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Rigid Objects in Equilibrium, Center of Gravity

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A lunch tray is being held in one hand. The mass of the tray itself is 0.200 kg, and its center of gravity is located at its geometrical center. On the tray is a 1.00-kg plate of food and a 0.250-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 held parallel to the ground. Assuming that the center of mass of all the items in tray together act at the geometrical centre of the tray, the four fingers will exert the force equal to total weight, 14.21 N. The thumb will apply the force to balnce due to the difference in the position of force exerted by the fingers at the rim of tray and the center of the tray. This force is determined to be F =r* 14.21/h. Note that the student has not provided the expert with values of r and height h, therefore, the answer is left in terms of r and h.

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I'm assuming that the center of mass of all the items in tray together act at the geometrical centre of the tray.
<br>
<br>total weight = (0.2+1.0+0.25)*9.8 = 1.45*9.8 = ...

Solution provided by:
Amrit Lal Ahuja, PhD (IP)
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  • BEng, Allahabad University, India
  • MSc , Pune University, India
  • PhD (IP), Pune University, India
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