Solve for aa = 3.38 m/s^2Pulley Quesiton Homework: Tension in Massless String

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SUMMARY

The discussion centers on calculating the tension in a massless string connecting two objects, one on an inclined ramp and the other on a horizontal surface. The inclined object has a mass of 0.98 kg and is at an angle of 32 degrees, while the horizontal object has a mass of 2.13 kg. An external force of 9.00 N is applied to the inclined object, leading to an acceleration of 3.38 m/s². The equations derived for tension include FT + 5.09 N - 9 N = 0.98a and Ft - 3.91 = 0.98a, which are essential for solving the problem.

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Homework Statement



The figure below shows an object on an inclined ramp of mass 0.98kg. The angle of the inclined surface is 32deg with the horizontal. The object on the ramp is connected to a second object of mass 2.13kg on a horizontal surface below an overhang that is formed by the inclined surface. Further, an external force of magnitude 9.00N 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?

[see picture attached]

Homework Equations


The Attempt at a Solution



FT+ 5.09N - 9N = 0.98a
Ft - 3.91 = 0.98a

-3.91 = -1.155a
 

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oneshot said:

Homework Statement



The figure below shows an object on an inclined ramp of mass 0.98kg. The angle of the inclined surface is 32deg with the horizontal. The object on the ramp is connected to a second object of mass 2.13kg on a horizontal surface below an overhang that is formed by the inclined surface. Further, an external force of magnitude 9.00N 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?

[see picture attached]

Homework Equations



The Attempt at a Solution



FT+ 5.09N - 9N = 0.98a
Ft - 3.91 = 0.98a

-3.91 = -1.155a
attachment.php?attachmentid=56994&d=1364015606.png


What's your question?

If you want your solution critiqued, you should show in more detail how you got those quantities.

Where does the angle of the incline come into play, etc. ?
 

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