why should i assume the vertical component of H to act in the opposite direction of the normal reaction force
#force components
glossy fog
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why should i assume the vertical component of H to act in the opposite direction...
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twilit trail
why should i assume the vertical component of H to act in the opposite direction...
What do you mean by vertical component
It says H is purely horizontal
glossy fog
It says H is purely horizontal
ye but it has a vertical component perpendicular to the slop and its horizontal component parralel to the slope
glossy fog
What do you mean by vertical component
elfin nova
OK so take components of h and the one perpendicular to the incline will be balanced by normal and the other component will be used to keep the particle in equilibrium
What's the issue?
glossy fog
What's the issue?
why is it
R = Wcos(x) + Hsin(x)
why is Hsin(x) in the opposite direction of R
elfin nova
why is it
R = Wcos(x) + Hsin(x)
why is Hsin(x) in the opposite direction of R
It's the direction of force imagine you are applying the horizontal force like that won't you feel the Normal force from the incline?
I can just telll you it's like that mathematically but mechanics is about visualizing
glossy fog
It's the direction of force imagine you are applying the horizontal force like t...
thx for the insight, i appreciate it
elfin nova
Another way to think would be to shift your axis
Assume the incline is the x axis in a Cartesian plane
crisp fjord
thx for the insight, i appreciate it
Newton's first law. An object accelerates if and only if it experiences net force. Is P accelerating?
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glossy fog
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glossy fog
I have like a full explanation planned.
still down to here it
crisp fjord
Okay, so. Like I said, Newton's first law states that an object accelerates if and only if it experiences net force. Is P accelerating?
...this really only works if you participate.
@glossy fog
crisp fjord
Right, therefore it must not be experiencing net force. However, it is experiencing force.
Thus the forces it is experiencing must cancel out.
Now, suppose H wasn't there. What would be the behavior of P?
glossy fog
slide down
crisp fjord
What do you mean "slide down"? You're suggesting it would move, which would require it to accelerate, but what direction would it move/accelerate in?
glossy fog
accelerate towards earth
crisp fjord
...straight down?
glossy fog
no , at an angle
crisp fjord
What angle?
glossy fog
alpha ?
crisp fjord
The angle the inclined plane is at.
The motion of P would be parallel to the surface of the plane.
But of course we know that the force of weight goes straight down.
And yet P is accelerating at an angle.
Which means that P must be experiencing a second force, besides its weight.
That would be the normal force exerted by the plane.
glossy fog
The angle the inclined plane is at.
that's what i meant by alpha (don't know how to the sign)
crisp fjord
that's what i meant by alpha (don't know how to the sign)
I was confirming and adding the important context, not correcting.
We could also deduce the existence of the normal force from Newton's third law.
crisp fjord
Every force has an equal and opposite counterforce.
glossy fog
lemme give it a try
the normal reaction force is the equal and oppsoite counterforce because the surface is experiencing the force of the objects weight
crisp fjord
Well, a component of P's weight.
A slightly more precise statement of Newton's third law is that if A exerts a force F on B, then B exerts a force -F on A.
In any case, like we said, if P is stationary, which it is, that means it's not accelerating, which means it's experiencing 0 net force, which means that all of the forces it definitely is experiencing must cancel out.
twilit trail
We could also deduce the existence of the normal force from Newton's third law.
the normal force isn't actually due to newton's third law
i think it's a common misconception
crisp fjord
the normal force isn't actually due to newton's third law
...isn't it?
twilit trail
it's due to the electromagnetic interaction of particles
the forces described in newton's third law act on two different bodies
and are the same type of force
e.g the normal reaction of the ball and the normal reaction on the slope is a force pair of newton's third law
crisp fjord
e.g the normal reaction of the ball and the normal reaction on the slope is a fo...
...yes. That's what I said. The ball exerts some force on the slope, so the slope exerts an equivalent force in the opposite direction on the ball.
twilit trail
...yes. That's what I said. The ball exerts some force on the slope, so the slop...
are you talking about the weight/force that the ball exerts?
the weight of the ball is due to the weight it exerts on earth
so the weight and normal contact force can't be related by newton's third law
the normal contact force just occurs because the atoms from both objects are repelling rather than an external force
crisp fjord
the normal contact force just occurs because the atoms from both objects are rep...
It feels like you're not really bothering to engage with the material at the level I'm presenting it. I'm not concerned with what the fundamental nature of any given force is, and neither was Newton, and neither is the entire field of classical mechanics. We deduce the existence of a force the particle is exerting perpendicular to the surface of the slope as a component of its weight, then we deduce the existence of a reactive force from the Third Law. At no point did I say the third law causes the normal force, merely that we could use the law to deduce the existence of the normal force.
If you want to get technical, no force is "due to" any of Newton's laws. Newton's laws are descriptions of how forces actually behave, they don't cause any behavior.