In the diagram shown, the bblocks A and B are connected by light string and are at rest. If WA= 10N then find the tension in the string and WB

To solve the problem involving blocks A and B connected by a light string, we need to analyze the forces acting on both blocks. Given that block A has a weight (WA) of 10 N and the system is at rest, we can infer that the tension in the string and the weight of block B (WB) must be in equilibrium. Let's break this down step by step.

Understanding the Forces

When the system is at rest, the forces acting on each block must balance out. For block A, the only forces acting on it are its weight and the tension in the string. Similarly, for block B, the forces acting on it are its weight and the tension in the string pulling it upwards.

For Block A

Block A is hanging, and its weight (WA) acts downward. The tension (T) in the string acts upward. Since the block is at rest, we can write the equation of equilibrium:

• WA = T

Substituting the given weight of block A:

• 10 N = T

This tells us that the tension in the string (T) is 10 N.

For Block B

Now, let's analyze block B. The weight of block B (WB) acts downward, and the tension (T) in the string acts upward. Since the system is still at rest, we can write a similar equation for block B:

• WB = T

Since we already found that the tension T is 10 N, we can substitute this value into the equation:

• WB = 10 N

This means that the weight of block B is also 10 N.

Final Results

In summary, we have determined that:

• The tension in the string (T) is 10 N.
• The weight of block B (WB) is also 10 N.

This example illustrates the principle of equilibrium in a system of connected objects. When the forces are balanced, we can easily find the tension in the string and the weights of the blocks involved. If you have any further questions or need clarification on any part of this explanation, feel free to ask!