A 30.0 kg irregularly shaped rock with a density of 4700 kg/m^3 rests on the bottom of a swimming pool.?

Question

What is the normal force of the pool bottom on the rock (in Newtons)?

Answer 1

N = W - B; where W = mg the weight of the rock and B = rho Vg the weight of the water displaced by the rock (aka buoyancy). rho is the water density, m is the rock mass, and V is the volume of the rock and displaced water. N is the normal force (perpendicular) on the bottom of the pool...your answer.

Thus we have N = mg - rho*V*g; rho = M/V ~ 1,000 kg/m^3 the density of fresh water at 4 deg C; where M is the mass of the displaced water of volume V, which is the volume of the rock as well. g = 9.81 m/sec^2. Rock density Rho = m/V; so that V = m/Rho; where Rho = 4700 kg/m^3 the density of the rock. Notice that rhoVg = B, which is the weight (not mass) of the displaced water and this is equal to the buoyancy acting on the rock.

Putting this all together, we have N = mg - rho*V*g = [m - rho(m/Rho)]g = [1 - (rho/Rho)]mg = [1 - (1000/4700)]30*9.81 which is approximately 150 Newtons. You can do the math for a more precise answer.

Lesson learned, the net force N on the rock is just its weight minus the weight of the displaced water. Also, had Rho <= rho rather than greater than water density, the rock would have floated because there would have been a point in sinking where N = W - B = 0; buoyancy would equal the weight of the rock.

Answer 2

the conventional tension is the tension that the pool backside applies to the rock. The forces appearing on the rock look like this: F=(mass)x(accelleration) =(tension of gravity)+(tension from the water)-(general tension of pool backside against rock) The tension of the backside of the pool against the rock is unfavorable because of the fact it rather is directed in the alternative direction because of the fact the different 2 forces. If the rock isn't shifting, then the accelleration is 0. yet whilst the accelleration is 0, then the tension is 0, because of the fact something cases 0 is 0 (so, F=ma is 0!) Then, we've this: F=0=(tension of gravity)+(tension from the water)-(general tension of pool backside against rock) utilizing a touch algebra, we see that, (general tension of pool backside against rock)=(tension of gravity)+(tension from the water) It seems that we are able to dismiss the tension of the water, so then we get: (general tension of pool backside against rock)=(tension of gravity) that's easy adequate to remedy, because of the fact that all of us be conscious of that the tension of gravity is comparable to (mass)x9.81m/s^2. So, then we get: (general tension of pool backside against rock)=(tension of gravity) =mg=(20kg)(9.81m/s^2)=196.2N the rationalization that we are able to dismiss the tension of all that water improve right into a touch puzzling for me to be sure on the initiating (it somewhat is been a jointly as because of the fact that i've got studied classical physics - i'm a physics substantial, and all my training as of previous due have been on quantum mechanics) yet then i found out something - we are all immersed in a fluid our entire lives - i.e. the ambience. because of the fact that we don't might desire to take the load of the ambience into consideration in physics questions, we additionally do no longer might desire to evaluate the load of the water in that undertaking.