Find Tension: Solve Homework on Frictionless Slope w/m1=3kg & m2=0.86kg

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Homework Help Overview

The discussion revolves around a physics problem involving two blocks on a frictionless slope, connected by a massless cord. The parameters include angles of inclination and masses of the blocks, with the goal of finding the tension in the cord.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to find acceleration and question the origin of a specific formula used for calculations. There are suggestions to analyze forces on each block separately and to draw free body diagrams (FBDs). Some participants explore the use of trigonometric functions to relate forces and acceleration.

Discussion Status

The discussion is active, with participants exploring different approaches to analyze the forces acting on the blocks. There is a focus on identifying the components of weight acting parallel to the slope and clarifying the role of tension. Guidance has been offered regarding the need to consider the forces acting on each block.

Contextual Notes

Participants note that the slope is frictionless, which influences the forces considered in the problem. There is an ongoing exploration of the relevant forces, including weight and tension, while clarifying misconceptions about the normal force.

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Homework Statement



In the figure below, assume that the slope is frictionless and that the two blocks are connected by a massless cord. Assume the following:
θ1 = 37°
θ2 = 45°
m1=3kg
m2=0.86kg.
What is the tension in the cord?.


Homework Equations



F=ma , w=mg, trig


The Attempt at a Solution



i know that you need to find acceleration. Which i did (a=3.04) and i found it by ((m1)(m2)g)/(m1+m2)

you'll have to use trig functions but I'm not sure which one to use.
And wouldn't you set it equal to each other?
 

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Rey4312 said:
i know that you need to find acceleration. Which i did (a=3.04) and i found it by ((m1)(m2)g)/(m1+m2)
Where did that formula come from?

In any case, try analyzing the forces on each block separately. Draw a FBD for each and apply Newton's 2nd law. Then combine the two equations to solve for the acceleration and the tension.
 
I was given it...

Could you use the trig function tangent?
so then your equations would be:
w tan(37)=3 a
and
w tan(45)=.86 a

but then you're only left with acceleration, and there's two different ones.
 
Rey4312 said:
I was given it...
It doesn't apply to this problem.

What forces act on m1 parallel to the surface?

What forces act on m2 parallel to the surface?
 
Doc Al said:
It doesn't apply to this problem.


oh okay...


the only force that i can think of would be friction, at least that is parallel
 
Rey4312 said:
the only force that i can think of would be friction, at least that is parallel
No. You are told that the slope is frictionless.

There are two other forces with components parallel to slope. What are they?
 
w=mg, tension?, or would it be the normal force (though i thought that was perpendicular)
 
Rey4312 said:
w=mg, tension?,
Right, those are the two forces with parallel components. What is the component of the weight parallel to the slope?
or would it be the normal force (though i thought that was perpendicular)
Yes, the normal force is perpendicular. So you won't need it.
 
W= mg (mass*gravity)
 
  • #10
Rey4312 said:
W= mg (mass*gravity)
What direction does the weight act?
 
  • #11
directly down parallel with the slope
 
  • #12
Rey4312 said:
directly down parallel with the slope
Directly down, yes. But that's not parallel to the slope. (The slope isn't vertical.) You need to find the component parallel to the slope.

You might want to read this: Inclined Planes
 
  • #13
the component parallel would be tension.
 
  • #14
Rey4312 said:
the component parallel would be tension.
The tension in the string is a different force, but yes it is parallel to the slope. You still need the component of the weight parallel to the slope. (Read the link I gave.)
 
  • #15
Doc Al said:
Read the link I gave.
Sorry didn't see that the first time..

so it would be net force?
 
  • #16
solved it... thanks for the help
 

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