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Slight confusion concerning Newton's first law

  • Thread starter Ripe
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  • #1
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Hi,

I've very recently begun physics self-teaching as I used to find it boring and thus avoided it as a subject but now find it increasingly interesting. Nevertheless, I'm having some problems regarding questions on Newton's First Law. One of the questions in the book I'm using says the following:

"A 455N crate is being pulled at constant velocity by a force (F) directed at 30 degrees to the horizontal. The frictional force on the crate is 1163N. What is the magnitude of the pulling force?"

Now, since it's being pulled at constant velocity, that should mean that the net force is 0. However, in order to get the same answer as in the mark scheme, one has to equate Fx (i.e. the x-axis component of F) to the same magnitude as the friction force and then use cos30=Fx/F to find the answer. The problem is that I don't see how the net force is 0 in this case. The fact that Fx is of the same magnitude, but opposite, to the friction force, wouldn't that mean that the resultant force is the Fy component, and thus not 0?

I would appreciate it if someone could shed some light on this matter!
 

Answers and Replies

  • #2
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There is another question which I'm having difficulties with: "A mass (M) is connected with a string to a smaller mass (m). "M" is resting on an inclined plane and the string goes over a pulley at the top of the plane so that "m" is hanging vertically. What must the angle of the plane be in order to have equilibrium?"

I'd appreciate it if I could get an explanation, and not simply the solution.
 
  • #3
HallsofIvy
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Hi,

I've very recently begun physics self-teaching as I used to find it boring and thus avoided it as a subject but now find it increasingly interesting. Nevertheless, I'm having some problems regarding questions on Newton's First Law. One of the questions in the book I'm using says the following:

"A 455N crate is being pulled at constant velocity by a force (F) directed at 30 degrees to the horizontal. The frictional force on the crate is 1163N. What is the magnitude of the pulling force?"

Now, since it's being pulled at constant velocity, that should mean that the net force is 0. However, in order to get the same answer as in the mark scheme, one has to equate Fx (i.e. the x-axis component of F) to the same magnitude as the friction force and then use cos30=Fx/F to find the answer. The problem is that I don't see how the net force is 0 in this case. The fact that Fx is of the same magnitude, but opposite, to the friction force, wouldn't that mean that the resultant force is the Fy component, and thus not 0?

I would appreciate it if someone could shed some light on this matter!
The Fy component of force is not directed horizontally, so is not relevant to this question.
As long as the Fy component of force is not greater than the weight of the crate, the floor will exert an upward "passive force" exactly offsetting the difference.
 
  • #4
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Thanks for the quick reply!

I'm aware that Fy is not the horizontal component, but the vertical one. Shouldn't the passive force be downward, not upward? In either case, if you calculate Fy, it is greater than the weight (roughly 600N vs 455N), and the net force should thus not be 0?
 
  • #5
HallsofIvy
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There is another question which I'm having difficulties with: "A mass (M) is connected with a string to a smaller mass (m). "M" is resting on an inclined plane and the string goes over a pulley at the top of the plane so that "m" is hanging vertically. What must the angle of the plane be in order to have equilibrium?"

I'd appreciate it if I could get an explanation, and not simply the solution.
The mass, m, has gravity acting on it so that, in order for it to remain stationary, not moving, the upward force on it must be mg. The whole point of the pully is that it changes the direction the force. In order to exert force mg on the string, the force on the mass, M, which is, of course, Mg, must have a component parallel to the inclined plane, equal to mg.
If the inclined plane makes angle [itex]\theta[/itex] with the horizontal, the force along the inclined plane due to the mass, M, is [itex]Mg sin(\theta)[/itex] so, to have everything motionless, we must have [itex]Mg sin(\theta)= mg[/itex] or [itex]M sin(\theta)= m[/itex].
 
  • #6
HallsofIvy
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Thanks for the quick reply!

