I have a hard time with kinematics involving inclined planes

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Homework Help Overview

The discussion revolves around a kinematics problem involving two blocks connected by a string over a frictionless pulley, situated on inclined planes. The inclines are frictionless and both are angled at 35 degrees.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to set up equations of motion for both blocks but encounters difficulties with the signs in their equations. Some participants suggest reviewing the direction of forces and acceleration, while others recommend creating a free body diagram to clarify the forces acting on each block.

Discussion Status

Participants are actively engaging in clarifying the setup of the problem, particularly regarding the signs of the forces involved. Some guidance has been offered regarding the conventions for positive and negative directions, and there is acknowledgment of the importance of a free body diagram. The original poster indicates a resolution to their problem after revisiting their approach.

Contextual Notes

There is an emphasis on maintaining consistent sign conventions and the necessity of visual aids like diagrams to aid understanding. The original poster's initial confusion highlights the complexity of interpreting forces in inclined plane scenarios.

devilish_wit
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Homework Statement


Two blocks of mass m1 = 3.00 kg and m2 = 6.00 kg are connected by a massless string that passes over a frictionless pulley (see the figure below). The inclines are frictionless.

Image - https://www.webassign.net/serpse9/5-p-049.gif
5-p-049.gif

The inclined plane is shaped like a triangle with both sides inclined at 35 degrees.

Homework Equations


F = ma

The Attempt at a Solution


For m1: Fnetx = ma ---> Fgx - T = ma ----> sin35(9.8m/s^2)(3kg) - T = 3kga
Equation for T: T = sin35(9.8m/s^2)(3kg) - 3kga

For m2: Fnetx = ma ---> Fgx - T = ma ----> sin35(9.8m/s^2)(6kg) - T = 6kga
Equation for T: T = sin35(9.8m/s^2)(6kg) - 6kga

Then I made them equal to each other in order to find acceleration but my answer ended up being wrong. Idk what I did wrong
 

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You are not paying attention to your signs. For m1 the tension and the acceleration are in the same directon and for m2 the tension and the acceleration are in opposite directions. Assume the convention the anything uphill is positive and anything downhill is negative, i.e. the positive direction is the direction of the acceleration. Keeping the convention consistently, for m2 the acceleration is downhill, so anything downhill is positive and anything uphill is negative.
 
A diagram is always helpful
 
kuruman said:
You are not paying attention to your signs. For m1 the tension and the acceleration are in the same directon and for m2 the tension and the acceleration are in opposite directions. Assume the convention the anything uphill is positive and anything downhill is negative, i.e. the positive direction is the direction of the acceleration. Keeping the convention consistently, for m2 the acceleration is downhill, so anything downhill is positive and anything uphill is negative.

I went back to my problem and finally got the correct answer. Thank you so much!
 
You should draw a free body fiagram, which contains all the forces acting on each body, and an appropriate coordinate system
 

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