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Fundamentals Of Physics
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Short Answer

The tension in a string holding a solid block below the surface of a liquid (of density greater than the block) is T0 when the container (Fig. 14-57) is at rest. When the container is given an upward acceleration of 0.250g, what multiple of T0 gives the tension in the string?

The tension in the string is 1.25T5.

See the step by step solution

Step by Step Solution

Listing the given quantities

The tension in the string when the container is at rest T0.

Understanding the concept of pressure

Newton’s second law states that the net force F on a body with mass m is related to the body’s acceleration a by,

F=ma

Using the condition for equilibrium Fnet =0,we can write an equation for tension T0 in the string in terms of weight and buoyancy force Fb when the container is at rest. Then we can write the force equation using Newton’s second law for a block moving up with acceleration a. From these two equations, we can write T in terms of T0

Formulae:

At equilibrium, Fnet =0

Force of buoyancy Fb=ρlVg

Calculating the tension in the string

Free body diagram when the container is not accelerating:

When the container is at rest, the block is at equilibrium.

At equilibrium,

Fnet =0

From Free body diagram,

FbmgT0=0ρVgmgT0=0T0=ρVgmg

Free body diagram when container is accelerating:

Applying Newton’s law, we get

FbmgT=maρV(g+a)mgT=maρVg+ρVamgT=ma

Putting ρVgmg=T0 from equation (i) in it, we get

T0+ρVaT=maT=T0+ρVama=T0+(ρVm)a

So,

T=T0+ag(ρVgmg)=T0+agT0=T01+ag For a=0.250gTT01+0.250gg=1.25T0

The tension in the string is 1.25T0.

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