Inclined plane and inclined force-acceleration

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SUMMARY

The discussion focuses on solving a physics problem involving a sled of mass 8 kg on a frictionless slope inclined at 35 degrees, pulled by a rope with a tension of 40 N at a 20-degree angle to the slope. Participants resolved forces into components and applied Newton's second law (F=ma) to find acceleration. The key challenge was determining the resultant force and ensuring the correct signs for forces acting on the sled. The conclusion indicates that the sled's acceleration is directed up the slope, but the tension may not be sufficient to overcome gravitational forces.

PREREQUISITES
  • Understanding of Newton's second law (F=ma)
  • Knowledge of force resolution into components
  • Familiarity with gravitational force calculations (W=mg)
  • Basic trigonometry for angle calculations
NEXT STEPS
  • Study the concept of force resolution in inclined planes
  • Learn about the effects of tension in systems involving pulleys and slopes
  • Explore the implications of negative acceleration in physics problems
  • Review examples of frictionless motion on inclined planes
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and inclined planes, as well as educators looking for examples of force analysis in real-world applications.

dejna007
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Homework Statement


Hi, here is my problem:
A sled of mass 8kg is on frictionless slope inclined at 35 degrees to the horizontal. It is pulled by a rope whose tension is 40N and makes angle 20 degrees with the slope.

a)Resolve all forces, acting on the sled into components parallel and normal to the slope.
b) Find the acceleration of the sled(take up the slope as positive direction)


Homework Equations


F=ma
W=mg


The Attempt at a Solution


Ok so I managed to do part a), where first I found components of tension:
Thorizontal=40cos20=37.6 N
Tvertical=40sin20=13.7 N

Components of downward force:
Because the angle between the normal to the slope and the normal to the horizontal is the same as the one between the slope and horizontal
-mgcos35=8x9.8cos35=64.2
-mgsin35=8x9.8sin35=45

For next part, to calculate acceleration all I know is to use F=ma and some of the components calculated in part a).
Can anyone give me a hint to part b) and explain what the resultant force is? or how do I find resultant force in this case?

Thank you!
 
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the sled is moving up the slope, so the resultant acceleration is logically up the slope right?

So use (in the x-direction) ma= upward forces -downward forces
 
Thanks for fast reply rock.freak667! My answer has a negative value and according to what you write it should be positive. I can't see where I made mistake. Could it be sign mistake when calculating downward force, the sign of acceleration due to gravity will be negative here?
 
dejna007 said:
Thanks for fast reply rock.freak667! My answer has a negative value and according to what you write it should be positive. I can't see where I made mistake. Could it be sign mistake when calculating downward force, the sign of acceleration due to gravity will be negative here?

well it seems that you would get a negative answer...you'll just need to just write in the correct direction when you write down the resultant acceleration.

Seems who ever is pulling the rope isn't pulling it hard enough.
 

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