Calculating Tension of Connected Blocks on a Horizontal Surface

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The discussion focuses on calculating the tension Tz in a system of four connected blocks on a horizontal surface, influenced by two external forces, F1 and F2. Using Newton's second law, it is established that the net force on each block must equal zero due to negligible friction. The calculations reveal that Tz is initially determined to be equal to F1 at 75.0N, but further analysis shows that Tz must also equal F2, resulting in a final value of 51.0N. The tensions in the strings connecting the blocks are derived through a series of equations based on the forces acting on each block. Ultimately, the correct tension Tz is confirmed to be 51.0N.
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Four blocks are on a horizontal surface.

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The blocks are connected by thin strings with tensions Tx, Ty, Tz. The masses of the blocks are A=44.0 kg, B=21.0 kg, C=24.0 kg, D=35.0 kg. Two forces, F1=75.0N and F2=51.0N act on the masses as shown. Assume that the friction between the masses and the surface is negligible and calculate the tension Tz.
 

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To calculate the tension Tz, we can use Newton's second law, which states that the sum of all forces acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the blocks are on a horizontal surface, so their acceleration is zero. Therefore, the sum of all forces acting on each block must also be zero.

We can start by analyzing the forces acting on block A. There are two forces acting on it: Tz pulling to the right and F1 pulling to the left. Since the net force on block A is zero, we can set up the following equation:

Tz - F1 = 0

Solving for Tz, we get Tz = F1 = 75.0N.

Next, we can look at block B. There are three forces acting on it: Tx pulling to the right, Ty pulling to the left, and F2 pulling to the right. Again, since the net force on block B is zero, we can set up the following equation:

Tx - Ty + F2 = 0

We can also use the fact that the tension in the string connecting blocks A and B is equal to Ty, so we can substitute Ty with Tz from our previous calculation. This gives us:

Tx - Tz + F2 = 0

Solving for Tx, we get Tx = Tz - F2 = 75.0N - 51.0N = 24.0N.

Now, let's look at block C. There are two forces acting on it: Tx pulling to the left and Tz pulling to the right. Again, since the net force on block C is zero, we can set up the following equation:

Tx - Tz = 0

Solving for Tz, we get Tz = Tx = 24.0N.

Finally, we can look at block D. There are two forces acting on it: Tz pulling to the left and F2 pulling to the left. Since the net force on block D is zero, we can set up the following equation:

Tz - F2 = 0

Solving for Tz, we get Tz = F2 = 51.0N.

Therefore, the tension Tz in the string connecting blocks C and D is equal to 51.0N.
 
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