Posted by Mary on Tuesday, April 24, 2007 at 4:34pm.
A lunch tray is being held in one hand, as the drawing illustrates. The mass of the tray itself is 0.280 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 0.210 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.
T = N (downward)
F = N (upward)
W1= weight of the tray 0.280 kg
W2= weight of the plate 1.0 kg
W3= weight of the cup 0.210 kg
X1= center of gravity of tray is at geographical center 0.200m
X2= center of gravity of plate is at geographical center with respect to its position on tray 0.240m
X3= center of gravity of cup is at geographical center with respect to its position on tray 0.380m
Xcg= (W1X1)+(W2X2)+(W3X3)/ W1+W2+W3
I have found the center of gravity for the whole system which is 0.2522m. How to I proceed to answer the question?
I dont know why you did that.
The solution is to sum the moments about the edge (or any point) and set to zero. From the description, I am not certain if the thumb or the fingers are on the edge.
You have that equation, and one other: the sum of the vertical forces is zero.
That will be enough to solve.
The thumb is on top of the tray extending 0.0600m from the edge of the tray inward and exerting T downward.
The other four finger and beneath the tray entending 0.100m from the edge of the tray inward and exeting a force F upwards.
I still don't understand how to tackle this question. I thought you had to fing the center of gravity for the tray, plate and cup then using that find T and F. Please help!!!!
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