Posted by Physics on Tuesday, July 14, 2009 at 1:26pm.
I've asked a similar qustion earlier and now I think I have drawn the correct free body diagram and go about solving the problem. The only thing is I do not know how to solve it.
a block (mass m1) on a smooth horizontal surface, connected by a thin cord that passes over a pulley to a second block (m2), which hangs vertically.
If m1 = 13.0 kg and m2 = 6.0 kg determine the acceleartion of each block. Ignore friction and the masses of the pulley and cord
ok well according to my free body diagram
acceleration of blcok one is equal to m1^-1 Ft = a
for block 2 acceleration = m2^-1(Ft + Fg)
I do not see how to solve for the acceleration of each block only knowing this and given only the masses of each block.
- Physics - Count Iblis, Tuesday, July 14, 2009 at 2:05pm
You have to use that the masses are connected by each other via a cord that has some fixed length. This means that the two accelerations are the same. The force exterted by the cord has an opposite sign in both cases, or you have to take the accelerations to be opposite to each other.
If you put:
Ft = m1 a
m2 a = m2 g - Ft
then you can solve for a.
- Physics - bobpursley, Tuesday, July 14, 2009 at 2:06pm
What is the net pulling force? Answer m1(g-a).
What is the acceleration of the system? Answer: netforce/mass= (m1(g-a))/(m1+M2)
Observation: the acceleration each block is the same
- Physics - Physics, Tuesday, July 14, 2009 at 3:01pm
ok how do I solve for a
m1 a - m2 g = m2 a
- QUESTION - QUESTION, Tuesday, July 14, 2009 at 3:06pm
I got this
a = m2^-1 (m1 a) -g
i'm stuck from there
- QUESTION - QUESTION, Tuesday, July 14, 2009 at 3:10pm
they cancle each other out look
(a = m2^-1 (m1 a) -g )a^-1 = 0 = m2^-1 m1 - g
0 = m2^-1 m1 - g
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