A uniform 100-N pipe is used as lever. Where the fulcrum (the support point) must be placed if a 500-N weight at one end is to balance a 200-N weight at the other end? What is the reaction force exerted by the support on the pipe?

Answer 1

Let the fulcrum be placed at a distance x from the 500-N weight and (100 - x) from the 200-N weight. We can set up the following equation to satisfy the balance condition:

500 * x = 200 * (100 - x)

Solving for x, we get:

500x = 20000 - 200x
700x = 20000
x ≈ 28.57 cm

So, the fulcrum must be placed approximately 28.57 cm from the 500-N weight (and 71.43 cm from the 200-N weight) to balance the lever.

Now let's find the reaction force exerted by the support on the pipe. To do this, we can analyze the forces acting on the pipe:

The total weight on the pipe = 500 + 200 + 100 = 800 N

Since the pipe is in equilibrium, the reaction force (R) exerted by the support must be equal to the total weight acting on the pipe:

R = 800 N