VCE Physics Question Thread!

[lzxnl]

hey...another question please 
a person stands next to a table. explain how they can apply forces to the table to:
1.increse the normal force of the floor acting on them
2. decrease the normal force of the floor acting on them

thankyou in advance

Hey mq123,

An increase in normal force can simply be achieved, if that person pushes down on the table.
Theory?
Well, as the person is pushing down the table, there would be an increase in static friction. The increase in static friction would also cause the normal force to increase, as . Lets say the amount of increase in static friction equals to a, then , which proves that an increase in Static Friction would also cause an increase in normal force.

A decrease in the normal force can be achieved, if the person puts the table on an incline.
Theory?
Well, as if you put any object on an incline, the normal force would decrease. This can be proved, if you use the formula , and as you may know that any object on an incline cant equal to 0 dgrees or 180 degrees, the value of cos would also decrease, as cos 0/360 and cos 180 = 1, and any lower/higher that 180 and higher than 0 degrees, would end with a lower value of 1.

For example:
Lets say the person puts the table on an incline of 45 degrees and the table's mass is 2 kg







however, if the table was not on an incline, you would use the formula:







Hopefully this helps 

Objections.
1. Question asks for normal force of floor on person. You have misinterpreted the question.

2. To increase the ground's normal force, which is the force holding you up, you could make yourself 'heavier' by lifting the table/pushing up on the table, as pushing up means you're pushed down into the ground.

2. Decrease normal force could be done by partially leaning on the table; table now helps support your body

[Syndicate]

Objections.
1. Question asks for normal force of floor on person. You have misinterpreted the question.

2. To increase the ground's normal force, which is the force holding you up, you could make yourself 'heavier' by lifting the table/pushing up on the table, as pushing up means you're pushed down into the ground.

2. Decrease normal force could be done by partially leaning on the table; table now helps support your body

doesn't it ask for normal force increase/decrease of floor on the table? as it said, that how can they apply forces to the table

*I was a little confused with this one, sorry if I am wrong  *

[natdogg]

a person stands next to a table. explain how they can apply forces to the table to:
1.increse the normal force of the floor acting on them

I believe 'them' suggests that they are talking about the 'person'.
ie. increase the normal force of the floor acting on the person

[Syndicate]

I believe 'them' suggests that they are talking about the 'person'.
ie. increase the normal force of the floor acting on the person

(them) but what if suggested both? because how can the use of person, be they, later? xD

Books need to clarify their questions more fluently O.o

[Maz]

Hey mq123,

An increase in normal force can simply be achieved, if that person pushes down on the table.
Theory?
Well, as the person is pushing down the table, there would be an increase in static friction. The increase in static friction would also cause the normal force to increase, as . Lets say the amount of increase in static friction equals to a, then , which proves that an increase in Static Friction would also cause an increase in normal force.

A decrease in the normal force can be achieved, if the person puts the table on an incline.
Theory?
Well, as if you put any object on an incline, the normal force would decrease. This can be proved, if you use the formula , and as you may know that any object on an incline cant equal to 0 degrees or 180 degrees. The value of cos would also decrease, as cos 0/360 and cos 180 = 1, and any lower/higher that 180 and higher than 0 degrees, would end with a lower value of 1.

For example:
Lets say the person puts the table on an incline of 45 degrees and the table's mass is 2 kg







however, if the table was not on an incline, you would use the formula:







Hopefully this helps 

Objections.
1. Question asks for normal force of floor on person. You have misinterpreted the question.

2. To increase the ground's normal force, which is the force holding you up, you could make yourself 'heavier' by lifting the table/pushing up on the table, as pushing up means you're pushed down into the ground.

2. Decrease normal force could be done by partially leaning on the table; table now helps support your body

doesn't it ask for normal force increase/decrease of floor on the table? as it said, that how can they apply forces to the table

*I was a little confused with this one, sorry if I am wrong  *

I believe 'them' suggests that they are talking about the 'person'.
ie. increase the normal force of the floor acting on the person

(them) but what if suggested both? because how can the use of person, be they, later? xD

Books need to clarify their questions more fluently O.o

thankyou all of you...yeah books do need to write their questions better 

[knightrider]

How would you do this question attached?

