Calculating Force T on Tray with Food and Coffee

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To calculate the force T exerted by the thumb on a tray with food and coffee, the mass of the tray is 0.187 kg, with additional weights of 1.31 kg for the food and 0.300 kg for the coffee. The tray's center of gravity is at its geometric center, and the lengths provided (L1 to L5) are likely distances relevant to the positions of the weights. The key to solving the problem is recognizing that the torque around the support point must equal zero for the tray to be in equilibrium. Understanding the relationship between the forces and torques acting on the tray is essential for determining the magnitude of force T.
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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?
 
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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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