# Three blocks and two strings

Three objects are connected on a table as shown in the http://www.cybertrails.com/~reedaz/pic1.JPG [Broken]. The table has a coefficient of kinetic friction of 0.350. The objects have masses of 4.00 kg, 1.00 kg, and 2.00 kg. The pulleys are frictionless. Determine the acceleration of each object and their directions. Determine the tension in the two cords.

I begin solving this by drawing a free body diagram for the center object that is on the table. I find the forces acting in the x direction are T1 = -39.2N, T2 = 19.6N, and Fk = (0.350)9.8 = 3.43N.

Sum F = -39.2 + 19.6 +3.43 = ma
m = 1.00 kg
so a = -16.17 m/s^2

This answer of a = -16.17 m/s^2 does not seem reasonable because it is greater than freefall acceleration. Where did I go wrong?

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The whole system must accelerate at the same rate or funky things happen. So when you calculated ma, you should have used the mass of the entire system.

Thanks zwtipp05, the answer I get seems more reasonable now.

Steve

I'm still uncertain how to calculate the tension of the cords in a problem like this. I know that if there was no motion that one cord would have a tension of 39.2N and the other of 19.6N. How do I calculate the tension when the system is accelerating?

Steve

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The system is accelerating, but within the system, the block is not accelerating with respect to the ropes.

Thanks again zwtipp05.

Steve