What is the acceleration and tension in this two-block system?

In summary, the two blocks shown in the figure are initially at rest. Despise the masses of the pulleys and the friction between different parts of the system. Get (a) the acceleration of each block and (b) the tension in cable.
  • #1
Apprentice123
92
0
The two blocks shown in the figure are initially at rest. Despise the masses of the pulleys and the friction between different parts of the system. Get (a) the acceleration of each block and (b) the tension in cable.

Answer:
(a) aA = 2,56 m/s^2; aB = 1,28 m/s^2
(b) 677 N
 

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  • #2
Apprentice123 said:
The two blocks shown in the figure are initially at rest. Despise the masses of the pulleys and the friction between different parts of the system. Get (a) the acceleration of each block and (b) the tension in cable.

Answer:
(a) aA = 2,56 m/s^2; aB = 1,28 m/s^2
(b) 677 N

People can't help you unless you post your attempted solutions. It's a place where people help you on homework, not do your homework for you.
 
  • #3
kNYsJakE said:
People can't help you unless you post your attempted solutions. It's a place where people help you on homework, not do your homework for you.

I'm not asking you to resolve. I would like an explanation of the exercise
 
  • #4
Apprentice123 said:
Despise the masses of the pulleys and the friction between different parts of the system.
I presume that you are to treat the pulleys as massless and the surfaces as frictionless.
Apprentice123 said:
I would like an explanation of the exercise
What don't you understand?

Hints: Identify the forces acting on each mass. (Draw a free body diagram for each.) Apply Newton's laws to each mass.
 
  • #5
Doc Al said:
I presume that you are to treat the pulleys as massless and the surfaces as frictionless.

What don't you understand?

Hints: Identify the forces acting on each mass. (Draw a free body diagram for each.) Apply Newton's laws to each mass.

My solution (not find the answer)

Block A

ZFx = m.a
T1 - Psin(30) = mA . aA


Block B
zFy = m.a
-P + T1 + T2 = mB . aB

And:
T2 = 2T1
2aB = aA


Solving the system. I find:
aA = 1,14 m/s^2
aB = 0,57 m/s^2
T2 = 1097,92 N
 
  • #6
Apprentice123 said:
Block A

ZFx = m.a
T1 - Psin(30) = mA . aA
OK, I presume P is the weight of block A. Better to use mAg instead of P.

Block B
zFy = m.a
-P + T1 + T2 = mB . aB
Don't use the same symbol, P, to represent the weight of block B. Also, since the pulleys are massless and frictionless, the rope has a single tension throughout. Just call it T. So what's the total upward force on block B?

And:
T2 = 2T1
See comments above.
2aB = aA
Be careful with signs. If block A moves up the incline, block B must move down.
 
  • #7
Doc Al said:
OK, I presume P is the weight of block A. Better to use mAg instead of P.


Don't use the same symbol, P, to represent the weight of block B. Also, since the pulleys are massless and frictionless, the rope has a single tension throughout. Just call it T. So what's the total upward force on block B?


See comments above.

Be careful with signs. If block A moves up the incline, block B must move down.

Block A

ZFx = mA . aA
T - mA . g . sin(30) = mA . aA


Block B

ZFy = mB . aB
-mB.g + 2T = mB . aB


aA = - 2aB


New is correct thank you very much
 
  • #8
I find:

Block A
T - Psin(30) = 90,72aA

Block B
-2T + P = 158,7aB


How can I relate aA and aB ?
 
  • #9
Apprentice123 said:
How can I relate aA and aB ?
See your last post:
Apprentice123 said:
aA = - 2aB
 
  • #10
Doc Al said:
See your last post:

Thanks.
if I had two pulleys in B the accelerations were aA = -4aB ?
 
  • #11
Apprentice123 said:
if I had two pulleys in B the accelerations were aA = -4aB ?
It would depend on how they are arranged.
 

FAQ: What is the acceleration and tension in this two-block system?

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A material point is a theoretical concept used in physics to represent a small, infinitesimal particle with mass but no size. It is often used to simplify the analysis of larger, more complex systems.

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