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anonymous
 4 years ago
a wooden block of mass 8 kg is tied to a string attatched to the bottom of the tank.In equilibrium the block is completely immersed in water. if relative density of wood is 0.8 and g=10m/s^2 then what is the value of tension in the string?
anonymous
 4 years ago
a wooden block of mass 8 kg is tied to a string attatched to the bottom of the tank.In equilibrium the block is completely immersed in water. if relative density of wood is 0.8 and g=10m/s^2 then what is the value of tension in the string?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[F _{g}= Mg = 8kg x 10m/s² = 80N\] that is the gravitational force which pulling the block down. bouncy force = the total weight of water displaced by the block, since u say that the wooden block is totally immersed into the water, The volume of water displaced = the volume of wooden block. We need the volume of wooden block, but we only have density and mass. \[Density = m/v\] \[v = m \times v/m\] so with tat equation, u can cross the m and u will get the v, volume. \[v = m \times v/m\] \[m \div m/v = m \times v/m = v\] then v= 10, idk what is the unit since u didn't state the unite for the density. now, volume of water displaced also = 10 and i guess the density of water is 1.0 \[v \times m/v = m\] \[10 \times 1.0=10\] mass of water displaced. \[F _{bouyancy} = 10 \times 10 =100N\] resultant force on the string(tension), lets consider upward force is positve , negaative is for downward force, 100N80N = 20N 20N of tension

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0as the relative density of wood is less than 1( relative density or water ), the water lifts the wooden block up to the surface. Hence there wont't be any tension in the string.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1http://openstudy.com/users/sachin%20vishaul#/updates/4f1ec1f6e4b04992dd24cb4b From the last time we worked this problem, there are three forces acting on the block of wood F_g = force of gravity (down) F_b = force of buoyancy (up) F_s = force from the string (down) The block is in equilibrium hence the sum of the three is zero. Use that to solve for F_s. The magnitude of F_s is the tension in the string. We have found the mass of water displaced is 10 kg, hence \[ F_b = +10g \] On the other hand the weight of the block is \[ F_g = 8g \] As \[ F_b + F_g + F_s = 0 \] we have \[ F_s = (F_b + F_g) = (10g  8g) = 2g =  19.6 \ N \] This is a downward force and that makes complete intuitive sense. We know the block is less dense that the water so without the string it would rise to the top of the water.
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