Question

A container partially filled with water is moved horizontally with acceleration a=g3 . A small wooden ball of mass m is tied to the bottom of the container using a string. The ball remains inside the water with the string inclined at an angle θ to the horizontal. Assuming that the density of wood is half the density of water, and if the tension in the string is xmg, find x.

(Answer upto two digit after the decimal point)

(Answer upto two digit after the decimal point)

Solution

As the container is accelerating the liquid surface will be inclined such that

tanα=ag=13

Hence sinα=1√10 and cosα=3√10

In this case the buoyant force on the ball will be normal to water surface. The magnitude of buoyant force

B=Vdisρwgeff

B=mρρw√g2+a2=2m√g2+g29

which gives B=2√103mg

The force acting on the ball are shown in FBD (w.r.t container), then

B=T+mg cosα+ma sinα

2√103mg=T+mg.3√10+mg3.1√10

T=√103mg

tanα=ag=13

Hence sinα=1√10 and cosα=3√10

In this case the buoyant force on the ball will be normal to water surface. The magnitude of buoyant force

B=Vdisρwgeff

B=mρρw√g2+a2=2m√g2+g29

which gives B=2√103mg

The force acting on the ball are shown in FBD (w.r.t container), then

B=T+mg cosα+ma sinα

2√103mg=T+mg.3√10+mg3.1√10

T=√103mg

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