Tension, FB diagrams and some formula questions.

[lozzajp]

Ok this will be a long post, sorry in advance. I am an online student of a university and the help I am getting from them is ridiculous, I am also using a book called quicksmart introductory physics, and for the most part it is quite rubbish.

1st Problem: (not homework questions, I am trying to figure out the example and how it works)
http://postimage.org/image/wfk1kp6jx/ scan of the page

If you can see that scan of the page, great. It is about two masses on a pulley/string and one mass is on an incline. I can work out the acceleration of the separate masses (g for straight down, g x Sinθ for the incline) and that F=ma.

I am having trouble understanding how to get the resulting acceleration of the masses (I seem to have no trouble when it is a matter of two objects straight down on a string/pulley with different masses). As well as I can not figure out the Tension either.

I really need a simple explanation of how to do this process, as the book just goes from easy enough physics to all this overdrive in this example.

2nd Problem:
Short story, 2 mass on a pulley both directed at the ground. resulting in a 20kg mass accelerating towards the ground at 5.88m/s/s and has 2m to fall.

The acceleration part i worked out easy enough, but the time to calculate it to move those 2m i can't get it.

i used s=ut+0.5a(t^2)
rearranged it to get t^2 = 2.944/2
t = 1.21s

however this is apparently wrong and the book answers says a formula t^2 = 2s/a
t = 0.82s
where did this come from?

the resulting velocity i got from a = v/t so i mistakingly went v = a/t which ironically gave me the right answer for v (but i had t wrong).
v = at provides the right answer if t is right.

So why doesn't my first formula work for t?

Problem 3:
car 1 is traveling at a constant 30m/s and is 50m ahead of a stationary car 2.
car 2 uniformly begins accelerating at 4m/s/s. when will the car catch up?

The book gives me no information at all how to work this out.
The solution gives 50 +30t = 0 + 1/2 x 4t^2
I can see this is a modified s = ut + 0.5at^2 but I have no idea how it works or how to get to it?


Thanks if i can get just a little help understanding this!

 

[azizlwl]

Just apply Newton's Laws of Motion.

One of the most important skills for the student of elementary
mechanics is to learn to examine an object's environment and identify all the forces
on the object:
What are the things touching the object and possibly exerting contact
forces, such as friction or air pressure? And what are the nearby objects possibly
exerting action-at-a-distance forces, such as the gravitational pull of the Earth or the
electrostatic force of some charged body?

 

[lozzajp]

I can manage that for easier problems, but if you look at what I am having trouble with is what equation to use at the right time and how to get there/know that its that one to use.

just to add, friction, pressure and electrostatic forces I have not encountered yet :( simply having trouble with these 3 elementary topics.

 

[azizlwl]

We look at m2.
The Earth is pulling it down.
Since m1 is connected to m2, m1 is trying to stop m2.
Here Newton's 3rd. m1 is pulling m2 and this equal to m2 pulling the m1.

m1 will try to slow down m2, and m2 will try to make m1 moves.

So at m2,the net force applied at m2 is the sum of gravitational force and force from m1.

Newton's law, F_net=ma

F_{by earth}-F_{pull by m1}=m2a

.........

On the inclined plane, 4 forces acting on the m1
1. gravitational
2. the plane N
3. Friction which is apposing the motion which equal to μN
4. Pulling force from m2.

 

[lozzajp]

ill look at above post over the next while and see what I can get from it thanks :).

just figured out problem 2 on my own, i was rearranging it wrong.
s= ut + (0.5) x a x t^2
s/a = ut + (0.5) x t^2
2s/a = ut + t^2

no initial velocity so ut = 0
t^2 = 2s/a

 

[lozzajp]

F_{by earth}-F_{pull by m1}=m2a

Do you mean m1+m2? I did the exercise again using my own way of things and ended up with the right answer.

force 1 (gravity and m2) - force 2 (m1 down the incline) / m1+m2 (as in f=ma or a=f/m)
=2.9m/s/s
which is the answer the book gives, I think i have the idea of tension figured out now too.

Still have no idea the way the book does it though :/

 

[azizlwl]

1. No problem here.
2a. Fbd or vector diagram of m1. Since it is smooth no friction only 3 forces..
2b. Vector diagram for m2. Since pulley is frictionless, tension are equal.
3. Since the vectors are in different direction, the x-axis is taken as parallel to the plane for easier calculation. So N in y-direction and T in x-direction.
Now on mg which we have to resolve to 2 component of x and y direction.

4. Now begin calculation from motion equation.
F_net=ma

Which part that you do not understand?