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Handout problem solutions:

4.
a) The FBD has buoyancy force upwards, and weight (mg) downwards.
b) Fb = rho*g*Vsub where Vsub is the volume of the ball. Since the volume of a sphere is (4/3)*pi*R^3, and we know the density of water, we can calculate that the buoyancy force is
410.5 N 41.1 N.
c) Using the same method, but the density of the ball (800 kg/m^3) instead of water, we get that the mass is 33.5kg 3.35 kg.
d) Fnet in y = Fb - mg = ma, so a = (Fb - mg)/m = 2.454 m/s^2

5.
a) tray is in equilibrium, mg = rho*g*Vsub
Vsub = area of tray A*depth d, where A = L*w

mg = rho*g*L*w*d, m = 0.3 kg

b) Fbmax = rho*L*W*h = 23.52 N

c) If the tray is just on the edge of sinking, then Fbmax - (m_total)*g = 0
m_total = m_tray + m_weights

Solving, m_weights = 2.1 kg

Dividing by 0.005 kg per weight, this means there are 420 weights that could be added.

6.
a) Gauge pressure = 8 atm * (101.3 kPa/atm) = 8.103 x 10⁵ Pa

b) Gauge pressure = P - Patm = rho*g*h = 8.103 x 10⁵ Pa

c) F = P*A = 1.61 x 10⁶ N directed upwards

d) Same formula, F = P*A = 1.621 x 10⁶ N directed downwards

e) The difference between the two forces is 10,460 N directed downwards, so there needs to be an upwards force of 10,460 N to hold the window in place.

7.
a) The mercury will flow into the water since the pressure at the bottom is so much greater than the pressure of the water (P = P0 + rho*g*h, rho of mercury is much greater.)

b) When the mercury drops, we can say the distance it drops is H.

The height of the mercury column is then (1.0 m - H). Since the volume is conserved, the height of the water column is (1.0 m + H).

rho_Water*g*(1.0 m + H) = rho_Mercury*g*(1.0 m - H)

Solving for H, H = 1.863 m.