arithmetic sign is wrong.

So friction = ma - mg?

Be careful here, those two m's do not stand for the same mass.

 

[kelseybrahe]

Nearly right. One arithmetic sign is wrong.

Be careful here, those two m's do not stand for the same mass.

Friction - weight = total mass + acceleration

Right, so... Weight is (mg) or 2.44N as in #24

So Friction = 1.154kg + 1.05m/s^2 + 2.44N ?

 

[jbriggs444]

Friction - weight = total mass + acceleration

Compare that equation to ΣF = m * a and re-read post #46. One of the operators in your equation is incorrect.

 

[haruspex]

Friction = 1.154kg + 1.05m/s^2 + 2.44N ?

Fundamental rule: you must only add and subtract terms having the same dimensionality. Adding a mass to an acceleration or to a force makes no sense at all.
You can multiply and divide terms of different dimensionality, but you should keep track of the dimensions: mass x acceleration = force, etc.

 

[kelseybrahe]

Fundamental rule: you must only add and subtract terms having the same dimensionality. Adding a mass to an acceleration or to a force makes no sense at all.
You can multiply and divide terms of different dimensionality, but you should keep track of the dimensions: mass x acceleration = force, etc.

Oh! I messed up the equation.
Okay.

Friction + weight = mass x acceleration?

 

[haruspex]

Oh! I messed up the equation.
Okay.

Friction + weight = mass x acceleration?

Yes. In the context of this question, what exactly is the expression for weight there, and what exactly the expression for mass?

 

[kelseybrahe]

Yes. In the context of this question, what exactly is the expression for weight there, and what exactly the expression for mass?

What do you mean by expression?
Weight = 2.44N
Mass = 1.154kg ?

 

[kelseybrahe]

What do you mean by expression?
Weight = 2.44N
Mass = 1.154kg ?

Or weight is 'mg' and mass is 'm'?

 

[David Lewis]

To re-cap:
You have correctly calculated the hypothetical acceleration without friction.
You have experimentally measured the actual acceleration with friction.
You have found the difference between the two accelerations (1.10 m/s²)
You know how much gravitational force is tending to accelerate the system
Now you want to find the frictional force

 

[haruspex]

What do you mean by expression?
Weight = 2.44N
Mass = 1.154kg ?

Yes.

 

[kelseybrahe]

Yes.

So, therefore,

Friction + 2.44N = 1.154kg x 1.05m/s^2

Friction = 1.154 x 1.05 - 2.44

And to find only the friction of the cart, I would replace the total mass by only the mass of the cart?

 

[haruspex]

Friction = 1.154 x 1.05 - 2.44

Yes. Note that this gives a negative result because you effectively chose the positive direction as being in the same direction as the acceleration.

And to find only the friction of the cart, I would replace the total mass by only the mass of the cart

By what logic? It worries me that you persist in making wild guesses instead of getting to grips with the laws of mechanics and the equations that come from them.

 

[kelseybrahe]

Yes. Note that this gives a negative result because you effectively chose the positive direction as being in the same direction as the acceleration.

By what logic? It worries me that you persist in making wild guesses instead of getting to grips with the laws of mechanics and the equations that come from them.

Is that a problem? Do I just ignore the negative sign?

Oh. I assumed that I could use the same equation but narrow it down to find only the friction acting on the cart by swapping the numbers we used for the whole system to be only the numbers for the cart.

 

[haruspex]

Is that a problem? Do I just ignore the negative sign?

It is not a problem. In general, if you are clear in how you define the positive directions for forces and accelerations, and write your equations accordingly, then the sign in your answer can tell you something useful. In the present case, you wrote the sum of forces as friction+mg. in reality, it is clear that they will oppose each other, so it is to be expected that you will get a negative sign for the friction. It is not an error.

There are two aspects of the whole question I'm not happy with.

It is not good practice to add forces that are acting in different directions as though they are just numbers (that is, scalars rather than vectors). Forces have direction. In this problem, gravity acts vertically on the suspended mass and friction acts horizontally on the other mass. The right approach is to let the tension in the string be T. So in the horizontal direction for the mass on the table we have an equation m_table a = T- friction; in the vertical direction for the suspended mass we have m_sus a = m_sus g - T. Eliminating T between these two gives the equation you found. Yes, it is a longer route, but it helps you understand what is going on, and can be applied in more general circumstances.

Secondly, it asks you to find the total frictional force when some of the friction may be in the axle of the pulley. That would not be a force, but a torque. However, you can equate it to a force by dividing by the radius of the pulley. A frictional torque τ at the axle will have the same effect as a frictional force τ/r at the periphery of the pulley.

I assumed that I could use the same equation but narrow it down to find only the friction acting on the cart by swapping the numbers we used for the whole system to be only the numbers for the cart.

Friction slows the acceleration no matter where it occurs. Knowing what the total effective frictional force is gives you no clue as to how it is distributed in the system. It could be entirely axial torque in the pulley.
To answer this part of the question you need to think up a variant of the experiment, or some extra observation you could make.

 

[kelseybrahe]

It is not a problem. In general, if you are clear in how you define the positive directions for forces and accelerations, and write your equations accordingly, then the sign in your answer can tell you something useful. In the present case, you wrote the sum of forces as friction+mg. in reality, it is clear that they will oppose each other, so it is to be expected that you will get a negative sign for the friction. It is not an error.

There are two aspects of the whole question I'm not happy with.

It is not good practice to add forces that are acting in different directions as though they are just numbers (that is, scalars rather than vectors). Forces have direction. In this problem, gravity acts vertically on the suspended mass and friction acts horizontally on the other mass. The right approach is to let the tension in the string be T. So in the horizontal direction for the mass on the table we have an equation m_table a = T- friction; in the vertical direction for the suspended mass we have m_sus a = m_sus g - T. Eliminating T between these two gives the equation you found. Yes, it is a longer route, but it helps you understand what is going on, and can be applied in more general circumstances.

Secondly, it asks you to find the total frictional force when some of the friction may be in the axle of the pulley. That would not be a force, but a torque. However, you can equate it to a force by dividing by the radius of the pulley. A frictional torque τ at the axle will have the same effect as a frictional force τ/r at the periphery of the pulley.

Friction slows the acceleration no matter where it occurs. Knowing what the total effective frictional force is gives you no clue as to how it is distributed in the system. It could be entirely axial torque in the pulley.
To answer this part of the question you need to think up a variant of the experiment, or some extra observation you could make.

Okay, thank you. This all makes sense.

Okay, so the friction could be spread across the system in any way and the way I tried to isolate that was thus incorrect and wouldn't work at all. Alright, that seems really obvious now. Thank you.

So contributing to the friction acting on the cart we've still got the vertical mass of gravity and the mass of the cart itself, plus the frictional torque's acting upon the string and the pulley etc. (but these are still insignificant and basically cancel out).
So, is there something else we have to consider? Or am I not going about this the right way?