Solve the Tray Forces Puzzle: Find T and F

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AI Thread Summary
The discussion focuses on solving for the forces T and F acting on a lunch tray held horizontally, with a mass of 0.2 kg, a 1 kg plate, and a 0.25 kg cup. The key equation involves balancing counterclockwise and clockwise torques, leading to the conclusion that T can be isolated by using the point of contact of the fingers as the pivot point. The user initially miscalculated T as 1656.25 lbs, indicating a misunderstanding of the force balance. Correctly applying the torque balance and then solving for the vertical forces will yield the accurate values for T and F. The conversation emphasizes the importance of choosing the right pivot point for torque calculations.
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Homework Statement


A lunch tray is being held in one hand, as the Drawing illustrates. The mass of the tray is .2 kg and its center of gravity is located at its geometric center (the tray is .4m long). On the tray is a 1 kg plate of food and a .25 kg cup of coffee. Obtain the force T exerted by the thumb and the force F exerted by the four fingers. Both forces act perpendicular to the tray, which is being held parallel to the ground.


Homework Equations


T=FX=Counterclockwise forces - clockwise forces = 0


The Attempt at a Solution


T= ((f)(.1)) - ( (t)(.06m) + (.2kg)(.2m)(10 m/s^2) + (1kg)(.24)(10) + (.25)(.38)(10m/s^2)) = 0

manipulated the equation so I got F= 625 + T

I Isolated T to be1656.25lbs of force, which is obviously wrong. Help!
 

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shotgunbob said:
T=FX=Counterclockwise forces - clockwise forces = 0
no, ccw torques balance cw torques. "up" forces balance "down" forces.

T= ((f)(.1)) - ( (t)(.06m) + (.2kg)(.2m)(10 m/s^2) + (1kg)(.24)(10) + (.25)(.38)(10m/s^2)) = 0

looks like you're balancing torques, but what have you picked as a pivot point? I'd use the point of contact of the fingers, since that would make the torque there zero, leaving one unknown (T).

After finding T, go to balance the up and down forces to find F.
 
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