Maximum value of F - SIN Waterloo Contest

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The discussion revolves around calculating the maximum force F that Superman can exert on a large block without causing a smaller block on top to slip. The scenario involves two blocks with given masses and a friction coefficient, alongside a cable tied to the smaller block at a specific angle. Participants are attempting to set up free body diagrams (FBDs) and static force equations for both blocks to solve the problem. There is confusion regarding the correct application of trigonometric functions and the inclusion of all horizontal forces acting on the larger block. Clear guidance is requested for writing out the static force equations to progress in the solution.
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Homework Statement



Superman is regaining his strength by pulling with a force F on a large block of mass M resting on a frictionless floor as shown. A small block of mass m rests on the upper surface of the large block, and there is friction at that surface, with static coefficient μ. Lois Lane has tied the small block to the wall with a light strong cable sloping up at an angle ∅. Everything is shown on the sketch, and things that look horizontal are. What is the maximum value of F before something slips? (Answers in Newtons)

∅- 60°
m=250kg
M=450kg
μ=0.5

Possible answers;

a) 1230

b) 1094

c) 946

d) 829

e) 656

*Check attachment for a picture.

Homework Equations



F=ma, Ff=Fnxμ,Fnet=F1+F2+F3

The Attempt at a Solution



I drew the FBDs for both masses, but I do feel i did something wrong.

for mass m, I had Tcos60° upwards, Tsin60° to the right, Ff to the left, and Fg facing down.

for mass M, I had Fg facing down, Fn facing up, and Fnm [Normal force of mass m] also facing down.

That's as far as I got.

*The picture is upside down, sorry!
 

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I think you have the cos and sin swapped.
You've left out the horizontal forces on mass M.
Please write out the static force equations for the two directions one each mass. You know what Fg is for each mass, so don't write it in that form (especially as having two forces with the same name is confusing).
 
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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