I'm aware that Fy is not the horizontal component, but the vertical one. Shouldn't the passive force be downward, not upward? In either case, if you calculate Fy, it is greater than the weight (roughly 600N vs 455N), and the net force should thus not be 0?
No, the downward force is the weight of the object. By "passive force" I mean a force exerted by something that is exactly enough to support whatever is pressing on it. If we set a 1 N weight on the ground, it isn't going to move- that's because the ground is pressing up on it with force 1 N. If we set a 1000 N weight on the ground, the ground will push up on it with force 1000 N. That is why we call it a "passive" force- it doesn't cause any object to move.
 
  • #7
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The mass, m, has gravity acting on it so that, in order for it to remain stationary, not moving, the upward force on it must be mg. The whole point of the pully is that it changes the direction the force. In order to exert force mg on the string, the force on the mass, M, which is, of course, Mg, must have a component parallel to the inclined plane, equal to mg.
If the inclined plane makes angle [itex]\theta[/itex] with the horizontal, the force along the inclined plane due to the mass, M, is [itex]Mg sin(\theta)[/itex] so, to have everything motionless, we must have [itex]Mg sin(\theta)= mg[/itex] or [itex]M sin(\theta)= m[/itex].
Thanks, now I understand it better. I, however, don't quite follow why the force along the string is Mgsin(theta) and not just Mg? Thanks for the patience with me!

Edit: Actually, it's the same thing, isn't it? If you say the hypotenuse is Mg and the side opposite to (theta) is mg, you get sin(theta)=Mg/mg or sin(theta)=M/m.
 
  • #8
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Thanks, now I understand it better. I, however, don't quite follow why the force along the string is Mgsin(theta) and not just Mg? Thanks for the patience with me!

Edit: Actually, it's the same thing, isn't it? If you say the hypotenuse is Mg and the side opposite to (theta) is mg, you get sin(theta)=Mg/mg or sin(theta)=M/m.
There are three forces acting in the vertical direction on the crate: the upward component of the cable tension, the downward force of gravity on the crate, and the upward contact force of the ground on the crate. The sum of these three forces must be zero, since there is no acceleration in the vertical. You can use this equilibrium condition to calculate the upward force of the ground on the crate.
 
  • #9
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There are three forces acting in the vertical direction on the crate: the upward component of the cable tension, the downward force of gravity on the crate, and the upward contact force of the ground on the crate. The sum of these three forces must be zero, since there is no acceleration in the vertical. You can use this equilibrium condition to calculate the upward force of the ground on the crate.
Which problem are you talking about? The one with the crate or the one with the two masses?
 
  • #10
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The problem is that I don't see how the net force is 0 in this case. The fact that Fx is of the same magnitude, but opposite, to the friction force, wouldn't that mean that the resultant force is the Fy component, and thus not 0?
....................................
Fy and Fx are orthogonal.
Thus Fy and Fx are independent.
Forces in Fx will not change any value of Fy and vice-versa.
 
  • #11
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Yes but what I'm saying is that if the Fx component cancels out the magnitude of the friction force and the Fy component has a magnitude of roughly 600N, how can the net force be 0? Or is my thinking wrong?
 
  • #12
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Yes but what I'm saying is that if the Fx component cancels out the magnitude of the friction force and the Fy component has a magnitude of roughly 600N, how can the net force be 0? Or is my thinking wrong?
You right about Fx, 2 forces, net zero.
Fy consisted 3 forces, net zero.

The vertical component of pulling force is FSinθ°,up
The weight mg, down
The Normal force, up

Since the object not sinking or rising, thus the net force is zero.

For second question, for non-inclined plane, mg is orthogonal to the string.

By making the mg the resultant of 2 component forces, we can use the component that is parallel to the plane to pull the string.
 
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  • #13
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I know that, logically, since the object is not rising/sinking the net force is zero. What I need in order to be able to understand it completely is some mathematical evidence to support that. The way I see it, Fy is roughly 600N (calculated it yesterday but I don't have it written down atm), W is 455N and shouldn't the normal force thus also be 455N? That clearly gives a net force upwards. What am I missing?
 
