Jack and jill lift upward on a 1.3kg pail of water. jack exerts a force of 7.0n and jill exerts a force of 11n. jill’s force is exerted at an angle of 28º with the vertical. at what angle should jack’s force be if the pail is accelerated straight upward?

1) The minimum vertical upward force has to be slightly greater than the weight of the pail of water.

2) Weight of the pail of water = mass * g = 1.3 kg * 9.8 m/s^2 = 12.74 N

3) Jill's upward vertical force = 11.0N cos(28°) = 9.71 N

4) Jack's upward vertical force = 7.0N cos (x)

5) Net upward vertical force = weight of the pail of water

=> 9.71N + 7.0N cos(x) = 12.74N

=> cos(x) = [12.74 N - 9.71N ] / (7.0 N) = 0.4329

6) x = arc cos(0.4329) = 64.4°

64.4° with the vertical