To remain horizontal and in equilibrium, the tray must satisfy two relationships: the sum of the vertical forces must be zero and the sum of the torques must be zero. The force W can be calculated by multiplying the mass of the tray by the acceleration due to gravity. The magnitudes of forces P and Q can be found by solving a set of equations. To keep force P the same, the glass should be positioned directly above the point of force Q.
(a) (i) Two relationships that must be satisfied for the tray to remain horizontal and in equilibrium are: 1) The sum of the forces in the vertical direction must be zero, meaning the upward forces must balance the downward forces, and 2) The sum of the torques (or moments) about any point on the tray must be zero, meaning the clockwise torques must balance the counterclockwise torques.
(ii) To calculate the magnitude of the force W, we can use the equation:
F = m * g, where F is the force, m is the mass, and g is the acceleration due to gravity. Plugging in the values, we get:
F = (0.12 kg) * (9.8 m/s^2) = 1.176 N.
(iii) The magnitudes of forces P and Q can be calculated by setting up the equations for force balance:
Sum of vertical forces: P + Q - W = 0
Sum of torques about any point: Q * 0.10 m - P * 0.25 m = 0
Solving these equations simultaneously, we get:
P = 0.51 N
Q = 0.235 N
(b) To have the same value of force P, the glass should be positioned at a point that maintains the equilibrium of forces on the tray. This can be achieved by placing the glass directly above the point where force Q is applied.