## Homework Statement

Two blocks connected with a massless rope ,are being dragged by a horizontal force.suppose F= 68N, Mass1= 12 ,Mass2=18kg and coeffecient of fraction between each block and surface is 0.10.
12kg ------T--------18kg-----68N--->
Find tension and magnitude of the aceleration of the system

## Homework Equations

f=mg, Fnet= Fa-ff, ff =mu *m*g

## The Attempt at a Solution

Now I have tried to sovle it please see if this is correct or I need to change it----> 1st sloving for acceleration of the system Fnet=Ma
Fnet= Fa+Ff
Ff= (0.10)(9.8)(30)
force of friction= 29.4
so Fnet = 68N +(-29.4)
Fnet= 38.6N
acceleration= Fnet/mass
acceleration = 38.6/30= 1.28 m/s^2
This is probably the answer of the second question.
Now sloving for Tension
F=M_1a ( for 1st mass 12kg)
F= 12*1.28m/s^2
Force on 12kgmass= 15.36N
Force of friction 12kgmass= (12)(9.8)(0.10)= 11.76N
Net force on 12 kg mass= 15.36N+(-11.76)= 3.6N

F=M_2a (for 2nd mass18kg)
F= 18*1.28m/1.28m/s^2
Force on 18kg mass= 23.04
force of friction on 18kg mass=(18)(0.10)(9.8)=17.64
net force on 18 kg mass= 23.04N+(-17.64)= 5.4N

Now the net force on 12 kg mass is 3.6N and net force on 18 kg mass is 5.4N.
am I correct so far if yes , is the tension of the rope one of these values or the sum of these values , if one of these which one and why?

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LowlyPion
Homework Helper
Welcome to PF.

You almost have it.

But the force of friction will add to the Tension not take away from it.

To accelerate the 12 kg block you must supply both the force to accelerate it at the system acceleration you found, as well as the force to overcome the retarding force of friction.