- #36
David Lewis
- 846
- 258
You can't multiply or divide by gravity. Please append measurement units to your numbers (9.8 what)?
kelseybrahe said:Will it be the sumF multiplied by gravity (9.8)?
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]haruspex said:No. I'll divide it by gravity?
David Lewis said:You can't multiply or divide by gravity. Please append measurement units to your numbers (9.8 what)?
kelseybrahe said: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?
Yes.kelseybrahe said:Is it w=mg?
No, that is not a sum of forces. What two forces have we narrowed it down to?kelseybrahe said:Will it just be (sum) F = ma - g?
haruspex said:No, that is not a sum of forces. What two forces have we narrowed it down to?
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 said:Friction and gravity/weight
haruspex said: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.
Nearly right. One arithmetic sign is wrong.kelseybrahe said:Friction + weight = total mass + observed acceleration
Be careful here, those two m's do not stand for the same mass.kelseybrahe said:So friction = ma - mg?
haruspex said:Nearly right. One arithmetic sign is wrong.
Be careful here, those two m's do not stand for the same mass.
Compare that equation to ΣF = m * a and re-read post #46. One of the operators in your equation is incorrect.kelseybrahe said:Friction - weight = total mass + acceleration
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.kelseybrahe said:Friction = 1.154kg + 1.05m/s^2 + 2.44N ?
haruspex said: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.
Yes. In the context of this question, what exactly is the expression for weight there, and what exactly the expression for mass?kelseybrahe said:Oh! I messed up the equation.
Okay.
Friction + weight = mass x acceleration?
haruspex said:Yes. In the context of this question, what exactly is the expression for weight there, and what exactly the expression for mass?
kelseybrahe said:What do you mean by expression?
Weight = 2.44N
Mass = 1.154kg ?
Yes.kelseybrahe said:What do you mean by expression?
Weight = 2.44N
Mass = 1.154kg ?
haruspex said:Yes.
Yes. Note that this gives a negative result because you effectively chose the positive direction as being in the same direction as the acceleration.kelseybrahe said:Friction = 1.154 x 1.05 - 2.44
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 said:And to find only the friction of the cart, I would replace the total mass by only the mass of the cart
haruspex said: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.
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.kelseybrahe said:Is that a problem? Do I just ignore the negative sign?
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.kelseybrahe said: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 said: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 mtable a = T- friction; in the vertical direction for the suspended mass we have msus a = msus 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.
Friction doesn't cancel friction.kelseybrahe said:but these are still insignificant and basically cancel out).
haruspex said:Friction doesn't cancel friction.
With essentially the same set-up, what could you easily vary that would give a different acceleration?
If the friction consists of friction on the cart and friction at the pulley, which of those changes would alter the contribution of one friction source but not the other?
No, why do you say so? If you were to increase the mass of the cart that would also reduce the acceleration of the falling mass.kelseybrahe said:Well the pulley is controlling the speed that the accelerating mass falls
haruspex said:No, why do you say so? If you were to increase the mass of the cart that would also reduce the acceleration of the falling mass.
With the same arrangement, a cart, a pulley etc., what could you easily vary that would affect the acceleration?
Right. Now, suppose the friction just comes from the cart and the pulley axle. How would those two frictions vary if you change the suspended weight, say? Would they change in exactly the same way, i.e. stay in the same ratio?kelseybrahe said:Affect the acceleration of the cart?
Well if you vary the mass of the weight pulling the cart, it will accelerate faster. If you reduce it, the cart will travel slower.
The weight of the cart itself. Increase the weight and it will travel slower, reduce the weight and it will travel faster.
haruspex said:Right. Now, suppose the friction just comes from the cart and the pulley axle. How would those two frictions vary if you change the suspended weight, say? Would they change in exactly the same way, i.e. stay in the same ratio?
There will, but why? Does that reason apply also to the friction on the cart, which has the same mass as before?kelseybrahe said:there will be greater friction on the pulley too, won't there?
I did not say the rationbetween the friction and the acceleration. I said the ratio between the two frictions.kelseybrahe said:And the same if it falls slower. So the ratio would stay the same.
They don't. It is not the increased speed that would cause the increase in friction. What does kinetic friction between two surfaces depend on?kelseybrahe said:it seems strange to say that things have an increase in encountered friction if they move faster.
haruspex said:There will, but why? Does that reason apply also to the friction on the cart, which has the same mass as before?
I did not say the rationbetween the friction and the acceleration. I said the ratio between the two frictions.
They don't. It is not the increased speed that would cause the increase in friction. What does kinetic friction between two surfaces depend on?
David Lewis said: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
Yes, but to be accurate it's the normal force that matters. Because the cart's mass stays the same, the weight stays the same, so the normal force stays the same.kelseybrahe said:Oh, right. So, the friction acting on the cart will stay the same because it's mass doesn't change, but the friction acting on the pulley will vary as the mass of the weight varies.
Yes. The tension in the string leads to a normal force between the pulley's axle and its support, and that affects the frictional torque.kelseybrahe said:Is this because of the tension from the string