Question

A rock of mass M with a density twice that of water is sitting on the bottom of an aquarium tank filled with water. The normal force exerted on the rock by the bottom of the tank is:________.a. Mg/2b. 2Mgc. zerod. impossible to determine from the information givene. Mg

Answer 1

Answer:

a) Mg/2

Explanation:

• if the rock is sitting on the bottom of the aquarium, this means that it is not accelerated in the vertical direction.
• In this direction , we have three forces acting on the rock:
• 1) Gravity, going downward.
• 2) Buoyant Force, going upward.
• 3) Normal Force, going upward.

• The gravity force, can be expressed in terms of the mass as the product of the density times volume, as follows:

       $F_{g} = \delta_{r} *V*g (1)$

• The buoyant force is equal to the weight of the water displaced by the rock, which can be expressed also in terms of the density, as follows:

       $F_{b} = \delta_{w} *V*g (2)$

• From the givens, we know that δr = 2*δw (3)
• Now, applying Newton's 2nd law in the vertical direction, as net force is zero, we have:

       $F_{g} =F_{b} + F_{n} (4)$

• Due to the normal force can take any value, we can write (4) as follows:

       $F_{n} = F_{g} -F_{b} (5)$

• Replacing Fg and Fb by (1) and(2), applying the condition (3), we get:

       $F_{n} =2* \delta_{w} *V*g - \delta_{w} *V*g = \delta_{w} *V*g (6)$

⇒    Fn = (m*g)/2