Help with Physics!!!

[yawho]

Maybe she's got the wrong idea.

I can assure you, there's a misunderstanding somewhere along the lines. I guarantee you that the Normal Reaction Force is less than the Weight force.

Some one must be wrong. I tend to side with my teacher. She answered the question but you didn't.

[rife168]

Maybe she's got the wrong idea.

I can assure you, there's a misunderstanding somewhere along the lines. I guarantee you that the Normal Reaction Force is less than the Weight force.

Some one must be wrong. I tend to side with my teacher. She answered the question but you didn't.

I agree with Paul, the magnitude of the normal force on an inclined plane varies with the angle of the slope. If it was just 10N then the object wouldn't ever slide down the slope.

I think that your teacher may be referring to the reaction force due to gravity. The Earth pulls on the object with a force of 10N, and in return the object pulls back on the earth with a force of 10 N in the opposite direction.

Either that or the plane and the object have a very high coefficient of friction, so the object does not slide down the plane. If it is at rest, then the net force must be zero, which means that there must be an opposing force of 10N, though this wouldn't be the normal reaction force, it would be the normal reaction + resistance force.

I think your teacher may have misunderstood your question or you have misunderstood her answer, or maybe you missed a bit of information in the original question.

[paulsterio]

Some one must be wrong. I tend to side with my teacher. She answered the question but you didn't.

No, I'm not saying that someone is wrong, I'm saying that there's a misunderstanding, are you sure you wrote the question correctly and are you sure she saw the question correctly...etc. there's a lot of things which can go wrong in between, I'm not saying anyone's right or wrong, I'm saying we're obviously not all on the same page.

Yes, I agree with fletch on this, however, I'd also like to add this to clarify a little.

You mentioned originally that the object is stationary on the plane. This means that the net force is equal to 0. Hence, if you resolve the Weight force into two components, one parallel to the slope and one perpendicular, the friction force will cancel out the one parallel to the slope, hence it won't move down the incline. However, the one perpendicular to the slope is the Normal Reaction. This is what the Normal force is defined as, the force perpendicular to the slope. It will vary depending on the size of the incline. The larger the incline is, the greater the force down the slope, the less the normal reaction. Since they have to add to give the weight force, they will always be inversely proportional. A decrease in the force down the slope will cause an increase in the normal reaction and vice versa.

The maximum Normal Reaction force possible is 10N, which occurs when there is no incline, it is flat. The minimum Normal Reaction is 0N, which occurs when the object is in freefall, i.e. when the slope is 90 degrees and hence the object just falls vertically downwards.

There is no reason to assume that the Normal Reaction force is 10N. That's all I'm saying.

[yawho]

Maybe she's got the wrong idea.

I can assure you, there's a misunderstanding somewhere along the lines. I guarantee you that the Normal Reaction Force is less than the Weight force.

Some one must be wrong. I tend to side with my teacher. She answered the question but you didn't.

I agree with Paul, the magnitude of the normal force on an inclined plane varies with the angle of the slope. If it was just 10N then the object wouldn't ever slide down the slope.

I think that your teacher may be referring to the reaction force due to gravity. The Earth pulls on the object with a force of 10N, and in return the object pulls back on the earth with a force of 10 N in the opposite direction.

Either that or the plane and the object have a very high coefficient of friction, so the object does not slide down the plane. If it is at rest, then the net force must be zero, which means that there must be an opposing force of 10N, though this wouldn't be the normal reaction force, it would be the normal reaction + resistance force.

I think your teacher may have misunderstood your question or you have misunderstood her answer, or maybe you missed a bit of information in the original question.

My teacher's explanation: The weight of the object is 10 N downward, so it exerts a force of 10 N on the plane downward, newton's third law, the plane exerts a force of 10 N on the object upward. The question asked for reaction force, not the normal component of the reaction force.

[Bhootnike]

The weight of the object is 10 N downward, so it exerts aforce of 10 N on the plane downward, newton's third law, the plane exerts a force of 10 N on the object upward. The question asked for reaction force, not the normal component of the reaction force.

the  bit where you say " so it exerts a force of 10N on the plane downward" , sounds odd
because if you picture the incline,

                                   /|
                               * /  | 
                                /    |
                               /___|
^ sort of haha, but yeah, i imagined that the force that the object exerts on the INCLINED plane be 10cosTHETA.
i dont see what;d you gain from finding out the reaction force that your teacher has mentioned, i associated reaction with the normal reaction mgsintheta.
weirdd question

[yawho]

The weight of the object is 10 N downward, so it exerts aforce of 10 N on the plane downward, newton's third law, the plane exerts a force of 10 N on the object upward. The question asked for reaction force, not the normal component of the reaction force.

the  bit where you say " so it exerts a force of 10N on the plane downward" , sounds odd
because if you picture the incline,

                                   /|
                               * /  | 
                                /    |
                               /___|
^ sort of haha, but yeah, i imagined that the force that the object exerts on the INCLINED plane be 10sinTHETA.
i dont see what;d you gain from finding out the reaction force that your teacher has mentioned, i associated reaction with the normal reaction mgsintheta.
weirdd question

My guess is the question tests your understanding of the action force of the object on the plane and the reaction force of the plane on the object in newton's 3 law.