[lzxnl]

Note that the vertical force components must cancel and sum to zero.
This tells you something about the normal reaction force. What's the component of the normal reaction that's vertical?

The net force is thus the horizontal component of the normal reaction force. This must therefore be the centripetal force requirement for circular motion.

[anastasiaaa]

anyone have the pdf file for the 3/4 physics cambridge textbook for this year?

[knightrider]

Note that the vertical force components must cancel and sum to zero.
This tells you something about the normal reaction force. What's the component of the normal reaction that's vertical?

The net force is thus the horizontal component of the normal reaction force. This must therefore be the centripetal force requirement for circular motion.

Thanks lzxnl 

"This tells you something about the normal reaction force. What's the component of the normal reaction that's vertical?"

Would this be the force they have labelled R

" vertical force components must cancel and sum to zero."

Why is this?

Also if you were to draw a triangle to find the centripetal force how would it look like and where would theta be placed?

[Syndicate]

Thanks lzxnl 

"This tells you something about the normal reaction force. What's the component of the normal reaction that's vertical?"

Would this be the force they have labelled R

" vertical force components must cancel and sum to zero."

Why is this?

Also if you were to draw a triangle to find the centripetal force how would it look like and where would theta be placed?

Hey knightrider,

Normal force is the force, which is points [vertically] upwards, thus it is R (generally, but not in all cases)

As Iznxl said that the vertical components must cancel [Subract] from one another, so you can get the net force.

Do remeber that the Net force = centripetal force = =

As this is a banked curve, the centripetal force would face towards the centre of the radius (so in the same direction as Friction), which is facing [horizontally] right.

Therefore, the correct answer would be Rcos(theta), as cos corresponds to the horizontal value of a graph.

EDIT: what do you mean by your last question? You already have a triangle made up in the question, then why make another one?

Hopefully this helps 

[knightrider]

Hey knightrider,

Normal force is the force, which is points [vertically] upwards, thus it is R (generally, but not in all cases)

As Iznxl said that the vertical components must cancel [Subract] from one another, so you can get the net force.

Do remeber that the Net force = centripetal force = =

As this is a banked curve, the centripetal force would face towards the centre of the radius (so in the same direction as Friction), which is facing [horizontally] right.

Therefore, the correct answer would be Rcos(theta), as cos corresponds to the horizontal value of a graph.

EDIT: what do you mean by your last question? You already have a triangle made up in the question, then why make another one?

Hopefully this helps 

Thanks Syndicate 

But the answers say Rsin(theta)  how did they get this ?

[lzxnl]

With a bit of geometry, you can show that the angle theta given is equal to the angle between the normal and the vertical.
Hence, the horizontal component of the normal reaction force is equal to R sin theta

The reason why the vertical force components must cancel is simple. Is your object moving up and down? No. So the net force must not have a vertical component.

[Syndicate]

Hey,

Sorry about the wrong answer  (I clearly should have worked out it, before giving out an answer, carelessly)

Luckily, I worked it out on a paper, look at the attachment below!

http://imgur.com/FkC2T4s

[knightrider]

With a bit of geometry, you can show that the angle theta given is equal to the angle between the normal and the vertical.
Hence, the horizontal component of the normal reaction force is equal to R sin theta

The reason why the vertical force components must cancel is simple. Is your object moving up and down? No. So the net force must not have a vertical component.

Thanks so much lzxnl 

"With a bit of geometry, you can show that the angle theta given is equal to the angle between the normal and the vertical."

What geometry do you use for this ?

Hey,

Sorry about the wrong answer  (I clearly should have worked out it, before giving out an answer, carelessly)

Luckily, I worked it out on a paper, look at the attachment below!

http://imgur.com/FkC2T4s

no worries thanks so much Syndicate 

[lzxnl]

Thanks so much lzxnl 

"With a bit of geometry, you can show that the angle theta given is equal to the angle between the normal and the vertical."

What geometry do you use for this ?


no worries thanks so much Syndicate 

You literally just write out all of the angles and go angle-hunting.
Think about it this way. The normal is 90 degrees to the incline. Vertical is 90 degrees to horizontal. It makes sense that the angle between the normal and the vertical is the same as the angle between the horizontal and the incline.