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September 30, 2014

September 30, 2014

Posted by **Mary** on Tuesday, April 24, 2007 at 9:00pm.

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

Xcg= [(0.280kg)(0.200m)]+[(1.0kg)(0.240m)]+[(0.210kg)(0.380m)]/(0.280+1.0+0.210)

Xcg= (0.056+0.240+0.0798)/1.49

Xcg= 0.3758/1.49

Xcg= 0.2522m

I have found the center of gravity for the whole system which is 0.2522m. How to I proceed to answer the question?

For Further Reading

Physics please check - bobpursley, Tuesday, April 24, 2007 at 5:47pm

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.

Physics please check - Mary, Tuesday, April 24, 2007 at 7:09pm

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!!!!

No, you dont. Sum all the torques (force x distance) about the edge of the tray. and set to zero. That gives you one equation with two unknowns (fthumb, ffinger)

Then, sum the vertical forces, that gives you another equation with the same unknowns, and set to zero.

Solve for the two unknowns.

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