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

• Hypnos_16
In summary, the forces T and F exerted on the tray by the thumb and four fingers respectively can be calculated using the equations F = (Mgdp) + (Mgdc) - Ft and Ff = mgdp + mgdc. The values for the masses and midpoints of the tray, plate, and coffee can be found using the given values for L1-L5.
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

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

## 1. What does "Ft = 0.202" represent?

"Ft = 0.202" is a formula used to calculate the force applied on a lunch tray by the thumb and finger. It represents the magnitude of the force in Newtons (N).

## 2. How is the force on a lunch tray calculated using this formula?

The formula takes into account the length of the lunch tray, the distance between the thumb and finger, and the force applied by each. The resulting value is the sum of the forces applied by the thumb and finger on the lunch tray.

## 3. Why is it important to calculate the force on a lunch tray?

Calculating the force on a lunch tray can help determine if the tray is being properly held and if it can hold the weight of the food without tipping over. It can also provide insight into the amount of pressure being applied on the tray and how it may affect the food and the tray's stability.

## 4. What units are used for the force calculation?

The force is calculated using the SI unit of Newtons (N). However, other units such as pounds (lbs) or kilograms (kg) can also be used as long as they are consistent throughout the formula.

## 5. Can this formula be applied to other objects besides lunch trays?

Yes, this formula can be applied to any object that is being held between the thumb and finger, as long as the length and distance between the thumb and finger are known. It can also be modified to take into account different types of forces, such as friction or gravity, depending on the situation.

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