Why Isn't the Normal Force on the Sled 250 N?

  • Thread starter Thread starter 123lar
  • Start date Start date
  • Tags Tags
    Friction Slope
AI Thread Summary
The normal force on the sled is not 250 N because it is influenced by the incline of the hill. The sled's weight, calculated as 250 N, is distributed into components due to the 15-degree slope. As the sled moves down the incline at a constant speed, the normal force acts perpendicular to the surface and is less than the full weight. The actual normal force can be determined by calculating the cosine component of the weight relative to the incline. Therefore, the normal force is a fraction of the sled's weight, not the full 250 N.
123lar
Messages
6
Reaction score
0

Homework Statement


The question is: An empty sled of mass 25 kg slides down a hill with constant speed of 2.4 m/s. The slope of the hill is inclined at an angle of 15 degrees

How come the normal force of the sled is NOT 250 N?

Homework Equations

 
Physics news on Phys.org
Assuming here g= 10 N/kg, the weight is 250N.

However since the mass is moving down an incline at 15 degrees, the normal force will be a component of the weight i.e. a fraction of the weight.
 
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