Inclined plane and inclined force-acceleration

AI Thread Summary
The discussion revolves around a physics problem involving a sled on a frictionless incline, where the sled's mass is 8 kg, the slope is inclined at 35 degrees, and it is pulled by a rope with a tension of 40 N at a 20-degree angle to the slope. The user successfully resolved the forces into components but is struggling with calculating the sled's acceleration using F=ma. They are confused about the resultant force direction and whether their calculations are correct, especially regarding the signs of the forces involved. A contributor suggests that the negative acceleration indicates insufficient pulling force from the rope. The user is encouraged to ensure the correct direction is noted when reporting the resultant acceleration.
dejna007
Messages
2
Reaction score
0

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!
 
Physics news on Phys.org
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top