A machine is lifted from the floor of a room using two ropes. The two ropes ensure that the horizontal components of the forces are balanced at all times. It is assumed that at all times the machine moves vertically upwards at a constant velocity.
The machine is located in a room with height `h` metres.
One of the ropes is attached to the point `P` on the machine and to the fixed point `C` on the ceiling of the room. The point `C` is a distance `d` metres to the left of `P`. Let the vertical distance from `P` to the ceiling be `ℓ` metres and let `\theta` be the angle this rope makes with the horizontal.
The other rope is attached to the point `P` and to the fixed point `F` on the floor of the room. The point `F` is a distance `2 d` metres to the right of `P`. Let `\phi` be the angle this rope makes with the horizontal.
Let the tension in the first rope be `T_1` newtons, the tension in the second rope be `T_2` newtons, the mass of the machine be `M` kilograms and the acceleration due to gravity be `g\ text{m s}^(-2)`.
- By considering horizontal and vertical components of the forces at `P`, show that
- `tan theta=tan phi+(Mg)/(T_(2)cos phi)` (3 marks)
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- Hence, or otherwise, show that the point `P` cannot be lifted to a position `{2 h}/{3}` metres above the floor. (2 marks)
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