  • #14
PhanthomJay
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Which problem are you talking about? The one with the crate or the one with the two masses?
In the future, please post separate questions in separate posts to avoid this confusion.:wink:

I know that, logically, since the object is not rising/sinking the net force is zero. What I need in order to be able to understand it completely is some mathematical evidence to support that. The way I see it, Fy is roughly 600N (calculated it yesterday but I don't have it written down atm), W is 455N and shouldn't the normal force thus also be 455N? That clearly gives a net force upwards. What am I missing?
I believe you have copied down the problem wrong. Otherwise, the crate loses contact with the table since the normal force cannot act down on the block, and the problem makes no sense. Perhaps the weight of the crate is 4550 N??
 
  • #15
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I believe you have copied down the problem wrong. Otherwise, the crate loses contact with the table since the normal force cannot act down on the block, and the problem makes no sense. Perhaps the weight of the crate is 4550 N??
The problem doesn't mention the normal force. The book I've got mentioned that there's always a normal force on two contacting bodies, so I assumed there'd be a normal force on the crate as well.

Could you elaborate on your post?
 
  • #16
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I know that, logically, since the object is not rising/sinking the net force is zero. What I need in order to be able to understand it completely is some mathematical evidence to support that. The way I see it, Fy is roughly 600N (calculated it yesterday but I don't have it written down atm), W is 455N and shouldn't the normal force thus also be 455N? That clearly gives a net force upwards. What am I missing?
Normal force is pressing force.
The weight is 455N, downward, but there is also upward force, making it less pressed.
If the upward force is equal to weight, then Normal force equal to zero.
Just like a putting an object on a scale and pulling up, the reading will depend on how strong you pull the object up.
 
  • #17
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If the upward force is equal to weight, then Normal force equal to zero.
You mean net force equals zero, right? If so, I'm already aware of this.
 
  • #18
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wouldn't that mean that the resultant force is the Fy component
.......................
I think what you're thinking that Fx is equal to Friction, so cancelled out.
Thus leave Fy which not zero, true.


You have to see other forces acting in y-direction.
Then you can find those forces that make net force in y-direction equal to zero.
 
  • #19
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wouldn't that mean that the resultant force is the Fy component
.......................
I think what you're thinking that Fx is equal to Friction, so cancelled out.
Thus leave Fy which not zero, true.


You have to see other forces acting in y-direction.
Then you can find those forces that make net force in y-direction equal to zero.
That's more or less the whole point of my thread. What are the forces acting in the y-direction?
 
  • #20
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The obvious are Fy and mg.
There should be another force that makes the net force equal to zero, since no vertical motion.
 
  • #21
PhanthomJay
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The problem doesn't mention the normal force. The book I've got mentioned that there's always a normal force on two contacting bodies, so I assumed there'd be a normal force on the crate as well.

Could you elaborate on your post?
If you run the numbers in the vertical y direction, you will or have found that the vertical component of the pulling force is about 644 N, and the weight of the crate is supposedly given as 455 N. That means that if the block remains in contact with the table, the normal force must be 189 N acting down on the crate, which you seem to already have discovered. Now unless the table is magnetized or coated with a wet not quite set epoxy glue , the normal force cannot act down on the crate, and therefore the crate loses contact with the table and the result is that there is net force acting up and the block accelerates up but the there can't be friction. The bottom line is that the problem is incorrectly worded with bad numbers. So, toss it out, but as an exercise, assume the crate weighs 4550 N , then answer the question with that substitution.
 
  • #22
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If you run the numbers in the vertical y direction, you will or have found that the vertical component of the pulling force is about 644 N, and the weight of the crate is supposedly given as 455 N. That means that if the block remains in contact with the table, the normal force must be 189 N acting down on the crate, which you seem to already have discovered. Now unless the table is magnetized or coated with a wet not quite set epoxy glue , the normal force cannot act down on the crate, and therefore the crate loses contact with the table and the result is that there is net force acting up and the block accelerates up but the there can't be friction. The bottom line is that the problem is incorrectly worded with bad numbers. So, toss it out, but as an exercise, assume the crate weighs 4550 N , then answer the question with that substitution.
Thanks, that was the kind of response I was waiting for! Indeed if W>Fy then everything makes sense as the normal foce would cancel out the difference. Thanks for the help.
 

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