Mechanics 1, string and tension Problem, Home work due

Hi, This is silver doing 12th year, I have clearly anyidea regarding inextensible string, could you please explain and help me in this problem. Thank you well in Advance.

A mass of 2kg lies on a rough plane whichis inclined at 30 degree to the horizontal. One end of a light inextensible string is attached to this mass and the string passes up the line of greatest slope and over a smooth pulley fixed at the top of the slope; a freely suspended mass of 5 kg is attached to its other end. The system is released from rest; as the 2kg mass accelerates up the slope, it experiences a constant resistance of motion of 14 Ndown the slope due to the rough nature of the surface. Find the tension in the string?


MY solution: force = mg= 2x9.8= 19.6N at 30 degree

Force parallel to the plane= 196x sin 30 = 98N
A system of mass 7kg is acclerated. Force applied is 14N down
Object B weigh 9.8x5= 49N
Net force = 98N-49N-14N
= 35 N


I have no idea. I would appreciate if you take time to explain this.

forces acting on the hanging mass ...



forces acting on the incline's mass ...



combining the equations ...



solving for the acceleration ...



calculate the acceleration, then use the first equation to determine the tension in the string,

... next time, please post physics questions in the "Math Topics" forum.

Thank you Skeeter. Thank you once agin for the speedy response. Sorry, wil post on Math tics

Hanging mass
T-Mg=Ma
5x 9.8 = 49N
T- 49N= Ma-------------> (1)
Force acting on incline mass
T-mg sin theta -f= Ma
-2x9.8x sin 30 -f = Ma---------> (2)
Mg- mgsin theta -f= (M+m)a
19.6- 9-8/7=a
1.4= a

Mg-T= Ma
49-T = 14
-T= -35
T= 35N


Please verify and correct me . Thank you very much for your timely help.

incorrect ...

So it is 4g- 14= 7a
4x 9.8-14= 7a
39.2-14= 7a
3.6= a

T-Mg= Ma
T- 5g = Ma
T- 5x9.8= 3.6 x5
T- 49= 18
T= 18+49
= 67 N

thank you very much for explaining. Still not sure about the answer. Once again thank you very much for your time and help.

no again ...

the net force equation was ...



not

So 5g -T= Ma
5x9.8-T= 3.6x5
49-T = 18
-T= -31
T= 31

Thank you very much for your immense help.