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

AI Thread Summary
To calculate the tension in a pulley system on a ramp, the forces acting on each mass must be accurately represented. The equations of motion for the two blocks, taking into account the direction of tension and gravitational forces, are crucial. It was noted that gravity acts down the incline while tension acts up, necessitating careful attention to the signs in the equations. The discussion emphasized using the sine function for the gravitational force component parallel to the ramp, rather than cosine. Correctly applying these principles leads to a more accurate calculation of tension in the system.
brunettegurl
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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
 

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brunettegurl said:
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! :smile:

(i can't see the diagram yet, but …)

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

one is minus a :wink:
 
Also: Gravity acts down the incline while tension acts up--they must have different signs.
 
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??
 
Hi brunettegurl! :smile:
Doc Al said:
Also: Gravity acts down the incline while tension acts up--they must have different signs.

Same thing applies to the other block. :wink:
 
so is my new equation wrong??
 
oops!

brunettegurl said:
so is my new equation wrong??

oops! must have misread it :redface:

yes, the first two equations are correct …

but shouldn't the final one have sin rather than cos?
 
why would it be sin instead of cos??
 
Last edited:
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.
 
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