How to Calculate Force and Angular Acceleration in a Rotating System?

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To calculate the force exerted by the knee extensors for a given angular acceleration, the user applies Newton's second law for rotation. The parameters include a 30-degree angle of force application, a distance of 3 cm from the axis of rotation, a mass of 4.5 kg for the lower leg/foot, and a moment arm of 23 cm. The initial calculation of force resulted in an incorrect value of 0.0072 N, prompting confusion about the transition from the torque equation to the force calculation. Clarification is sought on the correct application of trigonometric functions, particularly the cosine of the angle, in the calculations. Accurate application of these principles is essential for resolving the discrepancy in the force calculation.
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Hi, I have the following q's. I can get to through most of it but I seem to get confused with computing the last part and get the wrong answer... can anyone help?

Q. The knee extensors insert on the tibia at an angle of 30degree angle at a distance of 3cm from the axis of rotation at the knee. HOw much force must the kneww extensors exert to produce an angular accelartion at the knee of 1 rad/s (squared) if the mass of the lower leg/foot is 4.5 kg and k= 23cm?

Known:

d=0.03m
\alpha= 1 rad/s (squared)
m= 4.5 kg
K = 0.23m

Soltuion: Using 2nd Newton angular law of motion: T=I\alpha = Fd=mk(squared)\alpha

(F sin 30N) x (0.03m) = (4.5kg) x (0.23m)squared x (1 rad/s squared)
-0.03F=0.24
F= 0.24/0.03
F=0.0072N ... whichis wrong ? any ideas what I did wrong ;-(
 
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You wrote,

"F= 0.24/0.03
F=0.0072N "

how did you get from the top line to the next? A picture would help, you have to use the 30 degrees I'm guessing but why not cos(30)?

Good luck!
 
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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