A 5.2 kg box is on a frictionless 36 degree slope and is connected via a mass less string over a mass less, frictionless pulley to a hanging 2.1 kg weight.

What is the tension in the string if the 5.2 kg box is held in place, so that it cannot move?

If the 5.2 kg box is not moving, neither is the 2.1 kg weight.

The weight's weight of 2.1 * 9.8 = 20.6 N is balanced by the string tension.

To find the tension in the string when the 5.2 kg box is held in place, we can use Newton's second law of motion. Here's how to get the answer:

1. Draw a diagram: Draw a diagram illustrating the problem. Label the box as "m1" with a mass of 5.2 kg, the hanging weight as "m2" with a mass of 2.1 kg, and the tension in the string as "T".

2. Break down the forces: On the 5.2 kg box, the gravitational force acts downwards and is split into two components: one parallel to the slope (mg*sinθ) and one perpendicular to the slope (mg*cosθ). The tension in the string T acts in the opposite direction to the gravitational force parallel to the slope. On the 2.1 kg weight, there is only the downward gravitational force (mg).

3. Write the equations: Apply Newton's second law of motion in the x and y directions for both masses:
- For m1 (the 5.2 kg box):
- In the x-direction: T = m1 * g * sinθ (since the box is held in place, there is no acceleration in this direction).
- In the y-direction: mg*cosθ = m1 * g
- For m2 (the 2.1 kg weight):
- In the y-direction: mg = m2 * g

4. Substitute the values:
Substitute the known values into the equations. The acceleration due to gravity (g) is approximately 9.8 m/s^2, and the angle (θ) is 36 degrees.

- For m1:
- In the x-direction: T = (5.2 kg) * (9.8 m/s^2) * sin(36°)
- In the y-direction: (5.2 kg) * (9.8 m/s^2) * cos(36°)
- For m2:
- In the y-direction: (2.1 kg) * (9.8 m/s^2)

5. Calculate the tension:
Plug the values into the equations and calculate the tension in the string:
- T = (5.2 kg) * (9.8 m/s^2) * sin(36°)

Now, you can calculate the value of T.