1 pulley, 2 masses, 2 inclines

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Homework Help Overview

The problem involves calculating the tension in a cable connecting two masses on different inclined planes, with one mass weighing 5 kg on a 60-degree incline and the other weighing 6 kg on a 70-degree incline. The scenario assumes frictionless surfaces.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to calculate acceleration for both masses and the implications of different inclines on the tension in the cable. There is mention of using free body diagrams to analyze forces acting on each mass and the cable. Some participants question the assumption that the tension would be independent of mass due to the different inclines.

Discussion Status

Participants are exploring various approaches to the problem, including the necessity of drawing free body diagrams and applying Newton's laws. There is acknowledgment of the complexity introduced by the differing inclines and the need to correlate the tension forces on both sides of the pulley. Some guidance has been offered regarding the importance of identifying constraints and the forces acting on the masses.

Contextual Notes

Participants note that the tension in the cable should be the same on both sides of the pulley, and there is an emphasis on the importance of accurately representing the forces in the free body diagrams. The discussion reflects uncertainty regarding the correct application of concepts and calculations.

TheronSimon
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Homework Statement


Calculate the tension in the cable connecting the two masses. Assume all surfaces are frictionless.
in lieu of the picture here are the given specs:

on the left side a 5 kg mass with a 60' incline
on the right side a 6 kg mass with a 70'incline


Homework Equations


a=Fgsinθ
Fnet = ma

The Attempt at a Solution


my attempt was to calculate both sides for acceleration since its frictionless both would be non dependent on the weight of the mass
then i used
Fnet=ma
so Ft-Fg = ma
and after the math Ft = 41.45 and for the right side Ft= 114.66 but its the wrong answer so how do i complete this question?
 
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TheronSimon said:
my attempt was to calculate both sides for acceleration since its frictionless both would be non dependent on the weight of the mass
that is not true when you have diffferent inclines with the masses connected by a cable
then i used
Fnet=ma
you have to look at each mass separately using a free body diagram to identify the forces acting along the incline and apply Newton 2 to each mass separately
so Ft-Fg = ma
the tension force on each block is Ft directed along the incline and pulling away from the mass you are examining. But the gravity force along the incline is a component of Fg.
and after the math Ft = 41.45 and for the right side Ft= 114.66 but its the wrong answer so how do i complete this question?
Note when you redo this problem that the tension in a cable wrapped around an ideal pulley is the same on both sides of the pulley.
 
Im still lost :S
 
As mentioned by PhanthomJay,

the only analytical approach is to draw a Free Body Diagram(FBD) of both bodies and the string and identifying the constraints.

Guessing that the values shouldn't depend on mass, is risky and should be avoided at all costs.

When you draw the FBD of blocks, forces acting on them are their weight components parallel and perpendicular to incline.The normal reaction exerted by inclines and the tension force exerted by chord.

Now you draw the FBD of string which will help you corelate the tension forces exerted by strings on both bodies.(What is it? How does the mass of string affect this relation?)

Finally you identify the constraints.The only possible movement can be parallel to the inclined surfaces (body 1 will move parallel to incline 1, body 2 will move parallel to incline 2)and at all times no motion takes place perpendicular to the incline(body 1 has no motion perpendicular to incline 1 and body 2 has no motion perpendicular to incline 2).

Plug in the values.
Apply Netwons 2nd and Third Law relating action reaction forces.
What do you see?
 

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