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.

Re: Mechanics 1, string and tension Problem, Home work due

forces acting on the hanging mass ...

$\displaystyle Mg - T = Ma$

forces acting on the incline's mass ...

$\displaystyle T - mg\sin{\theta} - f = ma$

combining the equations ...

$\displaystyle Mg - mg\sin{\theta} - f = (M+m)a$

solving for the acceleration ...

$\displaystyle \frac{Mg - mg\sin{\theta} - f}{M+m} = a$

calculate the acceleration, then use the first equation to determine the tension in the string, $\displaystyle T$

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

Re: Mechanics 1, string and tension Problem, Home work due

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

Re: Mechanics 1, string and tension Problem, Home work due

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.

Re: Mechanics 1, string and tension Problem, Home work due

incorrect ...

$\displaystyle Mg = 5g$

$\displaystyle mg\sin(30^\circ) = 2g \cdot \frac{1}{2} = g$

$\displaystyle f = 14$

$\displaystyle \frac{5g - g - 14}{5+2} = a$

Re: Mechanics 1, string and tension Problem, Home work due

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.

Re: Mechanics 1, string and tension Problem, Home work due

no again ...

the net force equation was ...

$\displaystyle Mg - T = Ma$

not

$\displaystyle T - Mg = Ma$

Re: Mechanics 1, string and tension Problem, Home work due

So 5g -T= Ma

5x9.8-T= 3.6x5

49-T = 18

-T= -31

T= 31

Thank you very much for your immense help.