Long Questions 11 10pts3pts total One sunny day you went to

Long Questions 11. (10pts+3pts total) One sunny day, you went to the Oregon coast with your friend to enjoy some laid-back time on the beach. As always ®, you brought your chainsaw with you. While wandering at the beach, you stumbled upon a pine log with a diameter of 20 inches. You want to cut a piece of log that is just long enough to carry the two of you on the water, saving you lots of labor to roll the log to the water. Let\'s assume that your mass is 55 kg and the mass of your friend is also 55 kg. The density of the log is 0.50 × 103 kg/m3. The density of the ocean water is 1.OX 103kg/m3· inch = 2.54 cm. a. The shortest piece of log you need will carry you and your friend in a way where the whole piece of log is submerged in ocean and two of you sit right on the surface of the ocean. What is this shortest length of log? (6pts+3pts)

Solution

We need to keep the following in consideration that for a body submerged in water, the force acting on it is equal to the weight of the water displaced by it.

a.) We will use the above to determine the volume (and hence length of log) to keep the two men afloat.

Now, let the volume of the log be V m^3. As the log is completely submerged, the net weight of water displaced by it will be given as:

V(1000) = 110 + 500V

or V = 110 / 500 = 0.22 m^3

Now, let the required length of log be x m

that is, 0.254^2 x = 0.22

or x = 1.08599 metres is the required length of the log

b.) If the log itself is left in the water, it will remain afloat. We will again balance the buoyant force with its weight to determine the percentage of volume above the ocean

Let the u m^3 be submerged in water.

hence 1000 u = 500 V

or u/V = 1/2

Therefore, 50% of the log will remain above the water.

NOTE: As for part b, you can answer such questions even without any calculations. The ratio in which the volume is submerged and remains above is same as the ratio of the densities of the body to that of water.