A car placed on incline and a mass placed on the car

  • Thread starter Thread starter eren.kizildag
  • Start date Start date
  • Tags Tags
    Car Incline Mass
AI Thread Summary
A car of mass M is sliding down a frictionless incline with an angle theta, while a mass m is placed on it, experiencing friction between them. The system is released, and m moves parallel to the incline, prompting the need to determine the kinetic coefficient of friction. The discussion highlights the importance of both masses having the same acceleration in the x and y directions. Participants emphasize the necessity of calculating the downslope acceleration of the combined system to solve for the friction coefficient. The conversation reflects a collaborative effort to clarify the mechanics involved in the problem.
eren.kizildag
Messages
5
Reaction score
0

Homework Statement


The car of mass M is sliding on the incline. A mass of m is placed as shown. There is friction between m and M in the vertical surface of M . The incline is frictionless. System is released, and it is observed that m moves parallel to the incline. We are asked to find the kinetic coefficient of friction. (Angle is theta)


Homework Equations


Ff+masin(theta)=mg, N=macos(theta), Ff=Mk*N where Mk is the kinetic coefficient of friction. N is the reaction force between M and m.


The Attempt at a Solution


I have chosen the ordinary coordinate system. The masses must have same acceleration. (In both x and y direction). I also showed the friction force acting on m. And I wrote down reaction forces, but could not get the correct answer.

(I'm sorry, I am new at the forum, and I don't know how to use the forum properly. If I made a mistake about the body of the message, please forgive me.)
 

Attachments

  • Ads?z.jpg
    Ads?z.jpg
    2 KB · Views: 413
Physics news on Phys.org
Hello eren.kizildag. Welcome to Physics Forums.

Since m is not sliding on M, both M and m must be moving downslope together. Have you worked out the downslope acceleration of the pair as a function of g and θ ?
 
Yeah, but I got nothing.
 
eren.kizildag said:
Yeah, but I got nothing.

No acceleration downslope? Perhaps you could show your calculation?
 
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