Solving Part B: Acceleration of End B after String Cut

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The discussion revolves around calculating the initial acceleration of end B of a rod when the supporting string is cut. The rod, measuring 69.0 cm and weighing 1.70 kg, experiences a gravitational force of 16.6 N. The user attempts to find the acceleration using the formula a = Fnet/m, resulting in an acceleration of 9.8 m/s². However, the user expresses confusion about the problem, particularly regarding the changes after the string is retied to half its original length. Clarification on the dynamics involved in both scenarios is sought.
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Homework Statement



A rod of length 69.0 cm and mass 1.70 kg is suspended by two strings which are 43.0 cm long, one at each end of the rod.

http://schubert.tmcc.edu/res/msu/physicslib/msuphysicslib/20_Rot2_E_Trq_Accel/graphics/prob13a_1016full.gif

(A) The string on side B is cut. Find the magnitude of the initial acceleration of end B.

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The string on side B is retied and now has only half the length of the string on side A.

http://schubert.tmcc.edu/res/msu/physicslib/msuphysicslib/20_Rot2_E_Trq_Accel/graphics/prob13b_1016half.gif

(B) Find the magnitude of the initial acceleration of the end B when the string is cut.

Homework Equations





The Attempt at a Solution



A bit lost on even how to solve this one. Any assistance would be appreciated. Thanks.
 
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Assuming no air resistance
Fg = (m)(g)
Fg = (1.70kg)(9.8m/s^2)
Fg = 16.6

a = Fnet/m
a = 16.6/1.70
a = 9.8 m/s^2

Think that's right...?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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