[paulsterio]

My teacher's explanation: The weight of the object is 10 N downward, so it exerts a force of 10 N on the plane downward, newton's third law, the plane exerts a force of 10 N on the object upward. The question asked for reaction force, not the normal component of the reaction force.

Firstly, Bhootnike, N = mgcos(theta) not mgsin(theta)

Secondly, I can see where you're coming from yawho but I agree with Bhootnike, it's a weird and very ambiguous question in that why would it want you to find the force that is acting vertically upwards?

[Bhootnike]

My teacher's explanation: The weight of the object is 10 N downward, so it exerts a force of 10 N on the plane downward, newton's third law, the plane exerts a force of 10 N on the object upward. The question asked for reaction force, not the normal component of the reaction force.

Firstly, Bhootnike, N = mgcos(theta) not mgsin(theta)

Secondly, I can see where you're coming from yawho but I agree with Bhootnike, it's a weird and very ambiguous question in that why would it want you to find the force that is acting vertically upwards?

woops! haha
my bad, but yeahh, you guys know what i meant :p

[yawho]

My teacher's explanation: The weight of the object is 10 N downward, so it exerts a force of 10 N on the plane downward, newton's third law, the plane exerts a force of 10 N on the object upward. The question asked for reaction force, not the normal component of the reaction force.

Firstly, Bhootnike, N = mgcos(theta) not mgsin(theta)

Secondly, I can see where you're coming from yawho but I agree with Bhootnike, it's a weird and very ambiguous question in that why would it want you to find the force that is acting vertically upwards?

Why did you say it's a weird and very ambiguous question?

[Bhootnike]

my view -
ambiguous since reaction force is usually associated with the normal reaction force.
thats why we were all asking if there was an angle

weird because its not something you'd usually wanna work out. like in practical sense, what would you gain. by having an angle present and thus figuring out normal, you can figure out the component of the weight force of the object acting on the plan. and those types r the questions ive seen ! but yeah as you said, if its testing your understanding , i guess its legit!

 p.d.s.n.o.a.f.y.v.

[paulsterio]

Can we stop arguing about this question?

I agree with Bhootnike, it's a little random, like you're not even figuring out anything. The fact that it's called the "Reaction Force" is also ambiguous, it should have been the force upwards or something like that which is more correct.

[yawho]

my view -
ambiguous since reaction force is usually associated with the normal reaction force.
thats why we were all asking if there was an angle

weird because its not something you'd usually wanna work out. like in practical sense, what would you gain. by having an angle present and thus figuring out normal, you can figure out the component of the weight force of the object acting on the plan. and those types r the questions ive seen ! but yeah as you said, if its testing your understanding , i guess its legit!

 p.d.s.n.o.a.f.y.v.

In general, reaction force is not usually associated with normal reaction force in N3L

The 10N upward reaction force on the object explains in a simple way why the object is at rest on the plane.

[queendaisy]

Actually no the 10N upward reaction force does not explain why the object is at rest because this upward force acts on the earth, not on the object... Action-reaction pairs always always always act on different bodies. Therefore you can't add the the pairs and say that because they add to zero, the object is at rest.. because they act on different bodies!
Friction explains why the object is at rest on an inclined plane.
But yes, this question is silly and not worded very well, and took way too long to answer haha

[yawho]

Can we stop arguing about this question?

I agree with Bhootnike, it's a little random, like you're not even figuring out anything. The fact that it's called the "Reaction Force" is also ambiguous, it should have been the force upwards or something like that which is more correct.

The force upwards? Where is this force upwards coming from?

[yawho]

Actually no the 10N upward reaction force does not explain why the object is at rest because this upward force acts on the earth, not on the object... Action-reaction pairs always always always act on different bodies. Therefore you can't add the the pairs and say that because they add to zero, the object is at rest.. because they act on different bodies!
Friction explains why the object is at rest on an inclined plane.
But yes, this question is silly and not worded very well, and took way too long to answer haha

It does explain why the object is at rest. The sum of this upward reaction force and the downward weight force is zero.