# Ramp Question

1. Oct 18, 2005

### 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

Last edited: Oct 18, 2005
2. Oct 18, 2005

Hi!

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

3. Oct 18, 2005

### suspenc3

i could draw one..i dont know how to attach it though

4. Oct 18, 2005

### suspenc3

it wont let me attach it.it says its too big

5. Oct 18, 2005

### 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

6. Oct 18, 2005

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

7. Oct 18, 2005

### lightgrav

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

8. Oct 18, 2005

### suspenc3

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

9. Oct 18, 2005

### suspenc3

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

10. Oct 18, 2005

### 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!

11. Oct 18, 2005

### suspenc3

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

Last edited: Oct 18, 2005
12. Oct 18, 2005

### lightgrav

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

13. Oct 18, 2005

### suspenc3

therefore I cant substitute F (F=ma) for m1 a?

14. Oct 18, 2005

### 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

15. Oct 18, 2005

### 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 )

Last edited: Oct 18, 2005
16. Oct 19, 2005

### suspenc3

but y isnt perpendicular to x its at a 30 degree incline..thats why I used trig

17. Oct 19, 2005

### suspenc3

Hers a link to a picture

http://img418.imageshack.us/img418/9505/ramp4sl.png [Broken]

Last edited by a moderator: May 2, 2017