Answer to Question #83865 in Mechanics | Relativity for Hardy

Question

1.A piece of yellow pine wood (density = 650 kg/m3) is 5 cm by 5 cm square
and 2.2 m long. How many Newtons (N) of lead (density = 11,400 kg/m3)
should be attached to the bottom of the wood so that it will float vertically with 30 cm out of the water

2. A raft is made by tying together ten 4-m lengths of bamboo, each of 8 cm in diameter. Determine the maximum load that the raft can carry. You may assume that the bamboo is just immersed below the water surface
The density of bamboo may be taken as 232.3 kg/m3.

3.A rectangular gate 20 m wide is designed to separate water and another fluid. Given that the density of water and the fluid is 1000 kg/m^3 and 800 kg/m^3 respectively. The heights of water and the fluid retained by the gate is 10 m and 15 m respectively, calculate the resultant force acting on the gate.

Answer 1

1. Buoyancy pushes the wood out of the water as well as the lead, but the force of gravity pulls them down. So all we need is to calculate the balance between these forces:

For wood, where a, b, c - sides, l - length under water, r1 - density of the wood, r0 - density of water:

$$Fbw=ablgr0,$$ $$l=c-0.3,$$ $$Fgw=abcr1g.$$

These forces are unbalanced for

$$F=Fbw-Fgw=abg(r0l-cr1)$$

So the difference between buoyancy and force of gravity for the lead must be equal to this force

$$F$$

, where

$$r2$$

- density of lead:

$$F=Vgr2-Vgr0.$$

Thus volume of the lead:

$$V=abg(r0l-cr1)/(g(r2-r0)),$$

We must attach

$$P$$

newtons of lead:

$$P=Vgj=jabg(r0l-cr1)/(r2-r0)=12.6$$

N

2. As in the previous problem, but

$$n$$

is the quantity.

$$F=n*3.14d^2/4*l(r0-r1)*g$$ $$F=10*3.14*0.08^2/4*4(1000-232.3)*9.8=1511.9$$

or 154.3 kg.

3. This problem requires using integrals since the pressure is changing with changing height. Take very small height

$$dh$$

where the pressure

$$p$$

can be assumed constant. The pressure on the wall of breadth

$$b$$

will cause a force

$$dF=p*dh*b=r*g*b*h*dh,$$

integrate it:

$$F=gbr*H^2/2$$

Difference of these forces for different liquids is

$$Fd=gb(r1*H^2/2-r2*h^2/2)$$ $$Fd=9.8*20*(800*15^2/2-1000*10^2/2)=7840000$$