Finding Tension in Frictionless Incline System

[suspenc3]

a 1kg box on a 30degree frictionless incline is connected to a 3kg box on a horizontal frictionless surface. The pulley is frictionless and massless.
A)if the magnitude of F is 2.3N what is the tension in the connecting cord?

I found all of my formulas

M1 "X" direction-----T-M1g = M1a EQ 1
"Y" direction-----N-M1g = 0 EQ 2

M2 "X" direction-----(T-M2cos30) = M2a EQ 3
"Y"--------------(N-M2gcos60) = 0 EQ 4

anyways..is it as simple as substituting the F value given into eq 1?
i.e- T-M1g = F

 

[pizzasky]

Hi!

I'm having a little trouble visualizing the scenario in my mind. Is there a diagram available?

 

[suspenc3]

i could draw one..i don't know how to attach it though

 

[suspenc3]

it won't let me attach it.it says its too big

 

[suspenc3]

just picture a flat line with m1 on it..and another line attached to the right of line1 drawn at a -30 degree angle with respect to the x axis(line1).mass 2 lies on the inclined plane.
There is a pulley at the intersection point of the two lines

 

[pizzasky]

Hi!
Thanks for the explanation. In the original post, you said that "the magnitude of F is 2.3N". Which force is F?

 

[lightgrav]

where is this "F" (=2.3N) connected?

 

[suspenc3]

F is the sum of all forces...F=ma
in the diagram F above M1 and is pointed towards the pulley,

 

[suspenc3]

im trying to find T when F=2.3N..Sorry for not being able to explain it clearly..i need a diagram

 

[lightgrav]

Well, the tension in the connecting cord,
if there is NO other (external) Force, is
(4.9 N)(3kg/4kg) , which is NOT 2.3 N .

So either this set-up is not on Earth's surface
or there's an additional Force (horizontal?)
applied to the (1kg) mass.

If there's an extra Force ("by hand") here,
just add it into the sum of forces!

 

[suspenc3]

the formula is T = F + M1g..isnt it
by my calcs..that is how I would do it..but I am not sure if you can just substitute the force into the original equation..or if there is more work to be done

 

[lightgrav]

No, if the block m1 is on the horizontal surface, then
x: T ( + F_external) = m1 a
y: N - m1 g = 0 .

 

[suspenc3]

therefore I can't substitute F (F=ma) for m1 a?

 

[suspenc3]

would i have to do this:
isolate a in EQ 3:
(T-M2cos30)/M2 = a

sub into 1

(T-F = M1((T-M2cos30)/M2)

?
but that might make solving for T difficult

 

[lightgrav]

Make sure "x" is parallel the surface all along it,
and "y" perp. to surface on both sides of pulley.

the equations adding to "m a" should have NO cos30!

Usually, you would add Eq1 to Eq3, so T cancels:
easily solve for a = (4.9N + 2.3N)/4kg ,
then re-do eq.1 by itself to isolate T .
( T + 2.3N = 3kg a )

 

[suspenc3]

but y isn't perpendicular to x its at a 30 degree incline..thats why I used trig

 

[suspenc3]

Hers a link to a picture

http://img418.imageshack.us/img418/9505/ramp4sl.png