Calculating Force T on Tray with Food and Coffee

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SUMMARY

The discussion focuses on calculating the force T exerted by the thumb on a tray holding a plate of food and a cup of coffee. The tray has a mass of 0.187 kg, with additional weights of 1.31 kg for the food and 0.300 kg for the coffee. The lengths provided (L1 = 0.0610 m, L2 = 0.130 m, L3 = 0.224 m, L4 = 0.365 m, L5 = 0.387 m) are crucial for determining the torque around the support point of the tray, which must equal zero for the system to be in equilibrium.

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CaptFormal
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Homework Statement



The mass of the tray itself is 0.187 kg, and its center of gravity is located at its geometrical center. On the tray is a 1.31 kg plate of food and a 0.300 kg cup of coffee. Assume L1 = 0.0610 m, L2 = 0.130 m, L3 = 0.224 m, L4 = 0.365 m and L5 = 0.387 m. Obtain the magnitude of the force T exerted by the thumb. This force acts perpendicular to the tray, which is being held parallel to the ground.

http://schubert.tmcc.edu/enc/47/9987abe8e7e44d5087aff777d122d591341673a3566037a0f448fca2c8a436593db5631f9eac0f94694865573949bce715d4c2ba0fcc1555ce73cdf70aa0bc55.gif

Homework Equations





The Attempt at a Solution



Not sure how to start this one. Any suggestions?
 
Last edited by a moderator:
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I have no idea what all those lengths are. You probably need the fact that the torque around
the point where the tray is supported is 0 if it is in equilibrium.
 

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