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

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Homework Help Overview

The discussion revolves around calculating the tension in a pulley system involving two blocks sliding along a frictionless ramp. The original poster presents equations based on the forces acting on each block, including gravitational force and tension.

Discussion Character

  • Mathematical reasoning, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the setup of equations for each mass, questioning the signs of the forces involved. There is a discussion about whether the acceleration should be treated as the same for both blocks or if one should be negative. Additionally, there is a debate about using sine versus cosine for the components of gravitational force acting on the ramp.

Discussion Status

The conversation is active, with participants providing feedback on the equations presented. Some guidance has been offered regarding the signs of the forces and the use of trigonometric functions, but there is no explicit consensus on the correct approach yet.

Contextual Notes

Participants note the absence of a diagram, which may affect the clarity of the discussion. There is also a mention of potential misinterpretations of the equations and the need to clarify the components of gravitational force acting on the blocks.

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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