Why Does the Tension in the Lower Rope Not Depend on Mass M1?

  • Thread starter Thread starter tizzful
  • Start date Start date
  • Tags Tags
    Tension
Click For Summary
The discussion revolves around understanding the tension in two ropes supporting two hanging masses, M1 and M2, during upward acceleration. It is clarified that the tension in the lower rope (T2) only depends on the weight of M2, hence does not involve M1. For the upper rope (T1), the correct approach involves applying Newton's second law to account for the forces acting on M1, including its weight and the tension from the lower rope. The participants emphasize the importance of identifying all forces acting on each mass to derive the correct tension equations. The conversation highlights the distinction between equilibrium conditions and scenarios involving acceleration.
tizzful
Messages
14
Reaction score
0
Two Hanging Masses (TENSION) :)

Homework Statement


Two blocks with masses M1 and M2 hag one under the other. For this problem take the positive direction to be upward and use g for the magnitude of the acceleration due to gravity. The blocks are now accelerating upwads (due to the tension in the strings) with acceleration of magnitude a. find the tension in the lower and upper rope.
Tension.jpg


Homework Equations


F=ma
W=mg


The Attempt at a Solution



I found the tension in the lower and upper rope when it was stationary and therefore in equilibrium.
T2=M2g
T1=M1g
Then to find the tension when accelerating I just used
T2=Ma+Mg
T1=Ma+M1g+M2g

For some reason it says that the tension in rope 2 does not depend on the variable M1... So I'm just really lost now on which direction to take..
Thanks in advance
 
Physics news on Phys.org
Welcome to PF>

While acceleration T2 is the normal force arising only due to the weight of M2.
Hence T2 does not depend on variable M1.
 
Oh thank you! That clears up the first one but I'm still stuck on the T1 when accelerating. Would I just go T1=(M1+M2)a + (M1+M2)g?
 
Yes becoz the total downward force acting on T1 while acceleration is the sum of the weights of M1 and M2.
 
tizzful said:
I found the tension in the lower and upper rope when it was stationary and therefore in equilibrium.
T2=M2g
T1=M1g
That would be correct for M2 but not for T1 (three forces act on M1). In any case, the equilibrium situation is not relevant to this problem.
Then to find the tension when accelerating I just used
T2=Ma+Mg
T1=Ma+M1g+M2g
Carefully apply Newton's 2nd law to each mass. Start by identifying all the forces that act on each mass. (Hint: three forces act on M1.)

(The diagram is misleading because it does not show all the tension forces acting on the masses.)
 


first, solve for the T2.
T2=M1(g)
then, for T1,
since T1 is carrying the M2,
ΣF=ma
substitute ΣF to T2 + M1(g) -T1
so,
T2 + M1(g) -T1= ma (at rest)
T2 + M1(g) -T1= 0
T1=T2 + M1(g)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
16
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 33 ·
2
Replies
33
Views
3K
  • · Replies 7 ·
Replies
7
Views
8K
Replies
11
Views
3K