6 Replies
@Gyro Gearloose
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.The first one will show more weight ig. Look at the container. In the first one, the only force exerted by container is (M+m)g where m and M are masses of container and water respectively. In the second one, the force exerted by the container is however (M+m)g-Tension due to string so weight read is lesser.
If the string is cut then the weight becomes same in both as no tension force is there.
uh can you explain this using buoyant force and normal reaction?
Basically just look at the container. For the first one, the container is acted upon by force of weight of water and the mass suspended. The buoyant force for the system is an internal force and the tension applies a force of mg-rho.V.g So, the net force downward(=normal) on the container 1 will be rho.V.g+m'g, where V is volume of the water+ball system and m' is mass of container
For the second part the tension here applies an upward force of rho.V.g-mg. So here, the force instead is mass of water-rho.V.g+mg+m'g
Or a rather simpler way to see it: level of water in both is the same so pressure is same at bottom of both containers, so P*area is same. But in the second one tension tries to pull up the container so weight will be less.
hmm a lot better
cuz buoyant force is due to pressure difference ultimately