Physics: Post your doubts here!

[shinnyyy]

Actually my sir did solve tht qtn but i forgot the method
Its 'C' I thnk....its a fluke tbh...T will be greatest as u need force greater than doors 'W' to move it..no idea abt 'H'

Sorry still I didn't get that :/

 

[Liv2Stdy]

Oh I got it ..tysm
And can you plz explain me the other question aswell?

Hi, which one?

 

[shinnyyy]

There's another question from forces unit :
If a car is moving in a horizontal direction then in which direction the net force will act on the car.secondly how we'll draw the vector diagram for this question.

This one

 

[shinnyyy]

View attachment 55142 How we have to solve this question and what's the logic behind the ans.

And this one

 

[Liv2Stdy]

And this one

OK so I'm not a 100 percent sure of this but If I'd been given to solve this, I would do it this way.

First you need to break T and H into components and then construct two equations saw that everything you see on the diagram is balanced.
Then its more or less a guessing game.


As you can see, the vertical component of T is balanced by both the vertical component of H and W.
The value of W would be the vertical component of T minus Vertical component of H.....mind you these are the components of forces we are talking about so obviously the value of W will be the smallest.
While the horizontal components of the two forces are equal, the vertical component of H is still smaller (vertical component of T - W), so we can assume that the magnitude of H is smaller than T.

So in the increasing order, the forces would be W, H, T. Answer is C.

Is it correct?

 

[Liv2Stdy]

This one

Is this the full question?

 

[shinnyyy]

Is this the full question?

Yeah it is ..actually it was some mcq and I have copied it's wording but idk the concept behind that

 

[Liv2Stdy]

Yeah it is ..actually it was some mcq and I have copied it's wording but idk the concept behind that

Well, I can tell you this much, If the car is "moving" in a certain direction, it means it has overcome frictional forces and air resistance, so the resultant force would be in the direction of the movement of the car.
As for the vector diagram, I have no idea...atleast not yet!lol. I might need to sit on that for a while and see. Sorry.

 

[shinnyyy]

Wkhan860 can you plz check this and lemme know if I have done it correctly, though I have multipled it by s but idk why I did that :/
Thanks in advance for the assistance

 

[shinnyyy]

Physicist can u plz solve the question which I have posted above?

 

[shinnyyy]

OK so I'm not a 100 percent sure of this but If I'd been given to solve this, I would do it this way.

First you need to break T and H into components and then construct two equations saw that everything you see on the diagram is balanced.
Then its more or less a guessing game.
View attachment 55171

As you can see, the vertical component of T is balanced by both the vertical component of H and W.
The value of W would be the vertical component of T minus Vertical component of H.....mind you these are the components of forces we are talking about so obviously the value of W will be the smallest.
While the horizontal components of the two forces are equal, the vertical component of H is still smaller (vertical component of T - W), so we can assume that the magnitude of H is smaller than T.

So in the increasing order, the forces would be W, H, T. Answer is C.

Is it correct?

Yeah the ans is C. Thanks alot

 

[Wkhan860]

View attachment 55172 Wkhan860 can you plz check this and lemme know if I have done it correctly, though I have multipled it by s but idk why I did that :/
Thanks in advance for the assistance

Work=F*displacement
So in this case the total force applied is Fs
While the force opposing the motion is component of weight mgsin(â)......the work done by th opposing force will be mgsinâ*s
Thn use efficiency formula mgsinâ*s/Fs.....s gets cancelled...whch gives D

 

[shinnyyy]

Work=F*displacement
So in this case the total force applied is Fs
While the force opposing the motion is component of weight mgsin(â)......the work done by th opposing force will be mgsinâ*s
Thn use efficiency formula mgsinâ*s/Fs.....s gets cancelled...whch gives D

How a force can be equal to Fs
and why we have multiplied that component with s ?though I did that with heat and trial method but I have no concept for this.

 

[Wkhan860]

How a force can be equal to Fs
and why we have multiplied that component with s ?though I did that with heat and trial method but I have no concept for this.

Its nit the force tht is Fs
Its work....we r finding efficiency of work
Thts y Force *displacement

 

[shinnyyy]

Its nit the force tht is Fs
Its work....we r finding efficiency of work
Thts y Force *displacement

Ikr it's work but what's useful work output and what's the work which is input ?
I think you aren't getting what I'm asking

 

[Wkhan860]

Ikr it's work but what's useful work output and what's the work which is input ?
I think you aren't getting what I'm asking

Thy say the car us movng at constant velocity...whch means the weight component is equal to forward force...dat is work output
So mgsinâ*s
Work input is the total energy whch is F*s
I hope u get it now

 

[shinnyyy]

Thy say the car us movng at constant velocity...whch means the weight component is equal to forward force...dat is work output
So mgsinâ*s
Work input is the total energy whch is F*s
I hope u get it now

I didn't get it yet , sorry for that :/ maybe cz I'm sleepy and brains not working..
na ways I'll see that tomorrow and tysm.

 

[Liv2Stdy]

View attachment 55172 Wkhan860 can you plz check this and lemme know if I have done it correctly, though I have multipled it by s but idk why I did that :/
Thanks in advance for the assistance

 

[shinnyyy]

View attachment 55173

Why the energy output is mgsinâ .s rather then f.s ? That's where I'm confused

 

[Liv2Stdy]

Why the energy output is mgsinâ .s rather then f.s ? That's where I'm confused

Work done would be Fs if the car was travelling on a horizontal road but in this case it's not, the car is travelling at angle a and the weight also acts on the inclined plane. You need to break this weight into its components coz the weight also acts at an angle a (look at my diagram). Force F is in the horizontal direction. If you look at the components, mgsina=F.......so even though you exert a force F on the car, due to the weight only a component acts on the car to make it move. Hence, useful energy output becomes mgsina x s.