# Torque has to equal zero, but introduce F

• jgeogh01
In summary, the problem involves finding the forces exerted by the thumb and the four fingers on a lunch tray held in one hand, with a plate of food and a cup of coffee on it. By setting the net torque equal to zero and considering the equilibrium of forces in the x and y directions, the two unknown forces can be solved for. The lever arms can be chosen arbitrarily since the tray is in equilibrium.
jgeogh01

## Homework Statement

A lunch tray is being held in one hand, as the drawing illustrates. The mass of the tray itself is 0.180 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 0.235 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)

## Homework Equations

$$T$$=Fl
$$\Sigma$$Fx = 0
$$\Sigma$$Fy = 0
$$\Sigma$$T = 0

## The Attempt at a Solution

$$\Sigma$$T = 0
$$\Sigma$$T = -Tthumb + Tfingers - Tplate - Tcup
0 = (-Tthumb*lthumb) + (Ffingers*lfingers) - (Wplate*lplate) - (Wcup*lcup)

I'm not sure where to go from here since I don't have Tthumb or Ffingers.
Also I'm unsure how to find the lever arms because if the pivot point is where I think it is (at the end of the tray in the hand) would they just be the distances given?

If so, it would look like this

0 = (-Tthumb*.0600) + (Ffingers*.100) - (.240) - (.0893)

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jgeogh01 said:

## The Attempt at a Solution

$$\Sigma$$T = 0
$$\Sigma$$T = -Tthumb + Tfingers - Tplate - Tcup
0 = (-Tthumb*lthumb) + (Ffingers*lfingers) - (Wplate*lplate) - (Wcup*lcup)
Don't forget the weight of the tray itself.

I'm not sure where to go from here since I don't have Tthumb or Ffingers.
You'll need two equations to solve for those two unknowns. The net torque = 0 is one of them. What's another condition for equilibrium?
Also I'm unsure how to find the lever arms because if the pivot point is where I think it is (at the end of the tray in the hand) would they just be the distances given?
Since the tray is in equilibrium, the net torque about any point must be zero. So it doesn't matter which point you choose as the reference for computing torques.

Thanks for the help, I figured it out. After I factored in the weight of the tray and found T in terms of F it was easy.

## 1. What is torque and why does it have to equal zero?

Torque is a measure of the turning force on an object. It has to equal zero because in order for an object to be in rotational equilibrium, the sum of all the torques acting on it must be zero. This means that the object will not rotate unless a new force is introduced.

## 2. How is torque related to force?

Torque is directly proportional to force and the distance between the force and the axis of rotation. This means that a larger force or a longer distance from the axis of rotation will result in a greater torque.

## 3. Can torque ever be negative?

Yes, torque can be negative if the direction of the force is opposite to the direction of rotation. This means that the force is causing the object to rotate in the opposite direction.

## 4. How does introducing a new force affect torque?

Introducing a new force will change the total torque acting on an object. If the new force is applied at a distance from the axis of rotation, it will add to the existing torque and cause the object to rotate in a different direction or at a different speed.

## 5. What happens if torque does not equal zero?

If torque does not equal zero, the object will be in rotational motion. This means that it will either rotate in a constant direction or accelerate in a circular motion, depending on the magnitude and direction of the torque.

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