Inclined plane and inclined force-acceleration

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Homework Help Overview

The problem involves a sled of mass 8kg on a frictionless inclined plane at 35 degrees, being pulled by a rope with a tension of 40N at an angle of 20 degrees to the slope. The tasks include resolving forces into components and finding the sled's acceleration.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to resolve forces into components and is seeking clarification on calculating the resultant force and acceleration. Some participants discuss the direction of acceleration and the implications of negative values in their calculations.

Discussion Status

Participants are actively engaging with the problem, with one offering a perspective on the direction of resultant acceleration. There is an acknowledgment of potential sign errors in calculations, and the discussion is exploring the implications of these errors without reaching a consensus.

Contextual Notes

There is a mention of the sled moving up the slope and the need to consider the direction of forces and acceleration, which may influence the calculations. The original poster is also grappling with the implications of negative values in their results.

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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