Question

Exercise 11 A squarish object is suspended from the end of a long, uniform bar tilted at an angle 0 above the horizontal. The bar's lower end is anchored to a wall, and the upper end is attached to a horizontal cable anchored to the same wall. All parts of this system are in equilibrium. cable wall object (a) Create a complete FBD for the bar. The FBD should not show the hanging object, the wall, or the cable, just the forces acting on the bar due to these objects, horizontal (x) and vertical (y) axes, and a torque axis (dot) located at the lower end of the bar. Use Fw (force from the wall, not parallel to the bar), its components Fwx and Fw;F (cable tension), Fg (gravitational force), and Fo (force on bar from the wire suspending the object). Make the bar as long as possible, so it's easy to see where each force is applied. (b) Let the mass of the bar be 16.0 kg, the hanging object's mass be 11.0 kg, and = 37.0° Use your FBD and N2L for torques to calculate the tension force in the cable holding the bar. You won't be able to solve this using N2L for forces, going with torques is the only way. (c) Use N2L for forces to find the components of Fw and then prove that Fw is, as claimed, not parallel to the bar: calculate the direction of Fw, as an angle measured from the horizontal. Exercise 12 You're swimming in a sheltered saltwater (p= 1 023 kg/m) cove, and have a water polo ball with a mass of 436 g and a radius of 10.6 cm. a) How much force do you have to exert on the ball to keep it 80% submerged? b) Imagine you now steadily pull the ball completely beneath the surface and then all the way to the sea floor. Does the buoyant force acting on the ball increase steadily during this process? Explain without equations or math. c) The total pressure at the sea floor is 3.38 atmospheres. How deep is the water?

Answer 1

(a) The F.B.D have been drawn JO *b) Taking ton que about a we have, Assuming length of red=0 TE + TE-TE, 3 Fax & caso + Fox e Coso-F lsing → Ę - (En caso + F, COSO) /sing = Fp Coto + Fo Coto cable wall object 1908 Co+370 + 1189.8x204370 2 (a) Create a complete FBD for the bar. The FBD should not show the hanging object, the wall, or the orces acting on the bar due to these objects, horizontal (x) and vertical (y) axes, and a torque axis (dot) located at the lower end of the bar. Use Fw (force from the wall, not parallel to the bar), its components Fwx and Fwi Ft (cable tension), Fa (gravitational force), and Fo (force on bar from the wire suspending the object). Make the bar as long as possible, so it's easy to see where each force is applied. = 104.04 + 143.05 = 247.1 N Ans Ewy - F + Fu = 1679.8 + 11x 9.8 = 264.6 N fwz = F = 247.IN and Tano = 264.6 (b) Let the mass of the bar be 16.0 kg, the hanging object's mass be 11.0 kg, and @ = 37.0° Use your FBD and N2L for forgues to calculate the tension force in the cable holding the bar. You won't be able to solve this using N2L for forces, going with torques is the only way. (c) Use N2L for forces to find the components of Fw and then prove that Fw is, as claimed, not parallel to the bar: calculate the direction of Fw, as an angle measured from the horizontal twe247.1 Q = 46.95°n 47° Ans

@ V= volume of ball = 4TR = 47.3.14 (0.106) = 0.04704 m3 F I As 80% is submerged so v'= volume of ball submerged = 0.8V = 0.8 X.04704 m = 0.037632 m3 Exercise 12 You're swimming in a sheltered saltwater (p = 1 023 kg/m) cove, and have a water polo ball with a mass of 436 g and a radius of 10.6 cm. How much force do you have to exert on the ball to keep it 80% submerged? b) Imagine you now steadily pull the ball completely beneath the surface and then all the way to the sea floor. Does the buoyant force acting on the ball increase steadily during this process? Explain without equations or math. c) The total pressure at the sea floor is 3.38 atmospheres. How deep is the water? = 1023X-037632 x 9.8 = 377.3 N Wem ).436 9.8 4.3 N Now Fring = FB F : 377-3-4:3 = 373.0 N Ans Buoyant force increases steadily, be cause volume of water displaced by ball increases hence FB also increases. © © P = 3.38X 0 Ml m3 = P. + Pgh 7h = 3.38X10–16 9.8 x 1023 23.74 meter