How can I find the tension between two blocks ?

Question

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--->
a) find tension and find magnitude of the acceleration of the system.


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?

Answer 1

Block 1, mass = 12kg: Normal force = 117.6N

Coefficient of friction = 0.1, friction force = 11.76N



Block 2. Mass = 18kg: Normal force = 18*9.8 = 176.4

Coefficient of friction = 0.1. Friction force = 17.64N



Applied force = 68.0N

Net force on blocks = 68.0-(17.64+11.76)

Net force = 38.6N



This net force must accelerate the two blocks:

Net force = total mass * acceleration

Acceleration = Net force / total mass

Acceleration = 38.6/ (12+18)

Acceleration = 38.6/30

acceleration = 1.287m/s²



The 12kg block at the back has to be accelerated at this rate. What force must the 18kg block apply to the 12kg block?



Force = mass * acceleration



Force = 12*1.287

Force = 15.44N



The tension in the rope joining block 2 to block 1 is 15.44N

Answer 2

Imagine if that 12kg block was hanging from the 18kg block which you were holding in your hand. The tension in the rope would be equal to the weight of the block on the end of the rope. It is independent of the block in your hand. For example imagine if instead of 18kg, the block in your hand was 180kg. Nothing would change in the tension of the rope, it would still be the weight of the block.



Now if you held the 18kg block in your one hand and with your other hand pulled down on the 12kg block hanging from the rope, the tension in the rope would increase by the exact same amount of force that you pull down with. You pulling down on the rope is analogous to the friction between that block and the table.



So I don't know how to do this stuff but it would seem that the tension in the rope would have to be equal to the net force on the 12kg block which according to your work is 3.6N.



Hope that's right. Good luck!



Tank's answer of T=272.004N doesn't make any sense to me because:

a) F=ma. So in a frictionless situation the acceleration would be F/m. And 68N/30kg = 2.26m/s^2. So it just doesn't seem reasonable that adding friction that opposes the motion into the equation should increase the acceleration at all, let alone by a factor of 10.

b) It seems weird that there can be an internal force in the system so much larger than the external net force causing it. It would seem to me that the MAXIMUM tension possible in the rope would be 68N and that would assume that the rope is attached to a wall or some immovable (or just massive enough that a 68N force is negligible) object.

c) F=m*a so how could F also equal m*a*(friction coefficient)?



But again, i don't know for sure, but i think your work was right...

Answer 3

Equation to use :F =ma* (coefficient of friction)



F = (m1 + m2)a * (coefficient of friction)

68 = (12 + 18)a * (0.1)

a = 68 / 30(0.1) = 68/3 = 22.667 m/s^2



12kg ------T--------18kg-----68N--->



T = m1*a

T = 12 * 22.667 = 272.004N