Two blocks, A and B, of the masses 3,3kg and 2,1kg respectively, are joined by an inextensible string. The string is threaded through a frictionless pulley allowing block B to be suspended by it. The coefficient between block A and the surface of the table is 0,1.

Question

Figure 6: Diagram showing two blocks connected by a string.
Calculate the tension in the string that holds the blocks together.

Answer 1

To calculate the tension in the string, we need to consider the forces acting on each block.

For block A:
- Weight (mg) = 3.3 kg * 9.81 m/s^2 = 32.433 N (downward)
- Normal force (N) = mg = 3.3 kg * 9.81 m/s^2 = 32.433 N (upward)
- Frictional force (Ffr) = coefficient of friction * normal force = 0.1 * 32.433 N = 3.2433 N (opposite to the direction of motion)
- Net force (F) = T - Ffr = T - 3.2433 N

For block B:
- Weight (mg) = 2.1 kg * 9.81 m/s^2 = 20.661 N (downward)

Since the blocks are connected by a string, the tension in the string is the same for both blocks. Therefore, T is the tension in the string.

Now, considering the system as a whole:
- The acceleration of the system (a) will be the same for both blocks since they are connected by the string.
- For block A: T - Ffr = ma
- For block B: T = mba

From the diagram and the given information, it is clear that block B is free to move vertically downwards. Therefore, the net force acting on block B is its weight.

So, the equation for block A becomes:
T - 3.2433 N = 3.3 kg * a

The equation for block B becomes:
T = 2.1 kg * a

Equating T in both equations:
2.1 kg * a = 3.3 kg * a + 3.2433 N
2.1a = 3.3a + 3.2433
3.3a - 2.1a = 3.2433
1.2a = 3.2433
a = 3.2433 / 1.2
a = 2.70275 m/s^2

Now, substitute the value of a back into the equation for block A:
T - 3.2433 N = 3.3 kg * 2.70275 m/s^2
T - 3.2433 N = 8.924725 N
T = 8.924725 N + 3.2433 N
T = 12.168 Na

Therefore, the tension in the string that holds the blocks together is 12.168 N.