How Do You Calculate Tension in a Pulley System on a Ramp?

[brunettegurl]

Homework Statement

Two blocks,connected by a string over a pulley, slide along the frictionless ramp shown in the diagram. What is the tension in the string?

Homework Equations

\sumF=ma

The Attempt at a Solution

so i set up an equation for each mass
m1=3kg
\frac{Fg1+Ft}{m1}=a
m1=5kg
\frac{Fg2+Ft}{m2}=a

i equated the two since the a would be the same to solve for Ft
m2[(m1*g*cos30deg)+T1]=m1[(m2*g*cos60deg]+T1]

when i solve the equation and simplify for T i get 26.9N when i should be getting around 25 N..pls help

 

[tiny-tim]

m1=3kg
\frac{Fg1+Ft}{m1}=a
m1=5kg
\frac{Fg2+Ft}{m2}=a

i equated the two since the a would be the same to solve for Ft

Hi brunettegurl!

(i can't see the diagram yet, but …)

if I'm guessing right, the accelerations aren't both a …

one is minus a ​

 

[Doc Al]

Also: Gravity acts down the incline while tension acts up--they must have different signs.

 

[brunettegurl]

m1=3kg
\frac{-Fg1+Ft}{m1}=a
m1=5kg
\frac{Fg2-Ft}{m2}=a

i equated the two since the a would be the same to solve for Ft
m2[(-m1*g*cos30deg)+T1]=m1[(m2*g*cos60deg]-T1]
does that look better??

 

[tiny-tim]

Hi brunettegurl!

Also: Gravity acts down the incline while tension acts up--they must have different signs.

Same thing applies to the other block.

 

[brunettegurl]

so is my new equation wrong??

 

[tiny-tim]

oops!

so is my new equation wrong??

oops! must have misread it

yes, the first two equations are correct …

but shouldn't the final one have sin rather than cos?

 

[brunettegurl]

why would it be sin instead of cos??

 

[ideasrule]

Try splitting the force of gravity, a downward vector, into a component parallel to and a component perpendicular to the ramp. You'll see that the parallel component is mgsin(a), where a is the angle of inclination of the ramp.