Solve Forces & Inclines Homework: Find Coefficient of Friction

  • Thread starter Thread starter seizureboi
  • Start date Start date
  • Tags Tags
    Forces
AI Thread Summary
To find the coefficient of kinetic friction for a 3kg block being pulled up a 37-degree incline at constant speed with a tension of 45N, the equation T - mgsin(θ) - μNF = 0 is used. The normal force (NF) is calculated as NF = mgcos(θ), leading to the equation 45 - (3)(9.8)sin(37) - μ(3)(9.8)cos(37) = 0. After solving, the correct coefficient of friction (μ) is determined to be approximately 0.4. There was confusion regarding the need to include sin(θ) for the tension, as it is already parallel to the incline. The final conclusion confirms that the initial calculation of μ = 0.4 is correct.
seizureboi
Messages
8
Reaction score
0

Homework Statement



A 3kg block is pulled by a rope up a 37 degree incline at constant speed. If the tension in the rope is 45N and the rope is parallel to the incline, calculate the coefficient of kinetic friction between the block and the incline.

Homework Equations



a = acceleration (0 m/s2)
Ө = theta (37 degrees)
g = 9.8 m/s2
m = mass
T = Tension
μ = coefficient of friction
NF = Normal Force
friction = μNF
NF = mgcosӨ


The Attempt at a Solution



My attempt was setting up the equation: TsinӨ - mgsinӨ - friction = ma
OR 45sin(37) - (3)(9.8)sin(37) - μ(3)(9.8)cos(37) = 0
OR 27.1 - 17.7 = μ23.5
OR 9.4 = μ23.5
Finally, μ = .4

Just wondering if it's correct. =)
 
Physics news on Phys.org
TsinӨ - mgsinӨ - friction = ma

You don't need a sinӨ multiplying the tension, the force of the tension is given parallel to the plane so you wouldn't need to factor an angle in, make sense?
 
Cool, because I did try it that way and i got μ = 1.16 and I thought I was wrong so I concluded with adding a random sinӨ lol.
 
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