Question

1. Florence, who weighs 560 N, stands on a bathroom scale in an elevator. What will she see the scale read when the elevator is accelerating upward at 5 m/s2?(g = 9.8 m/s2)   Newtons

2.You swing a 3.41 kg bucket of water in a vertical circle of radius 0.900 m. At the top of the circle the speed of the bucket is 3.25 m/s. Find the tension in the rope tied to the bucket at the top of the circle

Answer 1

Scale already reads (560/ 9.8) = 57.14 kg when elevator is at rest.

Now, force on the man in the upward direction from accelerating up (force E) = mass of the man x acc. of the lift in the upward direction = 57.14 x5 N = 285.7 N.

Now force of gravity (Fg) = 560 N.

Plus the normal force (n) -which keeps him falling through floor of elevator - is also acting on him = 686 N (same as gravity in up direction).

So, F(y) = Elevator (Up) 285.7 (up) + Normal (up) 560 N - Gravity (down) 560 N.

Therefore net force in the upward direction (against bottom of scale-floor) = (+385 + 686 - 686) N = 285.7 N.

Hence the reading of the scale has extra = (285.7N). Scale is calibrated for "g = 9.8 m/ss".

So, (285.7 / 9.8m/ss) kg = 29.15 kg extra. So, 29.15 kg (extra) + 57.14kg (original) = 86.3 kg. This is why you feel heavier when elevator goes up.

2)

T = m * v^2/ R - m * g
T = m (v^2 / R - g)

T = 3.41 * {(3.25^2 / 0.9) - 9.8}

T= 6.6 N

so the tension in the rope at the top of the circle is 6.6 N