My physics IB (international baccalaureat)

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In summary, the body has a normal force (N) and a weight (W) due to the gravitational force of the Earth. If the body is on a horizontal plane, the weight is the only force acting on it, and the object will move with a constant speed. If the body is on an inclined plane, the weight and the normal force are both acting on it, and the object will move at different speeds according to the angle of inclination. There is a friction force between the surface and the object, which opposes the motion.
  • #1
alex77
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So i decided to choose my ib on physics but there is a small problem,i didn't understand why an object has a normal force and a weight.If that object is on a horisontal surface you can move that object?Is there a friction force on a horisontal plane?Where and what does it do?I know there was something to do with velocity in a horisontal plane.Is that when you push the object?If it is tied to another object you will have tension?
Why is the tension equal to the weight of the body and why in some cases the tension T=W+F?
Could someone explain me?
Is there any way i can understand this phenomen?Is there a software programme for horisontal plane and inclined plane?If there is please announce me.
 
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  • #2
It would probably help if you described concrete scenarios that illustrate the questions you are asking. Your post does not have enough information to understand where you are having difficulty.

You should be aware that we can't give answers to homework problems on PF; you can post them in the homework forum (if you fill out the template), but all we can do there is give you hints to help you solve the problem yourself.
 
  • #3
Uh this is not homework...This is something that could enlighten me for my baccalaureat exam...Definetly not homework...
Ok i could give you some images as examples:
I think that is enough
 

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  • #4
Have you learned about the concept of "free body diagrams," or did you feel that they are unnecessary? If you have learned about free body diagrams, please draw a few that illustrate your questions for us.
 
  • #5
I also found something that looks like this
What is this R?Is that the normal force?(1st pic)
Also why do they change the direction of the angles from sine to cos and from cos to sine in the horisontal plane?
Here is the picture (2nd pic)
In the inclined plane the angles were different
The 3rd pic
 

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  • #6
alex77 said:
i could give you some images as examples

These images help to illustrate the scenarios you're interested in; so what are you having difficulty with?
 
  • #7
I don't have imagination and i couldn't figure it out how that body or object could move..If you push it you change the angle?
What are these angles with sines and cosinus in these planes?If i have a wire and i want to tie that object to another object does that mean that the tension is equal to the force of the objects that i want to pull?
 
  • #8
alex77 said:
What is this R?Is that the normal force?(1st pic)

I don't see an R in the first picture.

alex77 said:
why do they change the direction of the angles from sine to cos and from cos to sine in the horisontal plane?
Here is the picture (2nd pic)In the inclined plane the angles were different
The 3rd pic

Have you studied trigonometry? Do you know how the sine and cosine are defined as ratios of sides of a right triangle?

Also, the picture of the inclined plane case is showing an angle ##\theta## in a triangle whose sides are not the force vectors you are trying to relate. To understand how the sine and cosine relate the force vectors in this case, you first have to figure out where the angle ##\theta## is in the triangle of forces.
 
  • #9
Here is that picture with that R letter..
 

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  • #10
alex77 said:
Here is that picture with that R letter..

Ok, then R is the normal force on ##m_2##.
 
  • #11
PeterDonis said:
I don't see an R in the first picture.
Have you studied trigonometry? Do you know how the sine and cosine are defined as ratios of sides of a right triangle?

Also, the picture of the inclined plane case is showing an angle ##\theta## in a triangle whose sides are not the force vectors you are trying to relate. To understand how the sine and cosine relate the force vectors in this case, you first have to figure out where the angle ##\theta## is in the triangle of forces.
That is the same case in the horisontal plane?The thing is i don't see any triangle in the horizontal plane.All i see is a square that is attached to a line.
 
  • #12
Plus what i understanded from a vid is that if you push that square you will gain velocity.In horizontal plane pushing=force?And if you push that object where can you see the friction force?
 
