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Problem 4.25

4.25. a)

Since there is no net acceleration, the sum of the forces in both the x and the y directions must be zero.

tex2html_wrap_inline383

tex2html_wrap_inline385

Solving for T in the equation for y and plugging that into the equation for x, we find that

tex2html_wrap_inline387

tex2html_wrap_inline389

b)

Now, the mass of the bucket has increased to tex2html_wrap_inline391 . However, the drag force tex2html_wrap_inline393 is not dependent on the mass of the bucket, only the surface area exposed to the air, which hasn't changed, therefore we can rewrite our forces in the x and y directions, but keep tex2html_wrap_inline393 from part a.

tex2html_wrap_inline397

tex2html_wrap_inline399

tex2html_wrap_inline401

tex2html_wrap_inline403

To find the mass of the water, take tex2html_wrap_inline391 and subtract the weight of just the bucket.

tex2html_wrap_inline407

Note that the values given for the length of the rope and for the velocity helicopter were eeded. Don't be confused by extra information.



Jason George Zeibel
Thu Oct 2 04:03:14 EDT 1997