- #1

Hypnos_16

- 153

- 1

## 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