Motion in 2 Dimensions - Inclined Plane Pulleys

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SUMMARY

The discussion focuses on calculating the tension in a string used to haul a 5 kg mass up a smooth inclined plane at a 30-degree angle with an acceleration of 1.5 m/s². The correct tension in the string is determined to be 32.5 N, derived from analyzing the forces acting on the mass, including gravitational components. The participants emphasize the importance of understanding free body diagrams and the distinction between the components of weight acting parallel and perpendicular to the incline.

PREREQUISITES
  • Understanding of Newton's Second Law (Fnet = ma)
  • Knowledge of trigonometric functions (sine and cosine) in right-angled triangles
  • Ability to draw and interpret free body diagrams
  • Familiarity with concepts of forces acting on inclined planes
NEXT STEPS
  • Study the derivation of tension in inclined plane problems
  • Learn how to construct and analyze free body diagrams for various scenarios
  • Explore the effects of friction on inclined plane dynamics
  • Investigate the application of Newton's laws in multi-body systems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of inclined plane problems and tension calculations.

TheKovac
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Homework Statement


a Mess of 5kg is being hauled up a smooth invlined place at 30 degree to the horizontal with an acceleration of 1.5ms^-2 by mean of a string parallel to the incline:
a) What is tension in the string

Homework Equations


Fnet= ma
Rules of right angled triangles (SOC CAH TOA)
Any more?

The Attempt at a Solution


a)
=>Fnet = ma
Fnet = 5*1.6
Fnet = 8ms^-2
=>cos(theta)= O / H
cos60= O / 8
Tension = 16N = Wrong!
Answer = 32.5 N - How??
 
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The weight has 2 components, one is perpendicular to the plane, the other is down the plane. The tension acts up the plane. So the resultant force up the plane is?

(Also,the component down the plane is always weight*sine of the angle of inclination. Not cos)
 
2 Components. interesting..
That means the Perpendicular component
=> Fnet = ma
=> Fnet = 5*9.8
=> Fnet down = 49N

What next?
 
Before you approach questions like these, you must have a great understanding of how to draw freebody diagrams. For this question, Fnet = Ft + (-Fg parallel) = ma. If you do not understand where these forces are coming from, let me know and I'll draw you the freebody diagram.
 
Here, I attached a freebody diagram for the following question given that there is no force of friction due to the emphasis of a smooth surface and no coefficient of friction was given. Can you identify which forces are involved?
 

Attachments

Well I know that there are two main forces acting upon the pulley.

/|\
| Resistance Forces/ Tension
___
|___|

|
\|/ Gravity = 9.8ms Accel

These are the two I can think of, is there any more?
 

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