Blocks connected by rope, Newton's Third Law


by StephenDoty
Tags: blocks, connected, newton, rope
StephenDoty
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#1
Mar14-08, 08:10 AM
P: 267
The figure shows two 1.0 kg blocks connected by a rope. A second rope hangs beneath the lower block. Both ropes have a mass of 250 g. The entire assembly is accelerated upward at by force. What is F? What is the tension at the top end of rope 1? What is the tension at the bottom end of rope 1? What is the tension at the top end of rope 2?


The force on block B:
T-(m of B)g=(m of B)a

The force on block A:
F-[(m of a + m of b)g] = (m of b)a

What do I do now?
Thank you.
Stephen
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Doc Al
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#2
Mar14-08, 08:20 AM
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To find F, consider the masses and ropes as a single system. What external forces act on the system? What's the acceleration?
StephenDoty
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#3
Mar14-08, 11:03 AM
P: 267
The figure shows two 1.0 kg blocks connected by a rope. A second rope hangs beneath the lower block. Both ropes have a mass of 250 g. The entire assembly is accelerated upward at by force. What is F? What is the tension at the top end of rope 1? What is the tension at the bottom end of rope 1? What is the tension at the top end of rope 2?


If we make it one system the external forces are: F and weight of the blocks
So F= (m of a + m of b + 2*m of rope)a
and w=(m of a + m of b + 2*m of rope)g

since F-w=(m of a + m of b +2*m of rope)a, right??
Now what??

Thank you.

Doc Al
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#4
Mar14-08, 11:26 AM
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Blocks connected by rope, Newton's Third Law


Quote Quote by StephenDoty View Post
If we make it one system the external forces are: F and weight of the blocks
Right.
So F= (m of a + m of b + 2*m of rope)a
No. (The net force will equal M*a, but that comes later.)
and w=(m of a + m of b + 2*m of rope)g
Right.
since F-w=(m of a + m of b +2*m of rope)a, right??
Right!
Now what??
What's the acceleration? That must be given. Use it to solve for F.


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