# Math Help - Mechanics 1, string and tension Problem, Home work due

1. ## 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.

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

forces acting on the hanging mass ...

$Mg - T = Ma$

forces acting on the incline's mass ...

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

combining the equations ...

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

solving for the acceleration ...

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

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

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

3. ## 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

4. ## 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.

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

incorrect ...

$Mg = 5g$

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

$f = 14$

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

6. ## 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.

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

no again ...

the net force equation was ...

$Mg - T = Ma$

not

$T - Mg = Ma$

8. ## 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.