CP CALC A deep-sea diver is suspended beneath the sur-face of Loch Ness by a 100-m-long cable that is attached to a boat on the surface (?Fig. P15.77?). The diver and his suit have a total mass of 120 kg and a volume of 0.0800 m3. The cable has a diameter of 2.00 cm and a linear mass density of µ = 1.10 kg/m. The diver thinks he sees something moving in the murky depths and jerks the end of the cable back and forth to send transverse waves up the cable as a signal to his companions in the boat. (a) What is the tension in the cable at its lower end, where it is attached to the diver? Do not forget to include the buoyant force that the water (density 1000 kg/m3) exerts on him. (b) Calculate the tension in the cable a distance x above the diver. In your calculation, include the buoyant force on the cable. (c) The speed of transverse waves on the cable is given by (Eq. 15.14). The speed there-fore varies along the cable, since the tension is not constant. (This expression ignores the damping force that the water exerts on the moving cable.) Integrate to find the time required for the first signal to reach the surface.
Solution 84CP Step 1 of 10: Given data, Density of water,w = 1000kg/m3 Volume of diver, V=0.08 m3 Mass of the diver, M=120 kg Acceleration due to gravity, g=9.8 m/s Mass per unit length of cable, =1.1 kg/m Diameter of cable, d=2 cm=0.02 m Using radius r=(d/2) Radius of cable , r=0.01 m Step 2 of 10: To calculate the tension in the cable at the diver end, To draw the free body diagram(FBD) for the diver inside water, Upward force called buoyant force acts on the diver which is proportional to the volume of object immersed in liquid as here the diver is immersed in water, which is given by F = gwV B Where dwnsity of water, g is acceleration due to gravity and V is the volume of the object immersed(volume of diver) Substituting w= 1000kg/m ,g=9.8 m/s and V=0.08 m 3 F = (1000kg/m )(9.8 m/s ) (0.08 m )3 B F B784 N Step 3 of 10: Downward force acts on the diver due to mass, that is weight W= Mg Substituting M=120 kg ,g=9.8 m/s 2 W= 120 kg(9.8 m/s ) W=1176 N Therefore, the resultant force due to this two forces will act on the diver due to cable, That is, F= W-F =1B76 N-784 N F= 392 N Therefore, the tension force on the cable at the diver end is 392 N. Step 4 of 10: Now if we consider a section of the cable of length x starting with x=0 at diver end as shown in the figure below, Weight which acts downward on the cable at distance x, W=mg m Using = x as m= x W= x g 2 Substituting g=9.8 m/s and =1.1 kg/m W=(1.1 kg/m)(9.8 m/s ) x W=10.78 x N Step 5 of 10: Buoyant force which acts upward due to volume of cable immersed in water, FB= w V Where dewsity of water, g is acceleration due to gravity and V is the volume of the object immersed(volume of cable) 2 Using V=r x F B gwr x 2 Substituting = 1000kg/m ,g=9.8 m/s and r=0.01 m w 3 2 2 F B (1000kg/m ) (9.8 m/s )(0.01 m) x F B3.079 x N