Ft = 0.202"Calculating Forces on a Lunch Tray: Thumb and Finger Forces

[Hypnos_16]

Homework Statement

The mass of the tray itself is 0.209 kg, and its center of gravity is located at its geometrical center. On the tray is a 1.11-kg plate of food and a 0.278-kg cup of coffee. Assume L1 = 0.0600 m, L2 = 0.108 m, L3 = 0.260 m, L4 = 0.366 m and L5 = 0.396 m. Obtain the force T exerted by the thumb (enter first) and the force F exerted by the four fingers. Both forces act perpendicular to the tray, which is being held parallel to the ground.

Okay this is kind of hard to explain without the picture

Basically it's a hand holding a tray,
L1 is to the thumb pushing down to counter the weight of everything else
L2 is to the finger under the tray the only force pushing up, (The Pivot Point i guess)
L3 is to the midpoint of the plate of food
L4 is to the midpoint of the cup of coffee
L5 is to the end of the tray

Homework Equations

i would just use F x d for all the masses and forces on both sides
but with all the variables i think i goofed something up
i was hoping someone could both check over my workings and help me if there's anything gone strange there.

The Attempt at a Solution

Tray
Mass = 0.209kg
Midpoint = 0.90m

Plate
Mass = 1.11kg
Midpoint = 0.152m

Coffee
Mass = 0.278kg
Midpoint = 0.258m

Tcw = Tccw
T = (Mgdt) + (Mgdp) + (Mgdc) - Ft
Ft = mgdt + mgdp + mgdc
Ft = (0.209)(9.81)(0.90) + (1.11)(9.81)(0.152) + (0.278)(9.81)(0.258)
Ft = (1.845) + (1.655) + (0.703)
Ft = 4.203 x 0.048
Ft = 0.202N

 

[member 553083]

Fcw = Fccw F = (Mgdp) + (Mgdc) - Ft Ff = mgdp + mgdcFf = (1.11)(9.81)(0.152) + (0.278)(9.81)(0.258)Ff = (1.655) + (0.703)Ff = 2.358 x 0.048Ff = 0.113N