Mechanics (brother of physics)

[Manobilly]

43/oct13 q4

 

[AlishaK]

43/oct13 q4

What year thou?

 

[Manobilly]

What year thou?

Oh sorry12!hey I am not. So good at solving these kinda questions can u give me some way or sme tip to solve them easily.

 

[Adorkableme]

Woot! I'm doing the same paper.
Hopefully, you got the acceleration as -1m/s^2. Now we gotta use newtons second law or simply F=ma. And to do that, we need to know the forces acting on the particle and as we're supposed to find coeff of fric, it's pretty obvio that we need to find the Friction force and the Normal Reaction force first. The weight of the particle would b 'mg'. now resolving it with one component parallel to the plane i.e; mgsin30 (theta from i), and the other component perpendicular to the inclined plane i.e; mgcos30. Therefore, the Normal Reaction force (N) is equal to the perpendicular component that is 'N=mgcos30',
Coeff of fric=Friction force(F)/N; now as the particle slides down, we know that the force (parallel component from weight) acting downwards must be more than the friction force acting upwards. so we can form an eqn of 'mgsin30-F=m*-1 (this nothing but F=ma) where F=coeff of fric*N, F=Coeff of fric* mgcos30, hence, subs F in the previous eqn: mgsin30-Kmgcos30=-m (K is coeff of friction here) solve it now. cancel all the m's. final step wd be: 5-8.66K=-1, 8.66K=6, K=0.693 (3 sfig).

Hopefully I was clear enough! Best of luck! Remember me in ur prayers.

Thanks a lot!!