Question

Can anyone help me do the magdeburg hemisphere question

In 1654, the German scientist and inventor Otto von Guericke found that if you make a sphere out of two brass hemispheres and create a near-vacuum inside (p_w very small), it is extremely hard to pull apart the hemispheres. (Von Guericke famously demonstrated that two teams of eight horses couldn't do it!) Assuming that the hemispheres have very thin walls, so that R in the figure may be considered both the inside and outside radius, let's find out how much force is necessary. a) Explain why the hemispheres stick together in the first place. (A short sentence or two is fine.) b) In what direction does the force act that holds the hemispheres together? c) For simplicity, we'll assume that the left side of the sphere is attached to a wall and that we apply a force F horizontally to the right as shown in the picture above. This force has to counteract the total leftward horizontal component of the force you described in b). That means we have to add up (integrate!) those components over the right hemisphere. Define a vertical ring on the sphere's surface, located at some angle theta (measured from the central plane) and having some small angular d theta. What is the outer surface area dA of this ring? d) Write an expression for the magnitude of the horizontal force dF_x, that acts on the ring you defined in part b). c) Integrate your answer for c) over the appropriate range of values to get the total horizontal force we must counteract to pull the pieces apart. f) Taking R as 34 cm, the inside pressure as 0.12 atm, and the outside pressure as 1.0 aim. find the force needed to pull apart the hemispheres. Express your answer in Newtons and in pounds.

Answer 1

Hemisphere are stick together due to pressure difference between inside and outside of hemisphere. Inside the hemisphere, pressure is very low(vaccum) and outside atmospheric pressure is there. The force is perpendicular to the surfaces of hemispherehorizontal component of this force holds them together. Radius of rings r = R sin theta Outer surface area = (2 pi r) times height = 2 pi R sin theta times R d theta = 2 pi R^2 sin theta d theta dF_x = df cos theta = (P_out-P_in)cos theta(2 pi R^2 sin theta d theta) F_x = (P_out - P_in) pi R^2 integral^pi/2_0 sin 2 theta d theta = (P_out-P_in)pi R^2[-1/2 cos 2 theta]^pi/2_0 =(P_out-P_in)pi^2 [(-1)-1] times (-1/2) = (P_out -p_in) times pi R^2