Oxford PAT 2010, Questions 12 – 16

2010_page7

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

  1. For question 15, how did u know for certain that by adding the velocities vectorially you would end up with a right angled triangle.
    I’m basically asking how you knew the direction of the planes velocity with respect to the ground.

    1. The angle between NW and N is 45 degrees. The airspeed is 141, and the wind is 100, so you can see that the angle is about 90 degrees because cos (45 degrees) is about 100/141. So this is a standard 1, 1, root 2 triangle.

      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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