  • #13
alex77 said:
Plus what i understanded from a vid is that if you push that square you will gain velocity.In horizontal plane pushing=force?And if you push that object where can you see the friction force?
Please choose one of these figures to start out with (not the inclined plane). Please say in your own words what you think is happening in the figure. Please identify the forces that are acting on the object (or objects) in the vertical direction and the forces that are acting on the object in the horizontal direction. Please write down what you think the force balance equation on the object is for the vertical direction and what the force balance equation on the object is for the horizontal direction.
 
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  • #14
Is this ok?Have i written the forces in this 2d plane correctly?
 

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  • #15
Sorry for the wrong letter.Here it is.
In my country G is weight and that is why i noted like that.It won't happen again.
 

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  • #16
alex77 said:
Sorry for the wrong letter.Here it is.
In my country G is weight and that is why i noted like that.It won't happen again.
I asked you to articulate in your own words what you think is happening in the figure. The first step in attacking any problem is doing this. Plus, I also asked you to "identify (in words) the forces that are acting on the object (or objects) in the vertical direction and the forces that are acting on the object in the horizontal direction. Finally, I asked you to "write down what you think the force balance equation on the object is for the vertical direction and what the force balance equation on the object is for the horizontal direction."

I am unwilling to help you until you do this. Maybe someone else will be more compromising.

Chet
 
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  • #17
Well i do know that the normal force is always opposite to the force of weight.I also know that if that object is pushed you will gain velocity.Also the tension force which is pulled is opposite to the friction force and it is equal to the net force F.
I guess in math the equation are like these:
N-W=0 and T-Ff=0
I'm not sure if this is how you decompound the forces in the assembly system...That is all i know.
 
  • #18
alex77 said:
Well i do know that the normal force is always opposite to the force of weight.I also know that if that object is pushed you will gain velocity.Also the tension force which is pulled is opposite to the friction force and it is equal to the net force F.
I guess in math the equation are like these:
N-W=0 and T-Ff=0
I'm not sure if this is how you decompound the forces in the assembly system...That is all i know.
Normal force is, in general, just the force exerted by a surface on a body, in the direction perpendicular to the surface. So, in general, it is not opposite in direction to the weight and it is not of the same magnitude as the weight of the body. Can you think of situations where the normal force is not equal and opposite to the weight?
In the diagram as you have drawn it, the external force F is pointing in a direction that would cause the string to go slack (i.e., zero tension). Can you see that? Please redraw the figure with the force F pointing to the right.

Your force balance on each of the two blocks in the figure is correct for the vertical direction. It is incorrect for each of the blocks in the horizontal direction. Please draw a separate free body diagram for each of the blocks, showing the forces acting on each. Then write a force balance equation in the horizontal direction for each of the blocks.
 
  • #19
Well i learned that the normal force is up and that weight is always down but this is because you have a 2d dimensional system with vectors,right?I mean in vectors you used to write the sums by adding the half diagonal of each side.Is this the same here because i don't see the relation between vectors and bodies.
When i wrote the equations N-W=0 and Ff-Fcosa=F i didn't knew that F is the result of the vectors.
How can you understand vectors with bodies?
 

Related to My physics IB (international baccalaureat)

1. What is the purpose of the physics IB (international baccalaureate)?

The purpose of the physics IB is to provide a rigorous and globally recognized program for high school students that focuses on developing a deep understanding of key physics concepts and skills.

2. How is the physics IB different from a regular high school physics course?

The physics IB curriculum goes beyond basic concepts and equations and emphasizes critical thinking, experimentation, and real-world applications. It also includes an internal assessment and an externally assessed exam.

3. How is the physics IB assessed?

The physics IB is assessed through a combination of internal and external assessments. Internal assessments include laboratory work and a research project, while external assessments consist of written exams. The final IB grade is based on a combination of coursework and exams.

4. What are the benefits of taking the physics IB?

The physics IB prepares students for higher education in physics or other related fields, as well as developing critical thinking and problem-solving skills. It is also recognized by universities globally, giving students a competitive edge in the college admissions process.

5. Is the physics IB only for students planning to pursue a career in science?

No, the physics IB is a well-rounded course that can benefit students in any field. It teaches transferable skills such as critical thinking, problem-solving, and data analysis, which are valuable in any career path.

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