# Blocks connected by rope, Newton's Third Law

by StephenDoty
Tags: blocks, connected, newton, rope
 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 Attached Thumbnails
 Mentor P: 40,689 To find F, consider the masses and ropes as a single system. What external forces act on the system? What's the acceleration?
 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.
Mentor
P: 40,689

## Blocks connected by rope, Newton's Third Law

 Quote by StephenDoty 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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