The figure below shows an object on an inclined

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Homework Help Overview

The problem involves an object on an inclined ramp connected to another object on a horizontal surface, with specific masses and an external force applied. The scenario is set within a frictionless context, and the goal is to determine the tension in the connecting string.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the forces acting on both blocks, questioning the correctness of their force diagrams and calculations. There is an exploration of how to apply Newton's second law to the system, particularly regarding the role of gravitational force on the second block.

Discussion Status

Some participants have offered guidance on redrawing diagrams and focusing on free body diagrams. There is an acknowledgment of the need to clarify the forces acting on each block, and while some calculations have been shared, there is no explicit consensus on the correctness of the approach or results.

Contextual Notes

Participants note the absence of a clear diagram and the potential confusion arising from the setup. The discussion also reflects on the assumptions of frictionless surfaces and the massless nature of the connecting string and pulley.

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Homework Statement



The figure below shows an object on an inclined ramp of mass 1.30 kg. The angle of the inclined surface is 25° with the horizontal. The object on the ramp is connected to a second object of mass 2.98 kg on a horizontal surface below an overhang that is formed by the inclined surface. Further, an external force of magnitude 10.73 N is exerted on the object on the ramp. We observe both objects to accelerate. Assuming that the surfaces and the pulley are frictionless, and the connecting string and the pulley are massless, what is the tension in the string connecting the two objects?

I AM DESPERATELY IN NEED OF HELP ! PLEASEE!


Homework Equations



SEE ATTACHEMENT


The Attempt at a Solution



1, For block 1, i don't know if my forces are correct, they seem to be wrong
For block 2, mg and T are in different directions, how do I use ma= net force?
Does the normal force on block 2 cancel out with the m2g so ma= Ft?

1.3 a = 10.73N - Ft - mgsin25
1.3a = 5.346 - Ft
2.98a = Ft
a = 1.249 m/s^2
 

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I cannot see the geometry of the situation from your diagram. eg. I see two blocks in the diagram, both with forces mg on them, one directly below the other, but no means to attach them. There are no pulleys in the diagram either.

This is probably the source of mistakes ... redraw the diagram more carefully and start again.
Draw free body diagrams for each mass.
 
For block 2, mg and T are in different directions, how do I use ma= net force?

You can forget/ignore mg on block 2. It's acting at 90 degrees to FT. There is no friction so for block 2...

FT = m2 * a
 
Last edited:
CWatters said:
You can forget/ignore mg on block 2. It's acting at 90 degrees to FT. There is no friction so for block 2...

FT = m2 * a


then is my calculation correct?
 
Only had a quick look but seems ok to me.
Don't forget the question also asked for the tension.
 
Ok, thanks!
 

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