VCE Physics Question Thread!

[Maz]

Tension is the force on the string pulling on the object. So, if it's a vertical circle, at the bottom of the circle the tension is pointing up, and at the top it's pointing down.
To do this question, you would treat the tension force like a normal reaction force and just solve net force = mv^2/r
Then assign the direction based on the direction requirement for circular motion

okay cool- thank you

[Syndicate]

hey- just a few questions related to tension? please...i don't really understand this...
what does it mean by 'vertical of tensions is equal to the gravitational force acting on the object'? its in my book but i don't understand it
2. is a word problem... A 0.20kg objects is whirled in a vertical circle on the end of the string of length 0.6m. the speed of the object is at a constant 2ms^'1. what is the tension of the string at the top and bottom of the table.
my book isn't really explaining tension much and I'm a little confused. id appreciate any help
Thankyou

By "vertical of tension", your book is simply try to state that it is gravity, as simply all the forces going downwards will equal to gravity(9.8 m/s/s). If you think about it, vertical is a straight line (from top to bottom), therefore, it must equal the gravitational force.

To do the following question, you must use the formula

where:
T = Tension (what you want to calculate)
W = Work (mass x gravity) [0.2 x 9.8 = 1.96 J]
m = mass [0.2 kg]
a = acceleration of the moving object [2m/s/s]
therefore ma = 0.2 x 2 = 0.4 kg.m/s/s



(at the top)

Now we can work out tension at the bottom of the vertical circle, by simply having a negative acceleration quanity, and knowing that, the speed is constant throughout the course, you will have an acceleration of 2m/s/s downwards => -2m/s/s



Therefore,  (at the bottom)

Hope this helps 

[Maz]

By "vertical of tension", your book is simply try to state that it is gravity, as simply all the forces going downwards will equal to gravity(9.8 m/s/s). If you think about it, vertical is a straight line (from top to bottom), therefore, it must equal the gravitational force.

To do the following question, you must use the formula

where:
T = Tension (what you want to calculate)
W = Work (mass x gravity) [0.2 x 9.8 = 1.96 J]
m = mass [0.2 kg]
a = acceleration of the moving object [2m/s/s]
therefore ma = 0.2 x 2 = 0.4 kg.m/s/s



(at the top)

Now we can work out tension at the bottom of the vertical circle, by simply having a negative acceleration quanity, and knowing that, the speed is constant throughout the course, you will have an acceleration of 2m/s/s downwards => -2m/s/s



Therefore,  (at the bottom)

Hope this helps 

thankyou

[Maz]

hello...quick question please can someone help?
why does a train track for a very fast train have to have curves of a big radius?  I'm supposed to use an equation to demonstrate...
is it something to do with having to overcome the centrifugal force and is the equation mv^2/r?
thankyou in advance 

[Syndicate]

hello...quick question please can someone help?
why does a train track for a very fast train have to have curves of a big radius?  I'm supposed to use an equation to demonstrate...
is it something to do with having to overcome the centrifugal force and is the equation mv^2/r?
thankyou in advance 

yes, you can have a look at centipetal force, which keeps an object inside the turn. Usually train tracks have banked curves, so the train doesnt move away from the center of the curvature. The formula seems right.

more imformation is available below:
http://www.diffen.com/difference/Centrifugal_Force_vs_Centripetal_Force

[Maz]

yes, you can have a look at centipetal force, which keeps an object inside the turn. Usually train tracks have banked curves, so the train doesnt move away from the center of the curvature. The formula seems right.

more imformation is available below:
http://www.diffen.com/difference/Centrifugal_Force_vs_Centripetal_Force

thankyou

[Adequace]

http://imgur.com/kMVtqqk

For the last question of the attached, the answer states "X is exerting the same force on Y as Y is exerting on X" then they equated ma=ma and ended up with (a of y)/(a of x) = 0.2

I don't understand how they exert the same force on eachother? Is Newton's third law applied here, if it is I don't understand how? What even is the force that is being exerted on each other?

Thanks, if you could clarify these questions that would be great.

[Syndicate]

http://imgur.com/kMVtqqk

For the last question of the attached, the answer states "X is exerting the same force on Y as Y is exerting on X" then they equated ma=ma and ended up with (a of y)/(a of x) = 0.2

I don't understand how they exert the same force on eachother? Is Newton's third law applied here, if it is I don't understand how? What even is the force that is being exerted on each other?

Thanks, if you could clarify these questions that would be great.

Hey Adequance,

The last question can be worked by simply solving for the accelaration of these masses. So the right formula, we must use in this situation is: and as we dont know the Fnet, we can work it out.

As there is no force, other than gravity, we can state that your Net Force is equal to 10 N (as requested above, by the book)

So, your acceleration of y would be which equals to 2 and your acceleration of x would be , which equals to 10

As the book has asked you to workout the ratio , you can now simply sub in the numbers above

which equals to 0.2

If you need anymore clarification, dont hesitate to ask 

if you sub in all the number, you will find out that they are exerting the same amount of force on each other, as Y = 5 x 2 [10] and X = 1 x 10 [10]

[Adequace]

Wouldn't the tension in the string balance out the weight force of both of the masses combined?

Thanks though

[Syndicate]

Wouldn't the tension in the string balance out the weight force of both of the masses combined?

Thanks though

Yes, it would, due to the masses being equally distributed on both sides. However, the question is asking for the acceleration of Y and X, while both masses are rotating with the frequency of 2 revolutions per second (2 complete rotations in 1 second), and because acceleration doesnt equal to 0 (in this case), the rod isnt balanced.

[Maz]

hello humans...
can someone please help me with this question?
you are having difficulty undoing a screw...should you use a screwdriver with a thicker handle or a longer handle?
I'm thinking the answer is thicker as torque= r*F*sin(theta)...and thinker would give a bigger r value? how do you discard the longer handle...
thankyou in advance

[Syndicate]

hello humans...
can someone please help me with this question?
you are having difficulty undoing a screw...should you use a screwdriver with a thicker handle or a longer handle?
I'm thinking the answer is thicker as torque= r*F*sin(theta)...and thinker would give a bigger r value? how do you discard the longer handle...
thankyou in advance

hey mq123,

Yes, you are correct, the handle of the screwdriver must be thicker to get a larger torque value. As for the longer handle screwdriver, it wouldnt directly increase/ decrease the torque and would stay exactly the same, as with the original screwdriver. It would only matter, if the situation was different. Example: Try to open a the lid from can.

What did you exactly mean by discard?

[Maz]

hey mq123,

Yes, you are correct, the handle of the screwdriver must be thicker to get a larger torque value. As for the longer handle screwdriver, it wouldnt directly increase/ decrease the torque and would stay exactly the same, as with the original screwdriver. It would only matter, if the situation was different. Example: Try to open a the lid from can.

What did you exactly mean by discard?

thankyou so much...yeah by discard i was basically just asking why the longer handle one wouldn't work- which you answered so thank you 

[Maz]

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

[Syndicate]

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