Oxford PAT 2010, Questions 12 – 16

2010_page7

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6 thoughts on “Oxford PAT 2010, Questions 12 – 16

      1. I thought to find the force around the pivot for a lever it is F x d? Therefore 1000N multiplied by 1.5m for this lever should be 1500N. I am aware I am very wrong, but why am I wrong?

        1. To solve this you need to understand where the pivot (aka fulcrum) is. With any lever, the pivot is the bit that doesn’t move when you use the lever.

          In this case, you’ve got a plank being lifted at one end, with a weight somewhere along it. So the other end of the plank is going to be stuck to the ground. This means that the other end of the plank is the pivot.

          So the moment of the weight of the concrete mass is 100 x 10 x 0.5 (the mass, times g, times the distance to the pivot). And the moment of the force that the builder applies is F x 2 (the force F, times the distance to the pivot). These moments must balance, therefore F = 100 x 10 x 0.5 / 2, which equals 250N.

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