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A uniform rod of weight 10N is balance at a point 75cm from the end B the pivot is removed to point 30cm from A. What force must be applied at A to balance the rod horizontally?

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The fulcrum F is moved so that AF:FB = 1:4
So, the mass of AB is 1/4 the mass of FB
Now, since we have AF=30 and FB=120,
So, 8*30 = 2*120

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