2. A uniform rod of weight 5N and length 1m is pivoted at a point 20cm from one of its ends. A weight is hung from the other end so that the rod balances horizontally. What is the value of the weight?

The weight is 7.5 N

Explanation:

Since the rod is uniform, its weight applies downwards at its middle (i.e 50 cm mark). The diagram attached to this answer shows the appropriate sketch.

Applying the principle of moment to the question, let the weight by represented by W.

Total clockwise moment = Total counterclockwise moment

x = x

5 x 0.3 = W x 0.2

1.5 = 0.2 W

W =

   = 7.5 N

The value of the weight is 7.5 N.

since you're a real machine, your ama is always less than your ima, because you can never totally eliminate friction.

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