VCE Physics Question Thread!

[layalasmith]

What is the reaction force to weight force acting on object in free-fall? What would the the two reaction-action pairs be in relation to Newton's third law?

[G-Fr3sh]

What is the reaction force to weight force acting on object in free-fall? What would the the two reaction-action pairs be in relation to Newton's third law?

I don't think that there is reaction force for an object in free fall due to no object pushing the object in the reverse direction **aside air resistance which is ignored in vce physics**. I could be wrong, try to email your teacher

[sprout]

I don't think that there is reaction force for an object in free fall due to no object pushing the object in the reverse direction **aside air resistance which is ignored in vce physics**. I could be wrong, try to email your teacher

^ There's no normal force for objects that are free falling (if there was, the object wouldn't be accelerating at 9.8m/s^2!)

[layalasmith]

I just don't know what to called the reaction force of the weight force (F on ball by Earth), for instance. Apparently a variation of this question was asked by VCAA and students performed poorly... Like would it be valid to call the reaction as force on Earth by ball? I'll confirm it though.

[S200]

Unless you are referring to when the ball hits the earth, there is no reaction force

[layalasmith]

Unless you are referring to when the ball hits the earth, there is no reaction force

So if I understand this correctly, when the ball is in air, there is no reaction force?

[sprout]

So if I understand this correctly, when the ball is in air, there is no reaction force?

There's no normal force.
I think you might be getting confused by newton's law (forgot which one) that states for every force there's an equal and opposite one. That's the force on the earth by the ball. In questions where the object is against a table etc, the normal force is this equal and opposite force, so that might be confusing you.

[S200]

So if I understand this correctly, when the ball is in air, there is no reaction force?

Indeed. The only possible opposing force is air resistance, which IS ignored in VCE Physics.

[skrt skrt]

Since the ball is falling due to Earth's force onto the ball, the ball would exert a force onto the Earth(reaction force) but since the Earth has such large mass the acceleration of the Earth would be negligible(a=fnet/mass of earth)

Hope that answers your question

[layalasmith]

YES I get it! Thank you all for the insightful answers, often combining all of the ideas suggested just click.

[IonianDeo]

Why is option A correct instead of B?

[Bri MT]

Why is option A correct instead of B?

Force = current * magnetic field strength * length of the wire

As the armature rotates T/4 none of these quantities change until at T/4 when the wires are in the gap of the split ring communtator

When the armature rotates from T/4 to 3T/4, the current is running through the section of wire AB in the opposite direction
  (- Force = - current * magnetic field strength * length of wire)

Hope this helps

[IonianDeo]

Force = current * magnetic field strength * length of the wire

As the armature rotates T/4 none of these quantities change until at T/4 when the wires are in the gap of the split ring communtator

When the armature rotates from T/4 to 3T/4, the current is running through the section of wire AB in the opposite direction
  (- Force = - current * magnetic field strength * length of wire)

Hope this helps

Thanks, it helps a lot.
So if the y-axis value was recording torque on the coil instead, would Graph B be the most appropriate answer option or is it completely different?

[Bri MT]

Thanks, it helps a lot.
So if the y-axis value was recording torque on the coil instead, would Graph B be the most appropriate answer option or is it completely different?

 torque = F perpendicular r
We've just discussed how F is behaving (& we know it's always vertical in this case), so let's look at r

r isn't AB -  the armature is rotating around the dotted line  and r would be .5*BC  or equivalently .5*AD
the component of BC/AD which is perpendicular to F = the component of BC/AD which is horizontal
So lets define theta as the angle between the armature and the horizontal plane
(if this explanation didn't make sense let me know)

In this case,   torque = Force * .5*BC*cos(theta)
As you can see, this is best represented by the graph B

[IonianDeo]

torque = F perpendicular r
We've just discussed how F is behaving (& we know it's always vertical in this case), so let's look at r

r isn't AB -  the armature is rotating around the dotted line  and r would be .5*BC  or equivalently .5*AD
the component of BC/AD which is perpendicular to F = the component of BC/AD which is horizontal
So lets define theta as the angle between the armature and the horizontal plane
(if this explanation didn't make sense let me know)

In this case,   torque = Force * .5*BC*cos(theta)
As you can see, this is best represented by the graph B

I didn't really understand what you meant by this:
"the component of BC/AD which is perpendicular to F = the component of BC/AD which is horizontal"