The figure below shows an object on an inclined

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The discussion revolves around calculating the tension in a string connecting two objects, one on an inclined ramp and the other on a horizontal surface. The first object has a mass of 1.30 kg and is subjected to an external force of 10.73 N, while the second object has a mass of 2.98 kg. Participants emphasize the importance of drawing accurate free body diagrams to visualize the forces acting on each block, particularly noting that the gravitational force on the second block acts perpendicular to the tension. The equations of motion are discussed, with a focus on how to relate the acceleration and tension in the system. Ultimately, clarity in the diagram and correct application of physics principles are crucial for solving the problem accurately.
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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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