# Dynamics of material point

1. Jun 27, 2009

### Apprentice123

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.

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

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2. Jun 27, 2009

### kNYsJakE

3. Jun 28, 2009

### Apprentice123

I'm not asking you to resolve. I would like an explanation of the exercise

4. Jun 28, 2009

### Staff: Mentor

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.

5. Jun 28, 2009

### Apprentice123

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. Jun 28, 2009

### Staff: Mentor

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?

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

7. Jun 28, 2009

### Apprentice123

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. Jul 2, 2009

### Apprentice123

I find:

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

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

How can I relate aA and aB ?

9. Jul 2, 2009

### Staff: Mentor

10. Jul 2, 2009

### Apprentice123

Thanks.
if I had two pulleys in B the accelerations were aA = -4aB ?

11. Jul 2, 2009

### Staff: Mentor

It would depend on how they are arranged.