Missing Mass: What is the Mass of Block 2 in a System of Three Blocks?

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The discussion revolves around calculating the mass of Block 2 in a system of three blocks on a frictionless surface. Given the masses of Block 1 (12 kg) and Block 3 (35 kg), along with the tension of 111 N between Block 2 and Block 3, participants explore the application of Newton's second law. A key point raised is the need for the second known tension to solve for Block 2's mass accurately. Clarification on the forces acting on Block 3 is also emphasized, indicating that multiple forces contribute to its motion. Understanding the complete system and drawing free body diagrams are suggested as useful strategies for solving the problem.
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Homework Statement



Three blocks are pulled along a frictionless surface. The masses of two of the blocks and the tensions in two of the cords are given. Find the mass of the third block (Block 2).

Block 1 ------- Block 2------Block 3 ------>
Block 1: m1 = 12 kg
Block 2: m2 = UNKNOWN
Block 3: m3 = 35 kg
Tension between Block 2 and Block 3: 111 N

Homework Equations



Newton's 2nd Law: F = ma

The Attempt at a Solution



111N = (m1+m2)a = m3a <-- is this true?

Block 2, m2 = 35-12? This seems to easy to be true.
 
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That does not seem correct. You stated that the tensions in two cords are known, but only give the tension between Block 2 and 3. What is the other known tension?

Once you have that, draw free body diagrams, it might help to clear things up.
 
You said two cord tensions are given, but seem to have left one of them out here?

Sherrylee said:
111N = (m1+m2)a = m3a <-- is this true?

111N = (m1+m2)a is correct.
But it is not equal to m3a ... there are two forces contributing to m3's motion: the 111N and ____ ?
 

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